MTH101 — Midterm Summary (Lectures 1–22)

📘 Lecture 1 — Coordinates, Graphs and Lines

📖 Overview: This lecture introduces the fundamental concepts underlying calculus, beginning with the real number system and its historical development. It establishes the critical connection between numbers and geometry through the coordinate line and explores the properties of inequalities and intervals, which are essential tools for solving problems in calculus.

🗂️ Topics Covered

The lecture covers the hierarchy of real numbers from natural numbers to irrationals, the definition and notation of sets and subsets, the construction and one-to-one correspondence of the coordinate line, the order properties and theorems of inequalities, and the various types of intervals (finite, infinite, open, closed) used in mathematics.

📝 Lecture Summary

What is Calculus?

Calculus is the study of the continuous rates of change of quantities. It examines how various quantities, such as distance, time, or water flow, change with respect to other quantities continuously.

Natural Numbers

The simplest numbers are the natural numbers: 1, 2, 3, 4, 5,... They were the first numbers used by humans for counting and come to us "naturally," hence the name.

Sets and Subsets

A set is a collection of well-defined objects, often denoted with curly brackets {}, e.g., {George Bush, Tony Blair, Ronald Reagan}. A subset is a portion of a set where every member of the subset is also a member of the larger set. In mathematical notation, A is a subset of B (A ⊆ B) if ∀x ∈ A ⇒ x ∈ B.

Integers

The integers include the natural numbers, their negatives, and zero: ..., -4, -3, -2, -1, 0, 1, 2, 3, 4,...

Rational Numbers

The rational numbers are formed by taking ratios of integers, with the exception that division by zero is not allowed, e.g., 2/3, 7/5, -5/2. Every integer is also a rational number because an integer p can be written as p/1.

💡 Why this matters: Division by zero is undefined because if x/0 = y, then x = 0·y ⇒ x = 0, leading to a contradictory or non-unique value, which is mathematically unsatisfactory.

Irrational Numbers

Irrational numbers cannot be expressed as a ratio of integers. They were discovered by Hippasus of Metapontum in the 5th century B.C., who showed geometrically that the hypotenuse of a right triangle with sides equal to 1 (√2) is irrational. Other examples include π and cos 19°. Rational numbers in decimal form either terminate or repeat in a pattern, while irrational numbers do not.

Real Numbers

The real numbers (or real number system) comprise both rational and irrational numbers together.

Coordinate Line

The coordinate line (or real line) is developed by: (1) designating a positive direction (right) and negative direction (left), (2) choosing an arbitrary origin marked as 0, and (3) choosing a unit of measurement. With this, each real number is associated with a unique point on the line, and each point corresponds to a unique real number—this is a one-to-one correspondence.

Order Properties

The order of the real number set defines the size of one real number relative to another. For any two real numbers a and b, if b - a is positive, then b > a (or a < b). A statement involving < or > is called an inequality. The inequality a ≤ b means a < b or a = b. The expression a < b < c means a < b and b < c.

🔑 Definition — Inequality: A statement involving the symbols <, >, ≤, or ≥, which compares the relative size of two quantities.

Theorem 1.1.1 — Properties of Inequalities

The following properties remain true if < and > are replaced by ≤ and ≥: (a) If a < b and b < c, then a < c. (b) If a < b and a + c < b + c, then a - c < b - c. (c) If a < b and ac < bc when c is positive, and ac > bc when c is negative. (d) If a < b and c < d, then a + c < b + d. (e) If a and b are both positive or both negative and a < b, then 1/a > 1/b.

Intervals

An interval is a set of real numbers that corresponds geometrically to a line segment on the coordinate line.

🔑 Definition — Closed interval [a, b]: {x : a ≤ x ≤ b}. Includes both endpoints a and b. 🔑 Definition — Open interval (a, b): {x : a < x < b}. Excludes both endpoints a and b.

Half-open intervals include one endpoint but not the other, e.g., [a, b) or (a, b).

Finite intervals have finite real numbers as endpoints.

Infinite intervals extend indefinitely in the positive or negative direction, using the symbols -∞ (negative infinity) and +∞ (positive infinity), e.g., [a, +∞), (-∞, b]. These symbols do not represent numbers.

Solving Inequalities

Solving an inequality means finding its solution set—all values of the unknown that satisfy the inequality. The operations in Theorem 1.1.1 will not change the solution set if one does not multiply both sides by zero or an expression involving an unknown.

📌 Example: Solve 3 + 7x ≤ 2x - 9 Step 1: Subtract 3 from both sides → 7x ≤ 2x - 12 Step 2: Subtract 2x from both sides → 5x ≤ -12 Step 3: Divide both sides by 5 → x ≤ -12/5 Solution set: (-∞, -12/5]

📌 Example: Solve 7 ≤ 2 - 5x ≤ 9 This is a combination of 7 ≤ 2 - 5x and 2 - 5x ≤ 9. Step 1: Subtract 2 from both sides → 5 ≤ -5x < 7 Step 2: Divide by -5 (inequality symbols reversed) → -1 ≥ x ≥ -7/5 Step 3: Rewrite with smaller number first → -7/5 < x ≤ -1 Solution set: (-7/5, -1]

⭐ Key Takeaways

The real number system is a hierarchy from natural numbers to integers to rational numbers to irrational numbers, all comprising the reals. The coordinate line establishes a one-to-one correspondence between real numbers and points on a line, enabling the geometric representation of algebraic concepts. Inequalities and their properties, particularly the critical rule about reversing the inequality sign when multiplying or dividing by a negative number, are foundational for solving problems in calculus. Intervals, whether open, closed, half-open, finite, or infinite, provide a precise way to describe sets of real numbers and are essential for expressing solution sets of inequalities.

🧠 Quick Revision Questions

1. What is the fundamental difference between rational and irrational numbers?
2. Why is division by zero mathematically undefined?
3. Explain the one-to-one correspondence between real numbers and points on a coordinate line.
4. What happens to an inequality when you multiply or divide both sides by a negative number?
5. How would you describe the difference in notation between an open interval (a, b) and a closed interval [a, b]?

📘 Lecture 2 — Absolute Value

📖 Overview: This lecture introduces the concept of absolute value, a fundamental tool in algebra used to represent the magnitude of a real number, regardless of its sign. It is essential for understanding distance on a coordinate line, solving equations and inequalities, and is a cornerstone for more advanced topics like the triangle inequality.

🗂️ Topics Covered

The lecture begins with the formal definition of absolute value and provides illustrative examples. It then covers solving equations involving absolute values, followed by the critical relationship between square roots and absolute values. Several key properties are proven, and the geometric interpretation of absolute value as distance is explained. The lecture concludes with the solution of absolute value inequalities and a proof of the important Triangle Inequality.

📝 Lecture Summary

Definition

The absolute value or magnitude of a real number a, denoted by |a|, strips away the sign of the number. It is defined as |a| = a if a ≥ 0, and |a| = -a if a < 0. This means the result is always a non-negative number.

🔑 Definition — Absolute Value: |a| = a if a is non-negative, and |a| = -a if a is negative. 📌 Example: |5| = 5 because 5 > 0. |-4/7| = 4/7 because -4/7 < 0. |0| = 0 because 0 ≥ 0.

Solving Absolute Value Equations

An equation like |expression| = c, where c ≥ 0, is solved by considering two cases: expression = c or expression = -c. This accounts for the fact that the absolute value could have come from either a positive or negative original expression.

📌 Example: Solve |x - 3| = 4.

1. Case 1: x - 3 = 4 → x = 7
2. Case 2: x - 3 = -4 → x = -1 The solution set is {7, -1}.

📌 Example: Solve |3x - 2| = |5x + 4|. Since the absolute values are equal, the expressions inside must be equal or opposites.

1. 3x - 2 = 5x + 4 → -2x = 6 → x = -3
2. 3x - 2 = -(5x + 4) → 3x - 2 = -5x - 4 → 8x = -2 → x = -1/4 The solution set is {-3, -1/4}.

Relationship between Square Roots and Absolute Values

A common error is to think √(a²) = a. This is only true for non-negative a. The correct relationship for all real numbers is √(a²) = |a|. For example, √(-4)² = √16 = 4, but a = -4, so 4 ≠ -4.

🔑 Definition — Square Root Property: For any real number a, √(a²) = |a|. This ensures the principal (non-negative) square root. 📐 Formula: √(a²) = |a| → The principal square root of a number squared is the absolute value of that number. 📌 Example: √(-5)² = √25 = 5 = |-5|. √(3)² = √9 = 3 = |3|.

Properties of Absolute Value

The absolute value function has several key algebraic properties that are used in simplifying expressions.

🔑 Definition — Property (a): |-a| = |a|. A number and its negative have the same absolute value. 📌 Example: |-4| = |4| = 4.

🔑 Definition — Property (b): |ab| = |a| |b|. The absolute value of a product is the product of the absolute values. 📌 Example: |(2)(-3)| = |-6| = 6 = |2|| -3| = (2)(3) = 6.

🔑 Definition — Property (c): |a/b| = |a| / |b|. The absolute value of a ratio is the ratio of the absolute values (provided b ≠ 0). 📌 Example: |5/4| = 5/4 = |5| / |4| = 5/4.

Geometric Interpretation of Absolute Value

On a coordinate line, the distance between two points with coordinates a and b is the absolute value of their difference.

🔑 Definition — Distance Formula: For points A (a) and B (b) on a coordinate line, the distance d between them is d = |b - a|. 📐 Formula: d = |b - a| → Distance is the absolute value of the difference of coordinates. 📌 Example: The distance between the point x=2 and x=5 is |5 - 2| = 3. The distance between x=-2 and x=1 is |1 - (-2)| = |3| = 3.

This interpretation extends to common expressions:

• |x - a|: The distance between x and a.
• |x + a|: The distance between x and -a.
• |x|: The distance between x and the origin (0).

Solving Absolute Value Inequalities

Inequalities of the form |x - a| < k and |x - a| > k represent intervals of distance k from point a.

• |x - a| < k means the distance from x to a is less than k. This is equivalent to -k < x - a < k, which solves to an open interval (a-k, a+k).
• |x - a| > k means the distance from x to a is greater than k. This is equivalent to x - a < -k OR x - a > k, which solves to two disjoint intervals (-∞, a-k) ∪ (a+k, ∞).

📌 Example: Solve |x - 3| < 4. This means the distance from x to 3 is less than 4.

1. Write the compound inequality: -4 < x - 3 < 4
2. Solve for x (add 3 to all parts): -1 < x < 7 Solution in interval notation: (-1, 7)

📌 Example: Solve |x + 4| ≥ 2. (Note: x+4 is distance from x to -4). This means the distance from x to -4 is greater than or equal to 2.

1. Write the two inequalities: x + 4 ≤ -2 OR x + 4 ≥ 2
2. Solve each: x ≤ -6 OR x ≥ -2 Solution in interval notation: (-∞, -6] ∪ [-2, ∞) 💡 Why this matters: This shows that absolute value inequalities describe intervals on the real line, a key skill in calculus and analysis.

The Triangle Inequality

The Triangle Inequality is a fundamental theorem stating that the absolute value of a sum is always less than or equal to the sum of the absolute values.

🔑 Definition — Triangle Inequality: For any real numbers a and b, |a + b| ≤ |a| + |b|. This theorem is central to many proofs in analysis. 📐 Formula: |a + b| ≤ |a| + |b| → The magnitude of a sum cannot exceed the sum of the magnitudes. 📌 Example: Let a = 2 and b = -3. Then |a + b| = |2 + (-3)| = |-1| = 1. And |a| + |b| = |2| + |-3| = 2 + 3 = 5. So 1 ≤ 5, which satisfies the inequality. The proof proceeds by combining the inequalities -|a| ≤ a ≤ |a| and -|b| ≤ b ≤ |b| and then considering the two possible signs of the sum a+b.

⭐ Key Takeaways

The absolute value of a number is its distance from zero, defined piecewise as |a| = a for a ≥ 0 and |a| = -a for a < 0. Crucially, the square root of a square is always the absolute value: √(a²) = |a|. For solving equations like |expression| = c, you must set up two equations: expression = c or expression = -c. For inequalities, |x - a| < k translates to -k < x-a < k, yielding an interval centered at a, while |x - a| > k translates to x-a < -k OR x-a > k, yielding two separate intervals. Finally, the Triangle Inequality, |a + b| ≤ |a| + |b|, is a non-negotiable fact that the absolute value of a sum cannot exceed the sum of the absolute values.

🧠 Quick Revision Questions

1. What is the formal definition of |a| for a real number a?
2. What is the correct simplification of √(a²)? Why is it not simply a?
3. Describe the algebraic steps to solve the equation |2x + 1| = 5.
4. How do you rewrite the inequality |x - 5| > 3 as a compound inequality without absolute value symbols?
5. State the Triangle Inequality and give a simple example with a = 3 and b = -4 to verify it.

📘 Lecture 3 — Coordinate Planes and Graphs

📖 Overview: This lecture introduces the rectangular coordinate system for plotting points and visualizing algebraic equations as geometric curves. It covers how to graph equations, find intercepts, and use symmetry to simplify graphing, laying the foundation for connecting algebra with geometry.

🗂️ Topics Covered

The lecture covers three main topics: graphs in the coordinate plane, including plotting ordered pairs and the rectangular coordinate system; intercepts where graphs cross the x-axis and y-axis; and symmetry about the x-axis, y-axis, and origin as a tool for simplifying graphing.

📝 Lecture Summary

Graphs in the coordinate plane

Just as points on a line correspond to real numbers, points in a plane correspond to pairs of real numbers. A plane is the intersection of two coordinate lines at 90 degrees, called the coordinate plane. Every point P in a coordinate plane can be associated with a unique ordered pair of real numbers (a,b) by drawing perpendicular lines through P to the x-axis and y-axis. To plot a point P(a,b) means to locate it in the coordinate plane. This idea enables us to visualize algebraic equations as geometric curves and, conversely, to represent geometric curves by algebraic equations.

The coordinate axes divide the plane into four regions called quadrants, numbered counterclockwise with Roman numerals. The coordinate plane and ordered pairs together are known as the rectangular coordinate system.

A solution of an equation in two variables x and y is an ordered pair of real numbers (a,b) that satisfies the equation when we substitute x=a and y=b.

The graph of an equation in two variables x and y is the set of all points in the xy-plane whose coordinates are members of the solution set of the equation.

📌 Example 1: The pair (3,2) is a solution of 6x − 4y = 10 because 6(3)−4(2)=10, which is true. However, (2,0) is not a solution since 6(2)−4(0)=18≠10.

📌 Example 2: Sketch the graph of y = x². By plotting points from a table of values and connecting them, we get a U-shaped curve. The curve is only an approximation to the graph of y = x².

💡 Why this matters: When graphing by plotting points, whether by hand, calculator, or computer, there is no guarantee the resulting curve has the correct shape.

Intercepts

Points where a graph intersects the coordinate axes are of special interest. Intersections with the x-axis have the form (a, 0) and intersections with the y-axis have the form (0, b). The number a is called an x-intercept of the graph and the number b a y-intercept.

📌 Example: Find all intercepts of an equation. For the equation given, solving yields the x-intercept and y-intercept. The process involves setting y=0 to find x-intercepts and setting x=0 to find y-intercepts.

🔑 Definition — Intercept: The x-intercept is the x-coordinate where a graph crosses the x-axis (where y=0). The y-intercept is the y-coordinate where a graph crosses the y-axis (where x=0).

Symmetry

Symmetry plays an important role in applied mathematics and engineering. Points (x,y) and (x,-y) are symmetric about the x-axis. Points (x,y) and (-x,y) are symmetric about the y-axis. Points (x,y) and (-x,-y) are symmetric about the origin.

Symmetry is a tool for graphing — by taking advantage of symmetries when they exist, the work required to obtain a graph can be reduced considerably.

Tests for Symmetry:

• Symmetric about y-axis: Substituting -x for x yields the original equation.
• Symmetric about x-axis: Substituting -y for y yields the original equation.
• Symmetric about the origin: Substituting -x for x and -y for y yields the original equation.

📌 Example 9: Sketch the graph of y = 1/8(x⁴ − x²). The graph is symmetric about the y-axis since substituting -x for x simplifies to the original equation. Because of this symmetry, we only need to calculate points for x ≥ 0 (right half), then obtain corresponding points for x ≤ 0 (left half) by using the symmetry.

📌 Example 10: Sketch the graph of x = y². Solving for y gives y = √x and y = -√x. The curve x = y² is symmetric about the x-axis because substituting -y for y yields the original equation. So we only need to graph y = √x (the portion above or touching the x-axis), then reflect it about the x-axis to complete the graph.

⭐ Key Takeaways

The coordinate plane uses ordered pairs to represent points, and the graph of an equation is the set of all points satisfying it. Intercepts are where graphs cross axes — find x-intercepts by setting y=0 and y-intercepts by setting x=0. Symmetry about axes or the origin reduces graphing work: test by substituting −x, −y, or both into the equation. Always remember that plotted points only approximate the true graph shape. These concepts connect algebraic equations to geometric curves, essential for visualizing mathematical relationships.

🧠 Quick Revision Questions

1. What is an ordered pair, and how does it correspond to a point in the coordinate plane?
2. How do you find the x-intercept and y-intercept of a graph from its equation?
3. What are the three tests for symmetry, and what does each substitution check?
4. In Example 9, why did we only need to calculate points for x ≥ 0?
5. What is the difference between the point (2,5) and the point (−2,5) in terms of symmetry?

📘 Lecture 4 — Lines

📖 Overview: This lecture introduces the concept of slope as a measure of steepness for lines in the coordinate plane. It explains how to calculate slope, interpret its meaning, and connect it to the angle of inclination, while also covering equations of lines and their applications in real-world contexts like temperature conversion.

🗂️ Topics Covered

This lecture covers the definition and calculation of slope for non-vertical lines, including examples with positive, negative, and zero slopes. It explains the angle of inclination and its relationship to slope via the tangent function. The lecture then discusses conditions for parallel and perpendicular lines using slopes. It derives equations of lines in point-slope, slope-intercept, and general forms, including vertical and horizontal lines, and concludes with applications demonstrating linear relationships.

📝 Lecture Summary

Slope

The slope of a line measures its steepness. For a particle moving left to right along a non-vertical line from point P1(x1, y1) to P2(x2, y2), the vertical change (y2 − y1) is the rise, and the horizontal change (x2 − x1) is the run.

🔑 Definition — slope m: If P1(x1, y1) and P2(x2, y2) are points on a non-vertical line, then the slope m is defined by: m = rise / run = (y2 − y1) / (x2 − x1)

Key observations about this definition:

• It does not apply to vertical lines because x2 = x1, causing division by zero. The slope of a vertical line is undefined (sometimes informally called infinite slope).
• Reversing the order of points reverses the signs of both numerator and denominator, so the ratio remains unchanged.
• Any two distinct points on a non-vertical line will yield the same slope value.

📌 Example: Find the slope of the line through:

• (6,2) and (9,8): m = (8−2)/(9−6) = 6/3 = 2
• (2,9) and (4,3): m = (3−9)/(4−2) = −6/2 = −3
• (−2,7) and (5,7): m = (7−7)/(5−(−2)) = 0/7 = 0

💡 Why this matters: Since rise = m · run, as a point travels left to right, there are m units of rise for each unit of run. The slope m is the rate of change of y with respect to x along the line. A positive slope means the line inclines upward to the right, a negative slope means it inclines downward to the right, and a zero slope means the line is horizontal.

Angle of Inclination

When equal scales are used on both axes, slope relates to the angle a line makes with the positive x-axis.

🔑 Definition — angle of inclination (φ): For a line L not parallel to the x-axis, this is the smallest angle measured counterclockwise from the positive x-axis to L. For a line parallel to the x-axis, φ = 0. In degrees: 0° ≤ φ ≤ 180°. In radians: 0 ≤ φ ≤ π.

Theorem 1.4.3: For a non-vertical line, the slope m and angle of inclination φ are related by: m = tan φ

If a line is parallel to the y-axis (vertical), φ = π/2, so tan φ is undefined, matching the undefined slope.

📌 Example: Find the angle of inclination for:

• Slope m = 1: tan φ = 1, so φ = π/4 (45°)
• Slope m = −1: tan φ = −1, and since 0 ≤ φ ≤ π, φ = 3π/4 (135°)

Parallel and Perpendicular Lines

Theorem 1.4.4: Let L1 and L2 be non-vertical lines with slopes m1 and m2.

• (a) Lines are parallel if and only if: m1 = m2
• (b) Lines are perpendicular if and only if: m1 · m2 = −1

For perpendicular lines, this can be rewritten as m2 = −1/m1, meaning the slopes are negative reciprocals of one another.

📌 Example: Show that points A(1,3), B(3,7), and C(7,5) are vertices of a right triangle.

• Slope AB: m1 = (7−3)/(3−1) = 4/2 = 2
• Slope BC: m2 = (5−7)/(7−3) = −2/4 = −1/2
• Check: m1 · m2 = (2)(−1/2) = −1 Since the product is −1, line AB is perpendicular to line BC, so triangle ABC is a right triangle with the right angle at B.

Equations of Lines

Lines Parallel to the Coordinate Axes: Theorem 1.4.5: The vertical line through (a,0) is: x = a. The horizontal line through (0,b) is: y = b.

📌 Example: The graph of x = −5 is the vertical line through (−5,0). The graph of y = 7 is the horizontal line through (0,7).

Lines Determined by Point and Slope: Theorem 1.4.6 (Point-Slope Form): The line passing through P1(x1, y1) with slope m is: y − y1 = m(x − x1)

📌 Example: Equation of line through (2,3) with slope m = −3/2. y − 3 = −3/2 (x − 2) → y = −3/2 x + 6

📌 Example: Equation of line through (−2,−1) and (3,4). m = (−1−4)/(−2−3) = −5/−5 = 1 Using (−2,−1): y − (−1) = 1(x − (−2)) → y + 1 = x + 2 → y = x + 1

Lines Determined by Slope and y-Intercept: Theorem 1.4.7 (Slope-Intercept Form): The line with y-intercept b and slope m is: y = mx + b

📌 Example: The line y = 2x − 5 has slope 2 and y-intercept −5.

📌 Example: Find the slope-intercept form through (3,4) and (−2,−1). m = (−1−4)/(−2−3) = −5/−5 = 1. Using (−2,−1): y + 1 = 1(x + 2) → y = x + 1

The General Equation of a Line: Theorem 1.4.8: Every first-degree equation in x and y (Ax + By + C = 0, where A and B are not both zero) has a straight line as its graph, and every straight line can be represented by such an equation.

📌 Example: Find slope and y-intercept of 8x + 5y = 20. Solve for y: 5y = −8x + 20 → y = −8/5 x + 4 Slope m = −8/5, y-intercept b = 4

Applications

Many physical phenomena follow linear paths. Bodies falling from rest in a gravitational field, light rays, and coasting objects move along lines. The equations of lines (linear equations) are used to study such motions. Slope is important for measuring steepness in practical situations like roadbeds, roofs, and stairs. It also describes how rapidly things are changing, which is essential for calculus.

📌 Example: Fahrenheit (F) and Celsius (C) are related linearly: F = mC + b. Given water freezes at C = 0, F = 32: 32 = 0m + b → b = 32 Water boils at C = 100, F = 212: 212 = 100m + 32 → 100m = 180 → m = 9/5 Therefore: F = 9/5 C + 32

⭐ Key Takeaways

The slope of a non-vertical line is defined as the ratio of rise to run and can be positive, negative, zero, or undefined (for vertical lines). Slope is related to the angle of inclination by m = tan φ. Parallel lines have equal slopes, while perpendicular lines have slopes that are negative reciprocals (product = −1). The three essential forms for linear equations are point-slope (y − y1 = m(x − x1)), slope-intercept (y = mx + b), and the general first-degree form (Ax + By + C = 0). Understanding slope as a rate of change is fundamental for interpreting how quantities vary and for later study in calculus.

🧠 Quick Revision Questions

1. What is the slope of a line passing through points (4, −1) and (−2, 5)?
2. Why is the slope of a vertical line undefined?
3. If two non-vertical lines are perpendicular, what is the relationship between their slopes?
4. Find the equation of a line with slope 3 that passes through the point (−1, 2).
5. What is the angle of inclination of a line with slope −√3? (Give answer in radians)

📘 Lecture 5 — Distance; Circles; Equations of the form y = ax²+bx+c

📖 Overview: This lecture introduces the distance formula for points in a coordinate plane and uses it to derive the equation of a circle. It also covers the midpoint formula and the graphical analysis of quadratic equations of the form y = ax²+bx+c, including their parabolic shapes and vertex properties.

🗂️ Topics Covered

The lecture covers the distance formula between two points in the plane, the midpoint formula, the standard form of the equation of a circle and its variations, degenerate cases of circles, and the graph of quadratic equations in x (parabolas) including vertex location, intercepts, and applications such as solving inequalities and projectile motion.

📝 Lecture Summary

Distance between two points in the plane

The distance between two points A and B on a coordinate line with coordinates a and b is |b - a|. To find the distance d between two arbitrary points P₁(x₁, y₁) and P₂(x₂, y₂) in the plane, a right triangle is formed with P₁ and P₂ as vertices. The horizontal side has length |x₂ - x₁| and the vertical side has length |y₂ - y₁|. By the Pythagoras Theorem, the distance is d = √[(x₂ - x₁)² + (y₂ - y₁)²].

🔑 Definition — Distance Formula: The distance d between two points (x₁, y₁) and (x₂, y₂) in a coordinate plane is given by d = √[(x₂ - x₁)² + (y₂ - y₁)²].

📐 Formula: d = √[(x₂ - x₁)² + (y₂ - y₁)²] → The square root of the sum of the squares of the differences in x and y coordinates.

📌 Example: Find the distance between (-2, 3) and (1, 7). Let (x₁, y₁) = (-2, 3) and (x₂, y₂) = (1, 7). Then d = √[(1 - (-2))² + (7 - 3)²] = √[3² + 4²] = √[9 + 16] = √25 = 5. The same result is obtained if the points are labeled in reverse order.

📌 Example: Show that points A(4, 6), B(1, -3), and C(7, 5) are vertices of a right triangle. Compute side lengths: d(A, B) = √[(1 - 4)² + (-3 - 6)²] = √[9 + 81] = √90; d(A, C) = √[(7 - 4)² + (5 - 6)²] = √[9 + 1] = √10; d(B, C) = √[(7 - 1)² + (5 - (-3))²] = √[36 + 64] = √100 = 10. Since [d(A, B)]² + [d(A, C)]² = 90 + 10 = 100 = [d(B, C)]², triangle ABC is a right triangle with hypotenuse BC.

The Midpoint Formula

For two points a and b on a coordinate line, the coordinate of the midpoint is the arithmetic average (a + b)/2. For points P₁(x₁, y₁) and P₂(x₂, y₂) in the plane, the midpoint M(x, y) has x as the midpoint of x₁ and x₂ on the x-axis, and y as the midpoint of y₁ and y₂ on the y-axis.

🔑 Definition — Midpoint Formula: The midpoint of the line segment joining two points (x₁, y₁) and (x₂, y₂) in a coordinate plane is midpoint (x, y) = ((x₁ + x₂)/2, (y₁ + y₂)/2).

📐 Formula: x = (x₁ + x₂)/2, y = (y₁ + y₂)/2 → Each coordinate of the midpoint is the average of the corresponding coordinates of the two endpoints.

📌 Example: Find the midpoint of the line segment joining (3, -4) and (7, 2). The midpoint is ((3 + 7)/2, (-4 + 2)/2) = (10/2, -2/2) = (5, -1).

Circles

If (x₀, y₀) is a fixed point (the center), the circle of radius r centered at (x₀, y₀) is the set of all points whose distance from (x₀, y₀) is exactly r. A point (x, y) lies on this circle if and only if √[(x - x₀)² + (y - y₀)²] = r, or equivalently (x - x₀)² + (y - y₀)² = r².

🔑 Definition — Standard Form of the Equation of a Circle: (x - x₀)² + (y - y₀)² = r², where (x₀, y₀) is the center and r is the radius.

📌 Example: Find an equation for the circle of radius 4 centered at (-5, 3). Here x₀ = -5, y₀ = 3, r = 4. Substituting: (x - (-5))² + (y - 3)² = 4² → (x + 5)² + (y - 3)² = 16. In expanded form: x² + y² + 10x - 6y + 18 = 0.

📌 Example: Find an equation for the circle with center (1, -2) that passes through (4, 2). The radius r is the distance between (4, 2) and (1, -2): r = √[(4 - 1)² + (2 - (-2))²] = √[3² + 4²] = √25 = 5. The equation is (x - 1)² + (y + 2)² = 25, or x² + y² - 2x + 4y - 20 = 0.

💡 Why this matters: The standard form allows immediate identification of a circle's center and radius from the constants in the equation. For example, (x - 2)² + (y - 5)² = 9 has center (2, 5) and radius 3; (x + 7)² + (y + 1)² = 16 has center (-7, -1) and radius 4; x² + y² = 25 has center (0, 0) and radius 5; (x - 4)² + y² = 5 has center (4, 0) and radius √5.

The circle x² + y² = 1, centered at the origin with radius 1, is called the unit circle and has special importance.

Other forms for the equation of a circle

By squaring and simplifying the standard form, we get x² + y² + dx + ey + f = 0, where d, e, and f are constants. Another version is Ax² + Ay² + Dx + Ey + F = 0, where A ≠ 0. If the equation is given in these forms, rewrite it in standard form by completing the square to find the center and radius.

📌 Example: Find the center and radius of x² + y² - 8x + 2y + 8 = 0. Group x and y terms and move the constant: (x² - 8x) + (y² + 2y) = -8. Complete the square: (x² - 8x + 16) + (y² + 2y + 1) = -8 + 16 + 1 → (x - 4)² + (y + 1)² = 9. Center: (4, -1), radius: 3.

📌 Example: Find the center and radius of 2x² + 2y² + 24x - 81 = 0. Divide by 2: x² + y² + 12x - 81/2 = 0 → (x² + 12x) + y² = 81/2. Complete the square: (x² + 12x + 36) + y² = 81/2 + 36 → (x + 6)² + y² = 153/2. Center: (-6, 0), radius: √(153/2).

Degenerate Cases of a Circle

An equation of the form (x - x₀)² + (y - y₀)² = k may not always represent a circle. The possibilities depend on k:

• k > 0: The graph is a circle with center (x₀, y₀) and radius √k.
• k = 0: The only solution is x = x₀, y = y₀; the graph is a single point (x₀, y₀).
• k < 0: No real solutions; the equation has no graph.

📌 Example: (x - 1)² + (y + 4)² = -9 has no graph because no real x, y make the left side negative. (x - 1)² + (y + 4)² = 0 has graph as the single point (1, -4).

🔑 Theorem: An equation of the form Ax² + Ay² + Dx + Ey + F = 0, where A ≠ 0, represents a circle, a point, or has no graph. The last two cases are called degenerate cases.

The graph of y = ax² + bx + c

An equation of the form y = ax² + bx + c (a ≠ 0) is called a quadratic equation in x. Its graph is a parabola. If a > 0, the parabola opens upward; if a < 0, it opens downward. The parabola is symmetric about a vertical line called the axis of symmetry, which cuts the parabola at the vertex. The vertex is the lowest point if a > 0 and the highest point if a < 0.

📐 Formula: The x-coordinate of the vertex is given by x = -b / (2a). The y-coordinate is found by substituting this x-value into the equation.

📌 Example: Sketch the graph of y = x² - 2x - 2. Here a = 1, b = -2, c = -2. Vertex x-coordinate: x = -(-2) / (2*1) = 1. y-coordinate: y = 1² - 2(1) - 2 = -3. Plot vertex (1, -3) and points on each side to graph the upward-opening parabola.

📌 Example: Sketch the graph of y = -x² + 4x - 5. Here a = -1, b = 4, c = -5. Vertex x-coordinate: x = -4 / (2*(-1)) = 2. y-coordinate: y = -(2)² + 4(2) - 5 = -4 + 8 - 5 = -1. Plot vertex (2, -1) and points on each side to graph the downward-opening parabola.

The y-intercept is found by setting x = 0. The x-intercepts are found by setting y = 0 and solving the quadratic equation ax² + bx + c = 0.

📌 Example: Solve the inequality x² - 2x - 2 > 0 graphically. The inequality is satisfied where the parabola y = x² - 2x - 2 is above the x-axis. Find x-intercepts by solving x² - 2x - 2 = 0 using the quadratic formula: x = [2 ± √(4 + 8)] / 2 = [2 ± √12] / 2 = 1 ± √3. The solution set is (-∞, 1 - √3) ∪ (1 + √3, ∞).

📌 Example: A ball thrown straight up with initial velocity 24.5 m/sec has height s = 24.5t - 4.9t² after t seconds. (a) Graph s vs. t. Vertex at t = -b/(2a) = -24.5 / (2*(-4.9)) = 2.5 sec, s = 24.5(2.5) - 4.9(2.5)² = 30.625 m. t-intercepts: t = 0 and t = 5 (from factoring s = 4.9t(5 - t)). (b) The ball rises to a maximum height of 30.625 m.

Graph of x = ay² + by + c

If x and y are interchanged, the equation x = ay² + by + c is a quadratic in y. Its graph is a parabola with its line of symmetry parallel to the x-axis. The y-coordinate of the vertex is y = -b / (2a).

⭐ Key Takeaways

The distance formula d = √[(x₂ - x₁)² + (y₂ - y₁)²] and the midpoint formula ((x₁+x₂)/2, (y₁+y₂)/2) are fundamental for coordinate geometry problems. The standard equation of a circle (x - x₀)² + (y - y₀)² = r² allows immediate identification of center and radius; completing the square is essential for converting expanded forms to standard form. Recognize degenerate cases where an equation represents a point or no graph. For quadratics y = ax² + bx + c, the vertex x = -b/2a is critical for graphing and solving inequalities, and the concept applies to real-world problems like projectile motion.

🧠 Quick Revision Questions

1. What is the distance between points (-3, 2) and (5, -4)?
2. Find the midpoint of the line segment joining (0, 8) and (-6, -2).
3. What is the standard form of the equation of a circle with center (2, -3) and radius 6?
4. Determine the center and radius of the circle given by x² + y² + 6x - 8y + 9 = 0.
5. For the parabola y = 2x² - 8x + 5, what are the coordinates of the vertex?

📘 Lecture 6 — Functions

📖 Overview: This lecture introduces the fundamental concept of a function, explaining how one quantity depends on another. It covers the notation for functions, their domains and ranges, and explores how functions can be defined piecewise or with reversed variable roles. Understanding functions is crucial for modeling and analyzing relationships in mathematics and science.

🗂️ Topics Covered

The lecture begins by defining a function as a dependence of one quantity on another, illustrated with examples like the area of a circle depending on its radius. It then covers the notation for functions introduced by Euler, including how to evaluate functions at specific values and with different variables. The concept of the domain of a function is explored, distinguishing between natural and restricted domains, followed by techniques for finding the range of a function, including by inspection and algebra. The lecture concludes with functions defined piecewise and the reversal of the roles of x and y.

📝 Lecture Summary

Function

A function describes a situation where a quantity y depends on another quantity x in such a way that each value of x determines exactly one value of y. For example, the equation y = 4x+1 is a function because each x value leads to a unique y value.

🔑 Definition — Function: If a quantity y depends on another quantity x in such a way that each value of x determines exactly one value of y, we say that y is a function of x.

📌 Example: For y = 4x+1:

• If x = 1, y = 5
• If x = 2, y = 9
• If x = 3, y = 13

Not a function example: y = ±√x. If x = 4, then y = +2 and y = -2, so a single x value does not lead to exactly one y value.

Notations for Functions

Swiss mathematician Euler introduced the notation y = f(x), read as "y equals f of x". This indicates that y is a function of x.

Key points:

• The variable alongside f is INDEPENDENT (usually x)
• The other variable is DEPENDENT (usually y)
• f(x) is read as "y function of x", NOT as "f multiplied by x"
• f does not represent a number; it is just for expressing functional relationship
• Any letter can be used instead of f: y = f(x), y = g(x), y = h(x)

📌 Example: y = f(x) = x² f(3) = (3)² = 9 f(-2) = (-2)² = 4

📌 Example: If φ(x) = 1/(x³-1) φ(5^(1/6)) = 1/((5^(1/6))³-1) = 1/(5^(3/6)-1) = 1/(5^(1/2)-1) = 1/(√5-1) φ(1) = 1/(1³-1) = 1/0 = undefined

📌 Example: If F(x) = 2x²-1 F(d) = 2d²-1 F(t-1) = 2(t-1)²-1 = 2t²-4t+2-1 = 2t²-4t+1

If two functions look alike in all aspects other than a difference in variables, they are the SAME function. g(c) = c²-4c and g(x) = x²-4x are the SAME function.

Domain of a Function

The domain of a function is the set consisting of all allowable values for the independent variable. The independent variable is not always allowed to take on any value; it may be restricted to take on values from some set. Domain is determined usually by physical constraints on the phenomenon being represented by functions.

📌 Example: A square with side length x cm is cut from four corners of a 10 cm square cardboard. The remaining area y = 100 - 4x². Since x denotes length, x cannot be negative, and its value cannot exceed 5 (otherwise cuts overlap). Thus x must satisfy 0 ≤ x ≤ 5, so the domain is [0, 5].

Natural Domain: If a function is defined by a formula and no domain is explicitly stated, the domain consists of all real numbers for which the formula makes sense and the function has a real value.

📌 Example: h(x) = 1/[(x-1)(x-3)] If x=1 or x=3, the bottom becomes 0, so these are not part of the domain. Thus domain = (-∞, 1) U (1, 3) U (3, +∞)

Restricted Domains: Domains can be altered by restricting them for various reasons, such as simplifying functions by canceling common factors.

📌 Example: h(x) = (x²-4)/(x-2) This has a real value everywhere except at x=2 (division by zero). Thus domain = all x except x=2. If we rewrite: h(x) = (x-2)(x+2)/(x-2) = x+2 Now h(2)=2+2=4, so h(x) is defined at x=2. To cancel the factor without altering the domain, write: h(x) = x+2, x ≠ 2

Range of a Function

The range of a function is the set of all possible values for f(x) as x varies over the domain.

Techniques for Finding Range:

1. By Inspection

📌 Example: Find the range of f(x) = x² Rewrite as y = x². As x varies over the reals, y is all positive reals. Range = [0, +∞)

📌 Example: Find the range of g(x) = 2 + √(x-1) Domain is natural domain [1, +∞) As x varies over [1, +∞), √(x-1) varies over [0, +∞) So y = 2 + √(x-1) varies over [2, +∞) Range = [2, +∞)

2. By some algebra

📌 Example: Find the range of y = (x+1)/(x-1) Natural domain: all real numbers except x=1. Solve for x in terms of y: y = (x+1)/(x-1) y(x-1) = x+1 xy - y = x+1 xy - x = y+1 x(y-1) = y+1 x = (y+1)/(y-1) y=1 is not in the range. Range = {y: y ≠ 1} = (-∞, 1) U (1, +∞)

Functions Defined Piecewise

Sometimes functions need to be defined by formulas that have been "pieced together", with different formulas for different parts of the domain.

📌 Example: Taxicab ride cost Cost is 1.75 rupees for any ride up to and including one mile. After one mile, rider pays an additional 50 paisa per mile. If f(x) is the total cost for x miles: f(x) = { 1.75, 0 < x ≤ 1 { 1.75 + 0.50(x-1), 1 < x }

Reversing the Roles of x and y

Usually x is independent and y dependent, but roles can be reversed for convenience.

📌 Example: x = 4y⁵ - 2y³ + 7y - 5 This is of the form x = g(y), where x is expressed as a function of y, treating y as the independent variable.

📌 Example: 3x + 2y = 6 Can be written as y = -3/2 x + 3 or x = -2/3 y + 2 The choice depends on how the equation will be used.

⭐ Key Takeaways

A function is a relationship where each input (x) determines exactly one output (y), and Euler's notation f(x) clearly identifies the independent and dependent variables. The domain of a function is the set of all allowable input values, which can be the natural domain dictated by the formula or a restricted domain due to physical constraints. The range is the set of all possible output values, which can be found by inspection or algebraic manipulation. Functions can be defined piecewise with different formulas for different parts of the domain, and the roles of independent and dependent variables can be reversed when convenient.

🧠 Quick Revision Questions

1. What is the definition of a function and what condition must be satisfied for a relationship to be considered a function?
2. If f(x) = 2x² - 1, what is f(t-1) expressed in simplified form?
3. What is the natural domain of the function h(x) = 1/[(x-1)(x-3)]?
4. Find the range of the function y = (x+1)/(x-1) by solving for x in terms of y.
5. How would you write a piecewise function for a taxicab ride that costs 1.75 rupees for the first mile and 50 paisa per mile thereafter?

📘 Lecture 7 — Operations on Functions

📖 Overview: This lecture explains how functions can be operated upon using arithmetic operations (addition, subtraction, multiplication, division) and a special operation called composition. Understanding these operations is fundamental for manipulating and combining functions in calculus and advanced mathematics.

🗂️ Topics Covered

The lecture covers arithmetic operations on functions with formal definitions and domain considerations, notation for powers of functions, composition of functions with domain restrictions, the non-commutative property of composition, decomposition of functions into simpler components, and classification of functions including constant functions, monomials, and polynomials.

📝 Lecture Summary

Arithmetic Operations on Functions

Like numbers, functions can be added, subtracted, multiplied, and divided. When we perform these operations on two functions, we define a new function. For example, if f(x) = x² and g(x) = x, then the sum is f(x) + g(x) = x² + x, which defines a new function called the SUM of f and g, denoted as (f + g)(x).

Definitions for Operations on Functions

Given functions f and g, we define their sum, difference, product, and quotient as follows:

🔑 Definition — Sum: (f + g)(x) = f(x) + g(x) 🔑 Definition — Difference: (f – g)(x) = f(x) – g(x) 🔑 Definition — Product: (f · g)(x) = f(x) · g(x) 🔑 Definition — Quotient: (f/g)(x) = f(x)/g(x), where g(x) ≠ 0

For the functions f+g, f–g, and f·g, the domains are defined as the intersection of the domains of f and g. For f/g, the domain is the intersection of the domains of f and g except for points where g(x)=0.

📌 Example: f(x) = 1 + √(x – 2) and g(x) = x – 1 (f + g)(x) = (1 + √(x – 2)) + (x – 1) = x + √(x – 2) • Domain of f is [2, +∞) • Domain of g is (–∞, +∞) • Domain of f+g is [2, +∞) ∩ (–∞, +∞) = [2, +∞)

📌 Example: f(x) = 3 and g(x) = √x (f · g)(x) = 3 · √x = 3√x • The natural domain of 3x is (–∞, +∞), but the domain of (f·g) must be the intersection of domains of f and g, which is [0, +∞) • We clarify this by writing (f·g)(x) = 3√x, x ≥ 0

💡 Why this matters: When multiplying or combining functions, the domain of the resulting function is NOT simply the natural domain of the simplified expression — it must respect the original domains of both functions.

Notation

f²(x) means f multiplied by itself twice: f²(x) = f(x) · f(x) fⁿ(x) means f multiplied by itself n times: fⁿ(x) = f(x) · f(x) … · f(x) (n times)

📌 Example: (sin x)² = sin²(x)

Composition of Functions

Composition is a new operation that has no analog with arithmetic operations. When two functions are composed, one function is assigned as a VALUE to the independent variable of the other.

🔑 Definition — Composition of f with g, written as (f ∘ g)(x), is defined as (f ∘ g)(x) = f(g(x))

The domain of (f ∘ g) consists of all x in the domain of g for which g(x) is in the domain of f. In order to compute f(g(x)), one needs to FIRST compute g(x) for an x from the domain of g, then needs g(x) to be in the domain of f to compute f(g(x)).

A helpful analogy: "Put the Sock on first, then the shoe" — you must evaluate the inner function first.

📌 Example: f(x) = x³ and g(x) = x + 4 (f ∘ g)(x) = f(g(x)) = (g(x))³ = (x + 4)³

📌 Example: f(x) = x² + 3 and g(x) = √x (f ∘ g)(x) = f(g(x)) = (g(x))² + 3 = (√x)² + 3 = x + 3 • Domain of g is [0, +∞) • Domain of f is (–∞, +∞) • Domain of (f ∘ g) is all x in [0, +∞) such that g(x) lies in (–∞, +∞), so domain is [0, +∞)

Important: Generally, (f ∘ g) ≠ (g ∘ f). Like the sock-shoe analogy: "Put Sock on then Shoe ≠ Put Shoe on first then Sock."

Decomposition of Functions

Sometimes we want to break up functions into simpler ones — this is called DECOMPOSING them into composition of simpler functions.

📌 Example: h(x) = (x + 1)² First we add 1 to x, then we square (x + 1). We can break up the function as: f(x) = x + 1 g(x) = x² h(x) = g(f(x))

There is more than one way to decompose a function.

📌 Example: (x² + 1)¹⁰ can be decomposed in different ways: • Way 1: (x² + 1)¹⁰ = [(x² + 1)²]⁵ = f(g(x)) where g(x) = (x² + 1)² and f(x) = x⁵ • Way 2: (x² + 1)¹⁰ = [(x² + 1)⁵]² = f(g(x)) where g(x) = (x² + 1)⁵ and f(x) = x²

📌 Example: T(x) = ³√x³ = f(g(h(x))) f(x) = ³√x g(x) = x³ h(x) = √x

Classification of Functions

🔑 Definition — Constant Functions: These assign the same number to every x in the domain. 📌 Example: f(x) = 2, so f(1) = 2, f(–7) = 2, etc.

🔑 Definition — Monomial in x: Anything that looks like cxⁿ, with c a constant and n any nonnegative integer. 📌 Example: 2x⁵, 3x⁵⁵ — these are monomials. ⚠️ Not monomials: 4x⁻⁴, 5x^(2/3) — because powers are not nonnegative integers.

🔑 Definition — Polynomial in x: Anything like f(x) = a₀ + a₁x + a₂x² + … + aₙxⁿ. 📌 Example: 4x⁴ + 3x² + 1, 17 – (4/3)x²

⭐ Key Takeaways

The four arithmetic operations (sum, difference, product, quotient) can be applied to functions, and for each operation, the domain of the resulting function is the intersection of the domains of the original functions — with the quotient additionally excluding points where the denominator equals zero. Composition of functions (f ∘ g)(x) = f(g(x)) is fundamentally different from arithmetic operations, requiring that the inner function's output falls within the outer function's domain; composition is not commutative, meaning (f ∘ g) ≠ (g ∘ f) in general. Functions can be decomposed into compositions of simpler functions, often in multiple ways. Finally, functions are classified as constant (same output for all inputs), monomial (cxⁿ where n is a nonnegative integer), or polynomial (sum of monomials).

🧠 Quick Revision Questions

1. What is the definition of (f + g)(x) and how is its domain determined?
2. Why must we restrict the domain of (f · g)(x) even when the simplified expression has a larger natural domain?
3. What is the composition (f ∘ g)(x) and how does its domain differ from that of arithmetic operations?
4. Give an example showing that (f ∘ g) is generally not equal to (g ∘ f).
5. What distinguishes a monomial from a polynomial, and what restriction exists on the exponent for a monomial?

📘 Lecture 8 — Graphs of Functions

📖 Overview: This lecture explains how to represent functions graphically and visualize their behavior. It covers how to use graphs of simple functions to construct graphs of more complicated functions through translations, reflections, and scaling, culminating in the vertical line test to determine if a graph represents a function.

🗂️ Topics Covered

The lecture begins by defining the graph of a function as the graph of the equation y = f(x), with examples including linear and piecewise functions like the absolute value. It then explores graphing functions by translations (shifting up, down, left, right) and reflections about axes, followed by vertical scaling. The lecture concludes with the vertical line test, a critical tool for identifying function graphs.

📝 Lecture Summary

Definition of Graph of a function

A graph of an equation is just the points on the xy-plane that satisfy the equation. Similarly, the graph of a Function f in the xy-plane is the GRAPH of the equation y = f(x).

🔑 Definition — Graph of a function: The set of all points (x, y) in the xy-plane that satisfy the equation y = f(x).

📌 Example: Sketch the graph of f(x) = x + 2. By definition, the graph of f is the graph of y = x + 2. This is just a line with y-intercept 2 and slope 1.

📌 Example: Sketch the graph of f(x) = |x|. The graph will be that of y = |x|. The absolute value function is PIECEWISE defined as: y = |x| = { x if x ≥ 0; -x if x < 0 } The top part is the function y = f(x) = x, a straight line through the origin with slope 1, but only defined for x ≥ 0. y = -x is a straight line through the origin with slope -1 but defined only for x < 0.

📌 Example: t(x) = (x² - 4)/(x - 2) This is the same as t(x) = x + 2, x ≠ 2. The graph will be of y = x + 2, x ≠ 2. t(x) is the same as in example 1, except that 2 is not part of the domain, which means there is no y value corresponding to x = 2. So there is a HOLE in the graph.

📌 Example: g(x) = {1 if x ≤ 2; x + 2 if x > 2} For x is less than or equal to 2, the graph is just at y = 1 (a straight line with slope 0). For x is greater than 2, the graph is the line x + 2.

Graphing functions by Translations

If the graph of f(x) is known, we can find the graphs of y = f(x) + c, y = f(x) - c, y = f(x + c), y = f(x - c), where c is any POSITIVE constant.

• If a positive constant c is added to f(x), the geometric effect is the translation of the graph of y=f(x) UP by c units.
• If a positive constant c is subtracted from f(x), the geometric effect is the translation of the graph of y=f(x) DOWN by c units.
• If a positive constant c is added to the independent variable x of f(x), the geometric effect is the translation of the graph of y=f(x) LEFT by c units.
• If a positive constant c is subtracted from the independent variable x of f(x), the geometric effect is the translation of the graph of y=f(x) RIGHT by c units.

📌 Example: Sketch the graph of y = f(x) = √(x - 3) + 2. This graph can be obtained by two translations:

1. Translate the graph of y = √x 3 units to the RIGHT to get the graph of y = √(x - 3).
2. Translate the graph of y = √(x - 3) 2 units UP to get the graph of y = f(x) = √(x - 3) + 2.

📌 Example: Sketch graph of y = x² - 4x + 5 by completing the square. Complete the square: Divide the co-efficient of x by 2 (which is -4/2 = -2), square this result (4) and add to both sides of the equation. y + 4 = (x² - 4x + 5) + 4 y = (x² - 4x + 4) + 5 - 4 y = (x - 2)² + 1 To graph this:

1. Graph y = x².
2. Shift it RIGHT by 2 units to get graph of y = (x - 2)².
3. Shift this UP by 1 Unit to get y = (x - 2)² + 1.

Reflections

• (-x, y) is the reflection of (x, y) about the y-axis.
• (x, -y) is the reflection of (x, y) about the x-axis.
• Graphs of y = f(x) and y = f(-x) are reflections of one another about the y-axis.
• Graphs of y = f(x) and y = -f(x) are reflections of one another about the x-axis.

📌 Example: Sketch the graph of y = ³√(2 - x). We can get the graph by REFLECTION and TRANSLATION:

1. First graph y = ³√x.
2. Reflect it about the y-axis to get graph of y = ³√(-x). (Negative numbers HAVE cube roots.)
3. Translate this graph RIGHT by 2 units to get graph of y = ³√(2 - x) = ³√(-(x - 2)).

Scaling

If f(x) is MULTIPLIED by a POSITIVE constant c, then the following geometric effects take place:

• The graph of f(x) is COMPRESSED vertically if 0 < c < 1.
• The graph of f(x) is STRETCHED vertically if c > 1.
• This is called VERTICAL Scaling by a factor of c.

📌 Example: y = 2 sin(x), y = sin(x), y = (1/2) sin(x). Using c = 2 and c = 1/2 produces corresponding graphs with appropriate VERTICAL SCALINGS.

Vertical Line Test

So far we have started with a function equation and drawn its graph. But not every curve or graph in the xy-plane is that of a function.

📌 Example: A graph which is not the graph of a function: If you draw a VERTICAL line through the point x = a, the line crosses the graph in two points with y values y = b, y = c, giving points (a, b) and (a, c). This cannot be a function by the definition of a function.

🔑 Definition — VERTICAL LINE TEST: A graph in the plane is the graph of a function if and only if NO VERTICAL line intersects the graph more than once.

📌 Example: x² + y² = 25 The graph of this equation is a CIRCLE. Various vertical lines cross the graph in more than 2 places. So the graph is not that of a function, which means that equations of circles are not functions (x as a function of y). 💡 Why this matters: A given graph can be a function with y independent and x dependent. This would happen if the graph passes the HORIZONTAL LINE test (for each y, there can be only one x). Also, y = x² gives g(y) = x = ±√y, so for each x, two y's, and it's not a function in y.

⭐ Key Takeaways

The graph of a function f is the set of all points (x, y) satisfying y = f(x). You must master translating graphs by adding constants to the function (shifts up/down) or to the input variable (shifts left/right) and reflecting graphs across axes. The vertical line test is the definitive method to determine if a given graph represents a function: if any vertical line intersects the graph more than once, it is not a function. Piecewise functions, such as the absolute value function, require careful graphing of each piece over its specified domain.

🧠 Quick Revision Questions

1. How do you obtain the graph of y = f(x - 3) + 2 from the graph of y = f(x)?
2. What is the vertical line test, and why is it used?
3. For a positive constant c, what is the geometric effect of adding c to the independent variable x of f(x)?
4. How does the graph of y = -f(x) relate to the graph of y = f(x)?
5. What happens to the graph of f(x) when it is multiplied by a positive constant c where 0 < c < 1?

📘 Lecture 9 — Limits

📖 Overview: This lecture introduces the fundamental concept of limits in calculus, motivated by two classic problems: finding the area under a curve and finding tangent lines. It explains how limits provide a precise foundation for both differential and integral calculus, covering definitions, notations, and cases where limits fail to exist.

🗂️ Topics Covered

The lecture begins with the area problem and tangent line problem as motivations for calculus, then explores how limits define tangent lines through secant lines and how limits define area through rectangular approximations. It covers the concept of limits in detail, including left-hand and right-hand limits, numerical evaluation, and cases where limits do not exist due to oscillations or unbounded behavior, concluding with limits at infinity.

📝 Lecture Summary

AREA PROBLEM

Given a function f, find the area between the graph of f and the interval [a, b] on the x-axis. The calculus that comes out of the tangent problem is called DIFFERENTIAL CALCULUS, and the calculus that comes out of the area problem is called INTEGRAL CALCULUS. Both are closely related, and the precise definition of "tangent" and "area" depends on the more fundamental notion of LIMIT.

Tangent Lines and LIMITS

In geometry, a line is called tangent to a circle if it meets the circle at exactly one point. However, this definition does not work for all curves. Consider a point P on a curve in the xy-plane. Let Q be another point on the curve. Draw a line through P and Q to get what is called the SECANT line for the curve. Now move point Q toward P. The secant line will rotate to a "limiting" position as Q gets closer and closer to P. The line that occupies this limiting position is called the TANGENT line at P.

🔑 Definition — Secant Line: A line through two points P and Q on a curve. 🔑 Definition — Tangent Line: The limiting position of the secant line as Q approaches P.

Area as a LIMIT

For most geometric shapes, the area can be found by subdividing the shape into finitely many rectangles and triangles. However, some regions cannot be broken into rectangles and triangles that will fill up the area between the curve and the interval [a, b] on the x-axis. Instead, we use rectangles to APPROXIMATE the area. We use same-width rectangles and add their areas. If we let the number of rectangles increase, the approximation becomes better, and the result is obtained as a LIMITING value as the number of rectangles increases.

LIMITS

Limits are a way to study the behavior of the y-values of a function in response to the x-values as they approach some number or go to infinity.

EXAMPLE Consider f(x) = sin(x)/x where x is in radians. Remember that π radians = 180 degrees. f(x) is not defined at x = 0. What happens as x gets very close to 0? We can approach 0 from the left (negative x-axis) and from the right (positive x-axis).

We write:

• lim (x→0⁺) sin(x)/x means "the limit of f(x) as x approaches 0 from the right" — called the RIGHT HAND LIMIT
• lim (x→0⁻) sin(x)/x means "the limit of f(x) as x approaches 0 from the left" — called the LEFT HAND LIMIT

The tables show that as x approaches 0 from both sides, f(x) approaches 1. We write this as:

• lim (x→0⁺) sin(x)/x = 1
• lim (x→0⁻) sin(x)/x = 1

When both the left hand and right hand limits match, we say that the LIMIT exists. We write this as:

• lim (x→0) sin(x)/x = 1

In general, we write the limit as x approaches x₀ as: x → x₀

🔑 Definition — Limit exists: When the left-hand limit and right-hand limit are equal.

💡 Why this matters: The existence of a limit at a point does not require the function to be defined at that point; it only requires the function values to approach a single value from both sides.

TABLE of Limit Notations and situations

Numerical evidence for calculating limits can mislead

EXAMPLE Find lim (x→0) sin(π/x). The table shows values of f(x) for various x, suggesting the limit is 0. However, the graph has NO LIMITING VALUE as it OSCILLATES between 1 and -1.

Existence of Limits

Functions don't always have a limit as x approaches some number. If this is the case, we say LIMIT DOES NOT EXIST OR DNE. Limits fail for many reasons, but usual culprits are:

• Oscillations
• Unbounded Increase or Decrease

Example 1 — Unbounded Increase: When the values of f(x) = y increase without bound from both the left and the right:

• lim (x→x₀⁺) f(x) = lim (x→x₀⁻) f(x) = lim (x→x₀) f(x) = +∞
• The +∞ classifies the DNE as caused by unbounded-ness towards +∞. It is NOT A NUMBER.

Example 2 — Unbounded Decrease: When the values of f(x) = y decrease without bound from both the left and the right:

• lim (x→x₀⁺) f(x) = lim (x→x₀⁻) f(x) = lim (x→x₀) f(x) = -∞
• The -∞ classifies the DNE as caused by unbounded-ness towards -∞. It is NOT A NUMBER.

Example 3 — One-sided limits don't match:

• lim (x→x₀⁺) f(x) = +∞ and lim (x→x₀⁻) f(x) = -∞
• The two-sided limits don't match, so the limit does not exist.

Limits at Infinity

So far we saw limits as x approaches some point x₀. Now we see limits as x goes to +∞ or -∞.

Example:

• lim (x→x₀⁺) f(x) = 4 and lim (x→x₀⁻) f(x) = -1

Note that when we find limits at infinity, we only do it from one side. The reason is that you can approach infinity from only one side! x → ∞ means x gets bigger and bigger, and x can do that only from one side depending on whether it goes to +∞ or -∞.

Example: For a function where the graph oscillates as x → +∞, if the oscillations decrease and settle down on y = -2, then the limit at infinity is -2.

🔑 Definition — Limit at Infinity: The value that f(x) approaches as x increases or decreases without bound.

⭐ Key Takeaways

The concept of a limit is foundational to calculus, defining both tangent lines (through secant lines approaching a limiting position) and areas (through rectangular approximations approaching a limiting value). A limit exists only when the left-hand and right-hand limits are equal; symbols like +∞ and -∞ are used to classify the type of failure when limits do not exist due to unbounded behavior. Functions can fail to have limits due to oscillations or unbounded increase/decrease, and numerical evidence can sometimes be misleading, as in the case of sin(π/x). In limits at infinity, we only approach from one side because infinity itself can only be approached from one direction.

🧠 Quick Revision Questions

1. What is the definition of a tangent line using the limit concept?
2. How do we use rectangles to approximate area under a curve, and what role does the limit play?
3. What condition must be satisfied for a two-sided limit to exist at a point x₀?
4. Why can numerical evidence for limits sometimes be misleading? Give an example from the lecture.
5. For limits at infinity, why do we only consider one-sided limits?

📘 Lecture 10 — Limits and Computational Techniques

📖 Overview: This lecture focuses on algebraic techniques for finding limits, moving beyond the graphical approach of the previous lecture. It introduces a theorem for computing limits of combined functions and applies it to polynomials and rational functions. The lecture also covers limits involving the function 1/x and limits of polynomials as x approaches positive and negative infinity.

🗂️ Topics Covered

The lecture begins with a table of limits for two basic functions (f(x)=k and g(x)=x), then introduces Theorem 2.5.1 for limit operations (sum, difference, product, quotient). It covers limits of polynomials and a theorem for direct substitution, limits involving 1/x, limits of polynomials as x approaches ±infinity, and limits of rational functions as x approaches a finite number a and as x approaches ±infinity, including cases with zero denominators.

📝 Lecture Summary

Limits of Two Basic Functions

We begin with a table of limits for two basic functions: f(x) = k, where k is a constant, and g(x) = x. The limit of a constant function as x approaches a is the constant itself, and the limit of the identity function g(x)=x as x approaches a is the value a. These serve as building blocks for more complex functions.

Theorem 2.5.1

Here we have a theorem that will help with computing limits. This theorem states that if L₁ = lim f(x) and L₂ = lim g(x) both exist, then: a) lim [f(x) + g(x)] = L₁ + L₂ (Limit of a sum is the sum of the limits) b) lim [f(x) - g(x)] = L₁ - L₂ (Limit of a difference is the difference of the limits) c) lim [f(x) · g(x)] = L₁ · L₂ (Limit of a product is the product of the limits) d) lim [f(x) / g(x)] = L₁ / L₂, provided L₂ ≠ 0 (Limit of a quotient is the quotient of the limits)

Parts a) and c) can be extended to any number of functions. For instance, lim [f₁(x) + f₂(x) + ... + fₙ(x)] = lim f₁(x) + lim f₂(x) + ... + lim fₙ(x). Also, if f₁ = f₂ = ... = fₙ = f, then lim [f(x)]ⁿ = [lim f(x)]ⁿ. From this last result, we can say lim (xⁿ) = [lim x]ⁿ = aⁿ as x→a. Another useful result is that a constant factor can be moved through a limit sign: lim [k·g(x)] = k · lim g(x).

Limits of Polynomials

Polynomials are functions of the form f(x) = bₙxⁿ + bₙ₋₁xⁿ⁻¹ + ... + b₁x + b₀. To find the limit of a polynomial as x approaches a number 'a', we can substitute the value 'a' directly into the polynomial.

Theorem 2.5.2 states that lim p(x) = p(a) as x→a. This is proven by applying Theorem 2.5.1 to the polynomial's terms.

📌 Example: Find lim (x² - 4x + 3) as x→5. Step 1: Apply the limit to each term: lim x² - lim 4x + lim 3. Step 2: Use the constant factor rule: (lim x)² - 4 lim x + lim 3. Step 3: Substitute x=5: (5)² - 4(5) + 3 = 25 - 20 + 3 = 8.

Limits Involving 1/x

By looking at the graph of f(x) = 1/x, we get the following results:

• lim (1/x) = +∞ as x→0⁺
• lim (1/x) = -∞ as x→0⁻
• lim (1/x) = 0 as x→+∞
• lim (1/x) = 0 as x→-∞

For any real number a, the function g(x) = 1/(x - a) is a translation of f(x)=1/x, and its limits behave similarly near the point x=a.

🔑 Definition — 1/x limit behavior: As x approaches 0 from the positive side, 1/x grows without bound in the positive direction (+∞). As x approaches 0 from the negative side, 1/x grows without bound in the negative direction (-∞). As x goes to positive or negative infinity, 1/x approaches 0.

Limits of Polynomials as x → +∞ and -∞

For polynomials of the form xⁿ:

• lim xⁿ = +∞ as x→+∞ for n=1, 2, 3, ...
• lim xⁿ = +∞ as x→-∞ for n=2, 4, 6, ... (even powers)
• lim xⁿ = -∞ as x→-∞ for n=1, 3, 5, ... (odd powers)

For the function 1/xⁿ (n positive integer):

• lim (1/xⁿ) = 0 as x→+∞
• lim (1/xⁿ) = 0 as x→-∞

📌 Example: lim (2x⁵) = +∞ as x→+∞. lim (-7x⁶) = -∞ as x→+∞.

As x→+∞ or -∞, the limit of a polynomial is determined by its highest power term. For p(x) = c₀ + c₁x + ... + cₙxⁿ, lim p(x) = lim cₙxⁿ as x→±∞. This is because factoring out xⁿ leaves all other terms approaching zero.

Limits of Rational Functions as x → a

A rational function is a function defined by the ratio of two polynomials. To find its limit as x approaches a, we can use part d) of Theorem 2.5.1: lim [f(x)/g(x)] = (lim f(x)) / (lim g(x)), provided lim g(x) ≠ 0.

📌 Example 1: Find lim (5x³ + 4) / (x - 3) as x→2. Step 1: Apply the quotient rule: (lim 5x³ + 4) / (lim x - 3). Step 2: Substitute x=2: (5(2)³ + 4) / (2 - 3) = (40 + 4) / (-1) = -44.

If both numerator and denominator approach 0 as x→a, they have a common factor of (x - a), which can be cancelled.

📌 Example 2: Find lim (x² - 4) / (x - 2) as x→2. Step 1: Factor the numerator: (x + 2)(x - 2) / (x - 2). Step 2: Cancel the common factor (x - 2): lim (x + 2) as x→2. Step 3: Substitute x=2: 2 + 2 = 4. (Note: This is valid because the limit only considers values close to 2, not equal to 2.)

If the denominator's limit is 0 but the numerator's limit is not, the limit is ±∞. To determine the sign, analyze the sign of the function near the point 'a'.

📌 Example 3: Find lim (2 - x) / [(x - 4)(x + 2)] as x→4⁺. Step 1: As x→4⁺, the numerator (2 - x) approaches -2 (negative). Step 2: The denominator (x - 4)(x + 2): as x→4⁺, (x - 4) → 0⁺ (small positive), and (x + 2) → 6 (positive). So the denominator approaches 0 from the positive side. Step 3: A negative number divided by a small positive number gives a large negative number. Result: lim = -∞ as x→4⁺.

Limits of Rational Functions as x → +∞ and -∞

To find limits of rational functions as x→±∞, divide the numerator and denominator by the highest power of x in the denominator.

📌 Example: Find lim (4x² - x) / (2x³ - 5) as x→-∞. Step 1: The highest power of x in the denominator is x³. Divide both numerator and denominator by x³. Step 2: This yields lim (4/x - 1/x²) / (2 - 5/x³). Step 3: As x→-∞, 4/x → 0, 1/x² → 0, and 5/x³ → 0. Step 4: The expression becomes 0 / 2 = 0.

🔑 Quick Rule: For rational functions as x→±∞, the limit is determined by the ratio of the highest power terms: lim (cₙxⁿ + ... ) / (dₙxᵐ + ...) = lim (cₙxⁿ) / (dₙxᵐ). If the degree of the numerator is less than the degree of the denominator (n < m), the limit is 0. If n = m, the limit is cₙ/dₙ. If n > m, the limit is ±∞.

📌 Example continued: Using the quick rule for lim (4x² - x) / (2x³ - 5) as x→-∞: the highest power in the numerator is x², and in the denominator is x³. The ratio is 4x² / 2x³ = 2/x, which approaches 0 as x→-∞. This matches the previous result.

⭐ Key Takeaways

The core of this lecture is learning that the limit of a sum, difference, product, or quotient of functions can be broken down into the corresponding operations on their individual limits, provided those limits exist. A critical skill is recognizing when this theorem applies directly (e.g., for polynomials and many rational functions) and when algebraic manipulation is needed, such as factoring to cancel common factors when both numerator and denominator approach zero. For infinite limits, as x approaches a finite number, we must analyze the sign of the function to determine whether the limit is +∞ or -∞, and as x approaches infinity, the highest power term dominates the behavior of a polynomial or rational function.

🧠 Quick Revision Questions

1. State the four parts of Theorem 2.5.1 for limit operations.
2. What is the limit of the polynomial p(x) = 3x² - 5x + 1 as x approaches 2?
3. For the rational function f(x) = (x² - 9) / (x - 3), what is the limit as x approaches 3?
4. What is the limit of f(x) = 2x³ - 5x as x approaches +∞?
5. Explain how to determine the limit of a rational function as x approaches +∞ by comparing the highest degree terms in the numerator and denominator.

📘 Lecture 11 — Limits: A Rigorous Approach

📖 Overview: This lecture moves beyond the intuitive understanding of limits to establish a formal mathematical definition. It introduces the epsilon-delta definition of a limit and explains how it rigorously captures the idea of “approaching.” The lecture also covers left-hand and right-hand limits and uses the formal definition to prove limit statements and demonstrate when a limit does not exist.

🗂️ Topics Covered

This lecture begins by discussing the need for a formal definition of a limit, moving from the intuitive concept of “approaches” to a precise mathematical statement involving intervals. It defines a limit using epsilon (ε) and delta (δ), where ε represents how close we want f(x) to be to L and δ represents how close x must be to a. The lecture then demonstrates how to apply this definition to prove a simple linear limit and to prove that a piecewise function has no limit at a point, using a contradiction argument.

📝 Lecture Summary

Formal Definition of Limit

The lecture begins by stating that the concept of “approaches” has been intuitive so far and does not use the theory of real numbers. To formalize the limit (\lim_{x \to a} f(x) = L), the idea of intervals is introduced.

The formal statement is rephrased as: For any number ε > 0, if we can find an open interval ((x_0, x_1)) on the x-axis containing the point a such that (L - \varepsilon < f(x) < L + \varepsilon) for each x in ((x_0, x_1)) except possibly (x = a), then (\lim_{x \to a} f(x) = L). So, f(x) is in the interval ((L - \varepsilon, L + \varepsilon)).

The symbol ε is a small positive number that signifies “f(x) being as close to L as we want it to be.” For any such ε, we can find an interval on the x-axis that confines a. The inequality (L - \varepsilon < f(x) < L + \varepsilon) can be written as (|f(x) - L| < \varepsilon). The interval on the x-axis is represented as ((a - \delta, a) \cup (a, a + \delta)), which is the same as (0 < |x - a| < \delta). This δ is the positive number we must find for any given ε.

🔑 Definition — Epsilon-Delta Definition of Limit: (\lim_{x \to a} f(x) = L) means that for any (\varepsilon > 0), there exists a (\delta > 0) such that if (0 < |x - a| < \delta), then (|f(x) - L| < \varepsilon).

📐 Formula: (0 < |x - a| < \delta \Rightarrow |f(x) - L| < \varepsilon) → This means if x is within a distance of δ from a (but not equal to a), then f(x) is within a distance of ε from L. 💡 Why this matters: This is the rigorous foundation for all of calculus, replacing the vague notion of "approaches" with precise inequalities.

📌 Example: Find (\lim_{x \to 2}(3x - 5) = 1). Given any positive number ε, we need to find δ such that (|(3x - 5) - 1| < \varepsilon) if (0 < |x - 2| < \delta). Here, f(x) = 3x - 5, L = 1, a = 2. We start with the condition on f(x): (|(3x - 5) - 1| < \varepsilon) This simplifies to (|3x - 6| < \varepsilon). Then, (|3(x - 2)| < \varepsilon). This gives (|3| \cdot |x - 2| < \varepsilon), so (3|x - 2| < \varepsilon). Therefore, (|x - 2| < \frac{\varepsilon}{3}). Now, we can choose δ such that the second part of the statement is true. If we let (\delta = \frac{\varepsilon}{3}), then if (0 < |x - 2| < \delta = \frac{\varepsilon}{3}), the condition (|x - 2| < \frac{\varepsilon}{3}) is satisfied, which means the first part (|(3x - 5) - 1| < \varepsilon) is also true. Thus, we have proven that (\lim_{x \to 2}(3x - 5) = 1).

Left-hand and Right-hand Limits

This section uses the formal definition to show that a limit does not exist for a specific function.

📌 Example: Show that (\lim_{x \to 0} f(x)) does not exist, where (f(x) = \begin{cases} 1 & \text{if } x > 0 \ -1 & \text{if } x < 0 \end{cases}). We assume the limit exists and is L. Then for any ε > 0, we can find δ > 0 such that (|f(x) - L| < \varepsilon) if (0 < |x - 0| < \delta). In particular, if we take ε = 1, there is a δ > 0 such that (|f(x) - L| < 1) if (0 < |x| < \delta). But both (x = \frac{\delta}{2}) and (x = -\frac{\delta}{2}) satisfy the condition (0 < |x| < \delta). This gives us: (|f(\frac{\delta}{2}) - L| < 1) and (|f(-\frac{\delta}{2}) - L| < 1). Since (\frac{\delta}{2}) is positive, (f(\frac{\delta}{2}) = 1). Since (-\frac{\delta}{2}) is negative, (f(-\frac{\delta}{2}) = -1). So we get (|1 - L| < 1) and (|-1 - L| < 1). This means (0 < L < 2) from the first inequality, and (-2 < L < 0) from the second inequality. This is a contradiction because L cannot be between 0 and 2 and between -2 and 0 at the same time. Therefore, the limit does not exist.

⭐ Key Takeaways

The epsilon-delta definition is the rigorous foundation for limits, replacing intuition with precise inequalities. The key is understanding ε as the desired closeness of f(x) to L and δ as the required closeness of x to a. For a limit to exist, for every ε > 0 there must be a corresponding δ > 0. The process of proving a limit involves working backwards from |f(x)-L| < ε to find a suitable δ in terms of ε. To prove a limit does not exist, one can assume it does and derive a contradiction using the formal definition.

🧠 Quick Revision Questions

1. What are the roles of ε and δ in the formal definition of a limit?
2. Write the precise epsilon-delta definition for (\lim_{x \to a} f(x) = L).
3. In the example proving (\lim_{x \to 2}(3x-5) = 1), what was the chosen δ in terms of ε?
4. Why does the limit of the piecewise function f(x) = {1 if x>0, -1 if x<0} not exist as x→0?
5. If for a given ε you find a δ that works, can any smaller δ also work? Explain briefly.

📘 Lecture 12 — Continuity

📖 Overview: This lecture formally defines the concept of continuity of a function, building from graphical intuition to a rigorous mathematical definition using limits. It covers properties of continuous functions, continuity of polynomials and rational functions, continuity of compositions, and the crucial Intermediate Value Theorem (IVT) and its corollary for finding roots.

🗂️ Topics Covered

The lecture begins by identifying causes of discontinuity from a graph and gives the three-part mathematical definition of continuity at a point. It then proves that polynomials are continuous everywhere, checks continuity for piecewise functions like |x|, and lists algebraic properties of continuous functions (sum, difference, product, quotient). The lecture also addresses continuity of rational functions, compositions of continuous functions, left- and right-hand continuity for endpoints of closed intervals, and finally introduces and applies the Intermediate Value Theorem.

📝 Lecture Summary

CONTINUITY of a function becomes obvious from its graph

A curve is discontinuous at point c if f(x) is not defined there. A break or discontinuity in the graph of f(x) at x = c occurs when:

• f is undefined at c.
• The lim f(x) as x→c does not exist.
• The function is defined at c and the limit exists, but the value of f(x) and the value of the limit differ at the point c.

Definition 2.7.1

A function f is continuous at a number c if the following three conditions are satisfied: (a) f(c) is defined (b) lim f(x) as x→c exists (c) lim f(x) as x→c = f(c)

If any of these conditions fail, f is discontinuous at c, and c is called the point of discontinuity. If f is continuous at all points in an interval (a, b), we say f is continuous on (a, b). A function continuous on the interval (-∞, +∞) is called a continuous function.

🔑 Definition — Continuous at a point: A function f is continuous at c if lim_(x→c) f(x) = f(c). This single equation implies conditions (a) and (b) are true. 📌 Example: Consider f(x) = (x² - 4)/(x - 2) and g(x) = { (x² - 4)/(x - 2) if x ≠ 2; 3 if x = 2 }. f is discontinuous at x = 2 because f(2) is undefined. g is discontinuous at x = 2 because lim_(x→2) g(x) = lim_(x→2) (x+2) = 4, but g(2) = 3. Since lim_(x→2) g(x) ≠ g(2), condition (c) of the definition fails. 📌 Example: Show that f(x) = x² - 2x + 1 is a continuous function (i.e., continuous at all real numbers). We must show that lim_(x→c) f(x) = f(c) for any real number c. Since f(x) is a polynomial, lim_(x→c) (x² - 2x + 1) = c² - 2c + 1 = f(c). The condition is met for all c, so f(x) is continuous.

Theorem 2.7.2

Polynomials are continuous functions. Proof: If P is a polynomial and c is any real number, then by Theorem 2.5.2, lim_(x→c) p(x) = p(c). Since c is any real number, p(x) is continuous everywhere.

📌 Example: Show that f(x) = |x| is continuous. Rewrite f(x) = |x| as a piecewise function: f(x) = { x if x ≥ 0; -x if x < 0 }.

• If c ≥ 0: f(c) = c. The limit lim_(x→c) f(x) = lim_(x→c) x = c = f(c) (using the first piece as x approaches a non-negative c).
• If c < 0: f(c) = -c. The limit lim_(x→c) f(x) = lim_(x→c) (-x) = -c = f(c) (using the second piece as x approaches a negative c). In both cases, lim_(x→c) f(x) = f(c), so |x| is continuous.

Properties of Continuous Functions

Theorem 2.7.3: If functions f and g are continuous at c, then: a) f + g is continuous at c b) f – g is continuous at c c) f . g is continuous at c d) f/g is continuous at c if g(c) ≠ 0 and is discontinuous at c if g(c) = 0

Proof (for product): Let f and g be continuous at c. Then lim_(x→c) f(x) = f(c) and lim_(x→c) g(x) = g(c). By limit rules, lim_(x→c) f(x)·g(x) = [lim_(x→c) f(x)][lim_(x→c) g(x)] = f(c)·g(c). Thus, f·g is continuous at c.

Continuity of Rational Functions

📌 Example: Where is h(x) = (x² - 9)/(x² - 5x + 9) continuous? The numerator and denominator are polynomials, so they are continuous everywhere. By property (d) of Theorem 2.7.3, h is continuous at all points c where the denominator g(c) ≠ 0. Solve x² - 5x + 9 = 0. Solving this gives x = 2 and x = 3. These are the only points of discontinuity. h is continuous for all x ≠ 2, 3.

Continuity of Composition of functions

Theorem 2.7.5: Let lim stand for one of lim_(x→c), lim_(x→c⁺), lim_(x→c⁻), lim_(x→+∞), lim_(x→-∞). If lim g(x) = L and the function f is continuous at L, then lim f(g(x)) = f(L). That is, lim f(g(x)) = f(lim g(x)).

📌 Example: Evaluate lim_(x→3) √(5 - x²). Let f(x) = √x (which is continuous for non-negative arguments) and g(x) = 5 - x². Then lim_(x→3) g(x) = 5 - 9 = -4. By Theorem 2.7.5, lim_(x→3) √(5 - x²) = √(lim_(x→3) 5 - x²) = √(-4) = 4 (taking the principal square root).

Theorem 2.7.6: If the function g is continuous at the point c and the function f is continuous at the point g(c), then the composition f ∘ g is continuous at c.

Continuity from the left and right

Graphically, a function can be discontinuous at the endpoints a and b of a closed interval [a, b] because a two-sided limit doesn't make sense there.

Definition 2.7.7: A function f is called continuous from the left at a point c if:

1. f(c) is defined.
2. lim_(x→c⁻) f(x) exists.
3. lim_(x→c⁻) f(x) = f(c).

A function f is called continuous from the right at a point c if:

1. f(c) is defined.
2. lim_(x→c⁺) f(x) exists.
3. lim_(x→c⁺) f(x) = f(c).

Definition 2.7.8: A function f is said to be continuous on a closed interval [a, b] if the following conditions are satisfied:

1. f is continuous on (a, b).
2. f is continuous from the right at a.
3. f is continuous from the left at b.

📌 Example: Show that f(x) = √(9 - x²) is continuous on the interval [-3, 3].

• For any c in (-3, 3), lim_(x→c) √(9 - x²) = √(9 - c²) = f(c) (by Theorem 2.5.1(e) and continuity of polynomials and the square root function). So f is continuous on (3, -3).
• At x = 3: lim_(x→3⁻) √(9 - x²) = √(lim_(x→3⁻) 9 - x²) = √(0) = 0 = f(3). So f is continuous from the left at 3.
• At x = -3: lim_(x→-3⁺) √(9 - x²) = √(lim_(x→-3⁺) 9 - x²) = √(0) = 0 = f(-3). So f is continuous from the right at -3. Thus, f is continuous on [-3, 3]. 💡 Why this matters: The endpoints are approached from within the domain of the function, which is the interval [-3, 3].

Intermediate Value Theorem (Theorem 2.7.9)

If f is continuous on a closed interval [a, b] and C is any number between f(a) and f(b), inclusive, then there is at least one number x in the interval [a, b] such that f(x) = C.

Theorem 2.7.10 (Corollary): If f is continuous on [a, b], and if f(a) and f(b) have opposite signs, then there is at least one solution of the equation f(x) = 0 in the interval (a, b).

📌 Example: The equation x³ - x - 1 = 0 cannot be solved easily by factoring. Let f(x) = x³ - x - 1. f is a polynomial, hence continuous. Evaluate f(1) = 1 - 1 - 1 = -1 and f(2) = 8 - 2 - 1 = 5. Since f(1) = -1 and f(2) = 5 have opposite signs (and 0 is between them), by the Intermediate Value Theorem (Theorem 2.7.10), there is at least one solution to f(x) = 0 in the interval (1, 2).

⭐ Key Takeaways

The three-part definition of continuity at a point (f(c) defined, limit exists, and they are equal) is the foundation for all other topics. Polynomials are continuous everywhere, and algebraic combinations (sums, products, etc.) of continuous functions are continuous where defined, which allows us to quickly determine continuity for rational functions. The composition of continuous functions is continuous. Finally, the Intermediate Value Theorem is a powerful existence theorem: it guarantees that a continuous function takes on every value between its endpoints, and its corollary is the essential tool for proving that an equation has a root in an interval without solving it.

🧠 Quick Revision Questions

1. What are the three conditions required for a function f(x) to be continuous at a point x = c?
2. Using the definition, explain why the function f(x) = (x² - 4)/(x - 2) is discontinuous at x = 2.
3. If f and g are continuous at c, and g(c) = 0, what can you conclude about the continuity of f/g at c?
4. A function h is defined only on the closed interval [0, 5]. What two additional conditions (besides being continuous on (0,5)) must it meet to be considered continuous on [0, 5]?
5. If a continuous function f on [0, 1] has values f(0) = 2 and f(1) = -3, does the Intermediate Value Theorem guarantee a solution to f(x) = 1 in (0,1)? Explain.

📘 Lecture 13 — Limits and continuity of Trigonometric functions

📖 Overview: This lecture establishes the continuity of sine and cosine functions, extends continuity to other trigonometric functions, and introduces the Squeeze Theorem as a powerful tool for evaluating limits. It proves two fundamental limits involving trigonometric functions that are essential for calculus and examines the behavior of sine and cosine as x approaches infinity.

🗂️ Topics Covered

The lecture covers continuity of sine and cosine functions with formal proofs, continuity of other trigonometric functions (tangent, cotangent, secant, cosecant), the Squeeze Theorem for finding limits, proof that lim sin(x)/x = 1 and lim (1-cos(x))/x = 0 as x approaches 0 using geometric arguments, and limits of sine and cosine as x approaches positive or negative infinity.

📝 Lecture Summary

Continuity of Sine and Cosine

Sine and cosine are initially defined as ratios in right triangles: cos(θ) = adjacent/hypotenuse and sin(θ) = opposite/hypotenuse. We now treat these as functions with angles measured in radians (using x instead of θ).

From the graphs, it is intuitively clear that:

• lim sin(x) = 0 as x → 0
• lim cos(x) = 1 as x → 0

Note that sin(0) = 0 and cos(0) = 1, so the function values match the limits at x = 0.

THEOREM 2.8.1

The functions sin(x) and cos(x) are continuous.

Recall the definition of continuity: A function f is continuous at c if: (a) f(c) is defined (b) lim f(x) as x → c exists (c) lim f(x) as x → c = f(c)

An equivalent definition using h = x - c: A function is continuous at c if: (a) f(c) is defined (b) lim f(h + c) as h → 0 exists (c) lim f(h + c) as h → 0 = f(c)

Proof of Continuity of sin(x)

We assume lim sin(x) = 0 and lim cos(x) = 1 as x → 0.

To show: lim sin(c + h) = sin(c) as h → 0

lim sin(c + h) = lim [sin(c)cos(h) + cos(c)sin(h)] = lim sin(c)cos(h) + lim cos(c)sin(h) = sin(c) lim cos(h) + cos(c) lim sin(h) = sin(c)(1) + cos(c)(0) = sin(c)

💡 Why this matters: This proof shows that the sum formula for sine allows us to break the limit into known limits, establishing continuity everywhere.

The continuity of cos(x) is proved similarly.

Continuity of other trigonometric functions

By Theorem 2.7.3, if f(x) and g(x) are continuous, then f(x)/g(x) is continuous except where g(x) = 0.

tan(x) = sin(x)/cos(x) is continuous everywhere except where cos(x) = 0, which gives: x = ±π/2, ±3π/2, ±5π/2, ...

Similarly:

• cot(x) = cos(x)/sin(x)
• sec(x) = 1/cos(x)
• cosec(x) = 1/sin(x)

All are continuous on appropriate intervals using continuity of sin(x) and cos(x) and Theorem 2.7.3.

Squeeze Theorem for finding Limits

We will prove two important limits:

• lim sin(x)/x = 1 as x → 0
• lim (1 - cos(x))/x = 0 as x → 0

As x → 0, both numerator and denominator go to 0, creating an indeterminate form. Sin(x) → 0 pushes the fraction toward 0, while x → 0 pushes the fraction toward infinity. To resolve this, we confine the function between two simpler functions.

THE SQUEEZING THEOREM

Let f, g, and h be functions satisfying g(x) ≤ f(x) ≤ h(x) for all x in some open interval containing point a, with the possible exception that the inequality need not hold at a.

If g and h have the same limit as x approaches a: lim g(x) = lim h(x) = L as x → a Then f also has this limit: lim f(x) = L as x → a

Example using Squeeze Theorem

Find: lim x² sin²(1/x) as x → 0

Since 0 ≤ sin²(x) ≤ 1 for all x, we have 0 ≤ sin²(1/x) ≤ 1. Multiplying by x²: 0 ≤ x² sin²(1/x) ≤ x²

Now lim 0 = 0 and lim x² = 0 as x → 0. By the Squeezing Theorem: lim x² sin²(1/x) = 0

THEOREM 2.8.3

lim sin(x)/x = 1 as x → 0

Proof: Let x be such that 0 < x < π/2. Construct angle x in standard position from the center of a unit circle.

From the geometric figure: 0 < area of ΔOBP < area of sector OBP < area of ΔOBQ

Area calculations:

• ΔOBP = (1/2)(1)sin(x) = (1/2)sin(x)
• Sector OBP = (1/2)(1)²x = (1/2)x
• ΔOBQ = (1/2)(1)tan(x) = (1/2)tan(x)

Therefore: 0 < (1/2)sin(x) < (1/2)x < (1/2)tan(x)

Multiplying by 2/sin(x): 1 < x/sin(x) < 1/cos(x)

Taking reciprocals: cos(x) < sin(x)/x < 1

This holds for 0 < x < π/2, and also for -π/2 < x < 0 (check exercise 4.9).

Now: lim cos(x) = 1 and lim 1 = 1 as x → 0

Using the Squeezing Theorem: 1 < lim sin(x)/x < 1 as x → 0

Since the middle term is between 1 and 1, it must equal 1.

THEOREM 2.8.4

lim (1 - cos(x))/x = 0 as x → 0

Students should prove this themselves.

Limits of sin(x) and cos(x) as x approaches ±∞

From the graphs of sine and cosine, the y-values oscillate between 1 and -1 as x → +∞ or -∞. Therefore: The limits do not exist (DNE).

⭐ Key Takeaways

The sine and cosine functions are continuous everywhere, which is proven using the sum formula for sine and known limits at zero. All other trigonometric functions are continuous on their domains where denominators are non-zero. The Squeeze Theorem is a critical technique for evaluating limits of functions that are "trapped" between two simpler functions with the same limit. The fundamental limit lim sin(x)/x = 1 as x → 0 is proved using geometric area comparisons of triangles and a sector in a unit circle. As x approaches infinity, sine and cosine oscillate without approaching any single value, so their limits do not exist.

🧠 Quick Revision Questions

1. What are the three conditions that must be satisfied for a function f to be continuous at a point c?

2. Why is tan(x) discontinuous at x = π/2, and what type of discontinuity is this?

3. State the Squeeze Theorem precisely and explain why it was needed to find lim sin(x)/x as x → 0.

4. How does the geometric proof of lim sin(x)/x = 1 use areas of triangles and a sector, and why is a unit circle used?

5. What happens to sin(x) and cos(x) as x → ∞, and why do their limits not exist?

📘 Lecture 14 — Tangent Lines and Rates of Change

📖 Overview: This lecture bridges geometry and physics by showing how the slope of a tangent line to a curve defines the instantaneous rate of change of a function. It formalizes the transition from secant lines (average rates) to tangent lines (instantaneous rates) using limits, which is the foundation of differential calculus.

🗂️ Topics Covered

The lecture covers the geometric concept of a secant line evolving into a tangent line as one point approaches another, then defines the slope of a tangent line as a limit. It introduces average velocity as the slope of a secant line on a distance-time graph and instantaneous velocity as the limit of average velocities. Finally, it generalizes these ideas to average and instantaneous rates of change for any function y = f(x), with a complete worked example for f(x)=x²+1.

📝 Lecture Summary

Secant Lines and Tangent Lines

We saw that a Secant line between two points was turned into a tangent line by moving one of the points towards the other one. The secant line rotated into a LIMITING position which we regarded as a TANGENT line. For points P(x₀, y₀) and Q(x₁, y₁) on a curve y = f(x), the secant line connecting them has slope m_sec = [f(x₁) − f(x₀)] / [x₁ − x₀]. If we let x₁ → x₀ then Q will approach P along the graph, and the secant line will approach the tangent line at P. This means the slope of the secant line will approach that of the tangent line at P as x₁ → x₀.

🔑 Definition — Tangent Line Slope: m_tan = lim_{x₁→x₀} [f(x₁) − f(x₀)] / [x₁ − x₀]

Average and Instantaneous Velocity

Average Velocity = distance travelled / Time Elapsed. This formula tells us that the average velocity is the velocity at which one travels on average during some interval of time. More interesting is Instantaneous velocity, the velocity that an object is traveling at a given INSTANT in time. When a car hits a tree, the damage is determined by the INSTANTANEOUS velocity at the moment of impact, not on the average speed during some time interval before the impact.

To define instantaneous velocity, we look at distance as a function of time, d = f(t). Average velocity over [t₀, t₁] is v_ave = [f(t₁) − f(t₀)] / [t₁ − t₀] = slope of the secant line joining (t₀, d₀) and (t₁, d₁). To find instantaneous velocity at t₀, let t₁ approach t₀. As t₁ gets very close to t₀, our approximate instantaneous velocity gets better. So v_inst = lim_{t₁→t₀} v_ave = lim_{t₁→t₀} [f(t₁) − f(t₀)] / [t₁ − t₀]. This is just the slope of the tangent line at (t₀, d₀). Remember that the limit here means that the two sided limits exist.

Average and Instantaneous Rates of Change

Velocity is the rate of change of position with respect to time. We can generalize: Rate of change of d with respect to t, where d = f(t). Other examples include rate of change of bacteria w.r.t time, rate of change of length of a metal rod w.r.t temperature, and rate of change of production cost w.r.t quantity produced. All have the idea of the rate of change of one quantity w.r.t another quantity. We consider the rate of change of y w.r.t x, where y = f(x). Average rate of change is represented by the slope of a certain Secant Line. Instantaneous rate of change is represented by the slope of a certain Tangent Line.

🔑 Definition — Average Rate of Change: If y = f(x), then the average rate of change of y with respect to x over the interval [x₀, x₁] is m_sec = [f(x₁) − f(x₀)] / [x₁ − x₀], the slope of the secant line joining (x₀, f(x₀)) and (x₁, f(x₁)).

🔑 Definition — Instantaneous Rate of Change: If y = f(x), then the instantaneous rate of change of y with respect to x at point x₀ is m_tan = lim_{x₁→x₀} [f(x₁) − f(x₀)] / [x₁ − x₀], the slope of the tangent line to the graph of f at x₀.

📌 Example: Let y = f(x) = x² + 1 a) Find average rate of y w.r.t x over [3,5]: m_sec = [f(5) − f(3)] / [5 − 3] = [26 − 10] / 2 = 8. So y increases 8 units for each unit increase in x over [3,5].

b) Find instantaneous rate at x₀ = −4: m_tan = lim_{x₁→−4} [(x₁² + 1) − 17] / [x₁ + 4] = lim_{x₁→−4} (x₁² − 16)/(x₁ + 4) = lim_{x₁→−4} (x₁ − 4) = −8. Negative instantaneous rate of change means it's DECREASING.

c) Find instantaneous rate at a general point x₀: m_tan = lim_{x₁→x₀} [(x₁² + 1) − (x₀² + 1)] / [x₁ − x₀] = lim_{x₁→x₀} (x₁² − x₀²)/(x₁ − x₀) = lim_{x₁→x₀} (x₁ + x₀) = 2x₀. The result of part b can be obtained from this general result by letting x₀ = −4.

💡 Why this matters: The general result m_tan = 2x₀ for f(x)=x²+1 shows that the instantaneous rate of change is itself a function of x₀. This concept directly leads to the derivative, which will be the central tool of differential calculus.

⭐ Key Takeaways

The slope of the tangent line at a point equals the limit of the slopes of secant lines as one point approaches the other. Average velocity is geometrically the slope of a secant line on a distance-time graph, while instantaneous velocity is the slope of the tangent line at that moment, defined as a limit. Both average and instantaneous rates of change for any function y=f(x) are defined by secant line slopes and tangent line slopes respectively. The instantaneous rate of change at a general point x₀ for f(x)=x²+1 is 2x₀, which can be used to find the rate at any specific point. A negative instantaneous rate of change indicates the function is decreasing at that point.

🧠 Quick Revision Questions

1. What is the formula for the slope of a secant line between points (x₀, f(x₀)) and (x₁, f(x₁))?
2. How is the slope of a tangent line at point P obtained from secant lines?
3. What is the relationship between average velocity and the slope of a secant line on a distance-time graph?
4. For f(x) = x² + 1, find the instantaneous rate of change at x₀ = 2 using the limit definition.
5. What does a negative instantaneous rate of change indicate about the function?

📘 Lecture 15 — The Derivative

📖 Overview: This lecture formally introduces the derivative as the limit of the slope of a tangent line, establishing it as a fundamental concept in calculus. It defines the derivative function, demonstrates how to compute it from first principles for various functions, and explores geometric and notational interpretations. The lecture also examines the conditions for differentiability and the crucial relationship between differentiability and continuity.

🗂️ Topics Covered

The lecture begins by redefining the tangent line slope using a limit with h approaching 0, leading to the formal definition of the derivative. It then computes derivatives for polynomial and radical functions, introduces Leibniz notation for differentiation, and discusses the existence of derivatives in relation to corners, vertical tangents, and discontinuities. The session concludes by proving that differentiability implies continuity, using the limit definition.

📝 Lecture Summary

The Derivative

The slope of a tangent line to the graph of y = f(x) is given by a limit. Let h = x₁ - x₀, so x₁ = x₀ + h and h → 0 as x₁ → x₀. The tangent formula is rewritten as:

$$m_{\text{tan}} = \lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h}$$

Definition 3.2.1

If P(x₀, y₀) is a point on the graph of a function f, then the tangent line to the graph of f at P is defined to be the line through P with slope:

$$m_{\text{tan}} = \lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h}$$

The equation of the tangent line at the point P(x₀, y₀) is:

$$y - y_0 = m_{\text{tan}}(x - x_0)$$

Example: Find the slope and equation of the tangent line to the graph of f(x) = x² at the point P(3,9).

Using x₀ = 3 and y₀ = 9:

$$m_{\text{tan}} = \lim_{h \to 0} \frac{f(3 + h) - f(3)}{h} = \lim_{h \to 0} \frac{(3 + h)^2 - 9}{h}$$

$$= \lim_{h \to 0} \frac{9 + 6h + h^2 - 9}{h} = \lim_{h \to 0} \frac{6h + h^2}{h}$$

$$= \lim_{h \to 0} \frac{h(6 + h)}{h} = \lim_{h \to 0} (6 + h) = 6$$

The equation of the tangent line is:

$$y - 9 = 6(x - 3) \Rightarrow y = 6x - 9$$

Definition 3.2.2

The function f' defined by the formula:

$$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$

is called the derivative with respect to x of the function f. The domain of f' consists of all x for which the limit exists.

Geometric interpretation: The derivative f' is the function whose value at x is the slope of the tangent line to the graph of the function f at x.

Rate of Change interpretation: If y = f(x), then f' is the function whose value at x is the instantaneous rate of change of y with respect to x at the point x.

Example: Let f(x) = x² + 1. Find f'(x).

$$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = \lim_{h \to 0} \frac{[(x + h)^2 + 1] - [x^2 + 1]}{h}$$

$$= \lim_{h \to 0} \frac{x^2 + 2xh + h^2 + 1 - x^2 - 1}{h} = \lim_{h \to 0} \frac{2xh + h^2}{h}$$

$$= \lim_{h \to 0} (2x + h) = 2x$$

The slope of the tangent line at ANY point x is f'(x) = 2x. So at x = 2, slope f'(2) = 4. At x = 0, slope f'(0) = 0. At x = -2, slope f'(-2) = -4.

Example: Derivative of a Linear Function

For a straight line y = mx + b, the tangent line coincides with the line itself, so the slope must be m.

$$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = \lim_{h \to 0} \frac{[m(x + h) + b] - (mx + b)}{h}$$

$$= \lim_{h \to 0} \frac{mx + mh + b - mx - b}{h} = \lim_{h \to 0} \frac{mh}{h} = \lim_{h \to 0} m = m$$

Example 3: Derivative of a Radical Function

Find the derivative with respect to x of f(x) = √x.

$$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = \lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}$$

$$= \lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x + h} + \sqrt{x}}{\sqrt{x + h} + \sqrt{x}} = \lim_{h \to 0} \frac{(x + h) - x}{h(\sqrt{x + h} + \sqrt{x})}$$

$$= \lim_{h \to 0} \frac{h}{h(\sqrt{x + h} + \sqrt{x})} = \lim_{h \to 0} \frac{1}{\sqrt{x + h} + \sqrt{x}} = \frac{1}{\sqrt{x} + \sqrt{x}} = \frac{1}{2\sqrt{x}}$$

Note that lim (x→0⁺) 1/(2√x) = +∞. This means that as x goes to 0 from the right side, the slopes of the tangent lines to the graph of y = f(x) approach +∞, meaning the tangent lines become VERTICAL.

Derivative Notation

The process of finding the derivative is called differentiation.

When the independent variable is x, the differentiation operation is written as:

$$\frac{d}{dx}[f(x)]$$

This is read as "the derivative of f with respect to x".

If we write y = f(x), then we can also say:

$$\frac{d}{dx}[y] = f'(x) = \frac{dy}{dx}$$

The symbol dy/dx should be regarded as a single symbol for the derivative of a function y = f(x).

If the independent variable is u, then:

$$\frac{dy}{du} = f'(u) \quad \text{and} \quad \frac{d}{du}[f(u)] = f'(u)$$

To denote the value of the derivative at a specific point x = x₀:

$$\left.\frac{d}{dx}[f(x)]\right|_{x=x_0} = f'(x_0)$$

For example:

$$\left.\frac{d}{dx}[\sqrt{x}]\right|{x=x_0} = \left.\frac{1}{2\sqrt{x}}\right|{x=x_0} = \frac{1}{2\sqrt{x_0}}$$

Existence of Derivatives

From the definition, the derivative exists only at points where the limit exists. If x₀ is such a point, we say that f is differentiable at x₀ or f has a derivative at x₀.

f is differentiable on an open interval (a,b) if it is differentiable at EACH point in (a,b). f is a differentiable function if it's differentiable on the interval.

Points at which f is NOT differentiable typically occur when the graph of f(x) has:

• Corners — the two-sided limits don't match when taking the limit of secant lines to get the slope of the tangents
• Vertical tangents
• Points of discontinuity

Relationship between Differentiability and Continuity

Theorem 3.2.3: If f is differentiable at a point x₀, then f is also continuous at x₀.

Proof: We show that lim [h→0] f(x₀ + h) = f(x₀), or equivalently, lim [h→0] [f(x₀ + h) - f(x₀)] = 0.

$$\lim_{h \to 0} [f(x_0 + h) - f(x_0)] = \lim_{h \to 0} \left[\frac{f(x_0 + h) - f(x_0)}{h} \cdot h\right]$$

$$= \lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h} \cdot \lim_{h \to 0} h = f'(x_0) \cdot 0 = 0$$

This theorem states that a function cannot be differentiable at a point of discontinuity.

Example: f(x) = |x|. Find f'(x).

Recall that |x| = { x if x ≥ 0, -x if x < 0 }.

f'(x) = { 1 if x > 0, -1 if x < 0 }. The derivative does not exist at x = 0, because the function has a corner there.

💡 Why this matters: This example shows that while differentiability implies continuity, the converse is NOT true — a continuous function may not be differentiable at all points, as demonstrated by the absolute value function at x = 0.

⭐ Key Takeaways

The derivative is formally defined as the limit of the difference quotient, f'(x) = lim [h→0] (f(x+h)-f(x))/h, which geometrically represents the slope of the tangent line at any point x. Using this definition from first principles, we can compute derivatives for power functions, linear functions, and radical functions, with the derivative of √x being 1/(2√x). Derivative notation includes f'(x), dy/dx, and d/dx[f(x)], all representing the same concept. A function is differentiable at a point only if the defining limit exists; non-differentiability occurs at corners, vertical tangents, and discontinuities. Crucially, differentiability at a point implies continuity at that point, though the reverse is not necessarily true.

🧠 Quick Revision Questions

1. State the formal limit definition of the derivative of a function f at a point x.
2. Using the definition, find the derivative of f(x) = x² + 1 and explain what f'(2) represents geometrically.
3. Why is the function f(x) = |x| not differentiable at x = 0, even though it is continuous there?
4. If a function has a vertical tangent at x = a, what does this imply about the value of the derivative f'(a)?
5. Prove that if f is differentiable at x₀, then f is necessarily continuous at x₀.

📘 Lecture 16 — Techniques Of Differentiation

📖 Overview: This lecture develops theorems that provide shortcuts for calculating derivatives, moving beyond the direct definition used in earlier lectures. These techniques are essential for efficiently differentiating constant functions, power functions, constant multiples, sums, differences, products, and quotients of functions, forming the foundation for more advanced calculus.

🗂️ Topics Covered

The lecture covers several key theorems for differentiation: the derivative of constant functions, the Power Rule for positive integers, the constant multiple rule, rules for sums and differences of functions, the product rule, the quotient rule, the reciprocal rule, and an extension of the Power Rule to all integers.

📝 Lecture Summary

Derivatives of Constant Functions

Theorem 3.3.1 states that if "f" is a constant function, meaning ( f(x)=c ) for all ( x ), then its derivative is zero. This result is geometrically obvious because the function ( y = c ) is a horizontal line with a slope of 0.

🔑 Definition — Derivative of a Constant: ( f'(x) = 0 ) 📐 Formula: (\frac{d}{dx}[c] = 0 ) → The derivative of any constant is always zero. 📌 Example: If ( f(x) = 5 ), then ( f'(x) = 0 ).

Theorem 3.3.2 (Power Rule)

If ( n ) is a positive integer, then the derivative of ( x^n ) is ( n \cdot x^{n-1} ). The proof uses the limit definition of the derivative and the binomial expansion of ( (x+h)^n ). Distributing the limit over the sum causes all terms except the first to become zero.

🔑 Definition — Power Rule: ( \frac{d}{dx}[x^n] = n \cdot x^{n-1} ) for positive integer ( n ). 📐 Formula: (\frac{d}{dx}[x^n] = n x^{n-1} ) → To differentiate a power, bring the exponent down as a coefficient and reduce the exponent by one. 📌 Example:

• (\frac{d}{dx}[x^5] = 5x^{4})
• (\frac{d}{dx}[x] = 1 x^{1-1} = 1 x^{0} = 1 \cdot 1 = 1)

Theorem 3.3.3 (Constant Multiple Rule)

Let ( c ) be a constant and ( f ) be a function differentiable at ( x ). Then the function ( c \cdot f ) is also differentiable at ( x ), and its derivative is ( c ) times the derivative of ( f ).

🔑 Definition — Constant Multiple Rule: ( \frac{d}{dx}[cf(x)] = c \frac{d}{dx}[f(x)] ) 📐 Formula: ( \frac{d}{dx}[cf(x)] = c f'(x) ) → A constant factor can be pulled out in front of the derivative. 📌 Example: ( \frac{d}{dx}[3x^{8}] = 3 \cdot \frac{d}{dx}[x^{8}] = 3 \cdot 8x^{7} = 24x^{7} )

Derivative of Sums and Differences of Functions

If ( f ) and ( g ) are differentiable functions at ( x ), then ( f+g ) is also differentiable. The derivative of a sum is the sum of the derivatives. This is proven using the limit definition of the derivative and separating the limit over the sum of two separate limits. The same rule applies for the difference of two functions.

🔑 Definition — Sum/Difference Rule: ( \frac{d}{dx}[f(x) \pm g(x)] = \frac{d}{dx}[f(x)] \pm \frac{d}{dx}[g(x)] ) 📐 Formula: ( (f \pm g)' = f' \pm g' ) → Differentiate each term separately. 📌 Example: ( \frac{d}{dx}[x^4 + x^3] = \frac{d}{dx}[x^4] + \frac{d}{dx}[x^3] = 4x^3 + 3x^2 )

Derivative of a Product

Theorem 3.3.5: If ( f ) and ( g ) are differentiable functions at ( x ), then their product ( f \cdot g ) is differentiable, and its derivative is given by the product rule.

🔑 Definition — Product Rule: ( \frac{d}{dx}[f(x) \cdot g(x)] = f(x) \frac{d}{dx}[g(x)] + g(x) \frac{d}{dx}[f(x)] ) 📐 Formula: ( (f \cdot g)' = f \cdot g' + g \cdot f' ) → The derivative of a product is the first function times the derivative of the second, plus the second function times the derivative of the first. 💡 Why this matters: It's a common mistake to think the derivative of a product is simply the product of derivatives. The product rule is absolutely necessary for differentiating multiplied functions.

Derivative of Quotient

Theorem 3.3.6: If ( f ) and ( g ) are differentiable functions at ( x ), and ( g(x) \neq 0 ), then ( f/g ) is differentiable at ( x ).

🔑 Definition — Quotient Rule: ( \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{g(x) \frac{d}{dx}[f(x)] - f(x) \frac{d}{dx}[g(x)]}{[g(x)]^2} ) 📐 Formula: ( \left( \frac{f}{g} \right)' = \frac{g \cdot f' - f \cdot g'}{g^2} ) → The derivative of a quotient is "low dee-high minus high dee-low over low squared." 💡 Why this matters: Note the order in the numerator is crucial; the derivative of the denominator function is subtracted.

Derivative of a Reciprocal

Theorem 3.3.7: If ( g ) is differentiable at ( x ), and ( g(x) \neq 0 ), then ( 1/g(x) ) is differentiable.

🔑 Definition — Reciprocal Rule: ( \frac{d}{dx}\left[\frac{1}{g(x)}\right] = -\frac{\frac{d}{dx}[g(x)]}{[g(x)]^2} ) 📐 Formula: ( \left( \frac{1}{g} \right)' = -\frac{g'}{g^2} ) → This is a special case of the Quotient Rule where the numerator is 1.

Theorem 3.3.8 (Generalized Power Rule)

If ( n ) is any integer (negative or non-negative), then ( \frac{d}{dx}[x^n] = n x^{n-1} ). This is proven using the reciprocal theorem, thereby extending the Power Rule from only positive integers to all integers. 💡 Why this matters: This single rule now covers all powers of ( x ), making it much more powerful and widely applicable.

📌 Example: ( \frac{d}{dx}[x^{-3}] = -3 x^{-4} ). This uses the Generalized Power Rule.

⭐ Key Takeaways

The most critical points for an exam are the definitions and correct application of each differentiation theorem. You must be able to use the Power Rule for any integer exponent, the Constant Multiple Rule to factor out constants, the Sum/Difference Rule to separate terms, and the Product and Quotient Rules for functions that are multiplied or divided. The Generalized Power Rule negates the need for a separate reciprocal rule for power functions. Memorizing the formulas and practicing their application is the key to success.

🧠 Quick Revision Questions

1. What is the derivative of a constant function like ( f(x) = 10 )?
2. State the Power Rule for differentiation and give an example.
3. What is the product rule for differentiating ( f(x) \cdot g(x) )?
4. Using the quotient rule, find the derivative of ( \frac{x^2}{x+1} ).
5. Apply the Generalized Power Rule to find ( \frac{d}{dx}[x^{-5}] ).

📘 Lecture 17 — Derivatives of Trigonometric functions

📖 Overview: This lecture derives the derivatives of all six trigonometric functions (sin, cos, tan, sec, cosec, cot) using both the limit definition and the quotient rule. It emphasizes the importance of radians for differentiation and applies these derivatives to a real-world shadow problem.

🗂️ Topics Covered

The lecture begins by deriving the derivative of sin(x) using the limit definition and fundamental trigonometric limits. It similarly derives cos(x) and then uses the quotient rule to find derivatives of tan(x), sec(x), cosec(x), and cot(x) by rewriting them in terms of sin and cos. Finally, a practical application calculates the rate of change of a building's shadow length with respect to the sun's angle of elevation, demonstrating unit conversion between radians and degrees.

📝 Lecture Summary

Derivative of f (x) = sin (x)

To differentiate sin(x), we use the definition of derivative: ( \frac{d}{dx}\sin(x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin(x)}{h} ). Using the identity (\sin(x+h) = \sin(x)\cos(h) + \sin(h)\cos(x)), the expression becomes (\lim_{h \to 0} \left[ \sin(x)\frac{\cos(h)-1}{h} + \cos(x)\frac{\sin(h)}{h} \right]). Since (\sin(x)) and (\cos(x)) are constants with respect to (h), and using the limits (\lim_{h \to 0} \frac{\sin(h)}{h} = 1) and (\lim_{h \to 0} \frac{\cos(h)-1}{h} = 0), the result is (\cos(x)).

🔑 Definition — Derivative of sin(x): (\frac{d}{dx}\sin(x) = \cos(x))

Derivative of f (x) = cos (x)

Using the definition (\frac{d}{dx}\cos(x) = \lim_{h \to 0} \frac{\cos(x+h) - \cos(x)}{h}) and the identity (\cos(x+h) = \cos(x)\cos(h) - \sin(x)\sin(h)), a similar calculation yields the derivative.

🔑 Definition — Derivative of cos(x): (\frac{d}{dx}\cos(x) = -\sin(x))

Derivative of f (x) = tan (x)

While the definition works, a simpler approach uses the identity (\tan(x) = \frac{\sin(x)}{\cos(x)}) and applies the Quotient Rule. So (\frac{d}{dx}\tan(x) = \frac{d}{dx}\left[\frac{\sin(x)}{\cos(x)}\right] = \frac{\cos(x) \cdot \frac{d}{dx}\sin(x) - \sin(x) \cdot \frac{d}{dx}\cos(x)}{\cos^2(x)} = \frac{\cos(x)\cos(x) - \sin(x)[-\sin(x)]}{\cos^2(x)} = \frac{\cos^2(x) + \sin^2(x)}{\cos^2(x)} = \frac{1}{\cos^2(x)}).

🔑 Definition — Derivative of tan(x): (\frac{d}{dx}\tan(x) = \sec^2(x)) 📌 Example: Using the quotient rule, we get (\frac{\cos^2(x) + \sin^2(x)}{\cos^2(x)} = \frac{1}{\cos^2(x)} = \sec^2(x)).

Derivative of f(x) = sec(x)

Rewrite (\sec(x) = \frac{1}{\cos(x)}). Apply the quotient rule (or chain rule): (\frac{d}{dx}\sec(x) = \frac{\cos(x)(0) - (1)[-\sin(x)]}{\cos^2(x)} = \frac{\sin(x)}{\cos^2(x)} = \frac{\sin(x)}{\cos(x)} \cdot \frac{1}{\cos(x)}).

🔑 Definition — Derivative of sec(x): (\frac{d}{dx}\sec(x) = \sec(x)\tan(x))

Derivative of f(x) = cosec(x)

Rewrite (\csc(x) = \frac{1}{\sin(x)}). Using the quotient rule: (\frac{d}{dx}\csc(x) = \frac{\sin(x)(0) - (1)[\cos(x)]}{\sin^2(x)} = \frac{-\cos(x)}{\sin^2(x)} = -\frac{\cos(x)}{\sin(x)} \cdot \frac{1}{\sin(x)}).

🔑 Definition — Derivative of cosec(x): (\frac{d}{dx}\csc(x) = -\csc(x)\cot(x))

Derivative of f(x) = cot(x)

Rewrite (\cot(x) = \frac{1}{\tan(x)}). Using the quotient rule: (\frac{d}{dx}\cot(x) = \frac{\tan(x)(0) - (1)[\sec^2(x)]}{\tan^2(x)} = \frac{-\sec^2(x)}{\tan^2(x)} = -\frac{1}{\cos^2(x)} \cdot \frac{\cos^2(x)}{\sin^2(x)} = -\frac{1}{\sin^2(x)}).

🔑 Definition — Derivative of cot(x): (\frac{d}{dx}\cot(x) = -\csc^2(x))

Example

Suppose a building 100 feet high casts a shadow of length (x). Let (\theta) be the angle of elevation of the sun. Find (\frac{dx}{d\theta}) when (\theta = 45^\circ), expressing the answer in feet/degree.

Solution: From the figure, (\tan \theta = \frac{100}{x}), so (x = 100 \cot \theta). We need (\frac{dx}{d\theta}). Differentiating with respect to (\theta) (in radians) gives (\frac{dx}{d\theta} = -100 \csc^2 \theta). At (\theta = 45^\circ = \frac{\pi}{4}) radians, (\frac{dx}{d\theta} = -100 \csc^2(\frac{\pi}{4}) = -100 (\sqrt{2})^2 = -200) feet/radian. To convert to feet/degree, use the relationship (1 \text{ degree} = \frac{\pi}{180} \text{ radians}). So (\frac{dx}{d\theta} = -200 \frac{\text{feet}}{\text{radian}} \cdot \frac{\pi}{180} \frac{\text{radians}}{\text{degree}} = -\frac{10\pi}{9}) feet/degree.

💡 Why this matters: Derivatives of trigonometric functions require radians for standard formulas; unit conversion is crucial when angles are given in degrees.

⭐ Key Takeaways

• The derivatives of sin and cos are fundamental: (\frac{d}{dx}\sin x = \cos x) and (\frac{d}{dx}\cos x = -\sin x).
• Use the quotient rule to derive tan, sec, cosec, and cot derivatives: (\frac{d}{dx}\tan x = \sec^2 x), (\frac{d}{dx}\sec x = \sec x \tan x), (\frac{d}{dx}\csc x = -\csc x \cot x), (\frac{d}{dx}\cot x = -\csc^2 x).
• These formulas only hold when angles are in radians; from first principles, the limits (\lim_{h\to 0}\frac{\sin h}{h}=1) and (\lim_{h\to 0}\frac{\cos h -1}{h}=0) assume radian measure.
• In applied problems, if the angle is given in degrees, you must convert the derivative using (\frac{\pi}{180}) radians per degree.
• The shadow problem demonstrates how a rate of change (e.g., shadow length wrt angle) can be found by differentiating a trigonometric function and then converting units.

🧠 Quick Revision Questions

1. Using the limit definition, what is the first step to differentiate (\sin x)?
2. Why does the quotient rule provide an easier method for finding (\frac{d}{dx}\tan x) compared to the limit definition?
3. Write the derivative of (\sec x) in terms of (\sec x) and (\tan x).
4. In the shadow example, why must (\theta) be in radians before differentiating (x = 100 \cot \theta)?
5. Convert (-200) feet/radian to feet/degree using the conversion factor (\pi/180) radians/degree.

📘 Lecture 18 — The Chain Rule

📖 Overview: This lecture introduces the Chain Rule, a fundamental theorem for finding the derivative of a composition of functions. It explains the formula, provides two approaches to apply it, and demonstrates its use through multiple examples, showing how to differentiate complex functions by breaking them into inner and outer parts.

🗂️ Topics Covered

The lecture covers the derivative of composition of functions (Chain Rule), the generalized derivative formula, a more generalized derivative formula, and an alternative approach to using the Chain Rule. It includes the theorem statement, several worked examples with trigonometric and polynomial functions, and a table of generalized formulas for reference.

📝 Lecture Summary

Derivative of Composition of Functions (Chain Rule)

Suppose we have two functions f and g and we know their derivatives. We can use this information to find the derivative of the composition (f ∘ g)(x) = f(g(x)) by a rule called the Chain Rule for differentiation. If we let y = f(g(x)) and set u = g(x), then y = f(u). The goal is to find dy/dx using the known derivatives dy/du = f'(u) and du/dx = g'(x).

🔑 Definition — Chain Rule (Theorem 3.5.2): If g is differentiable at the point x and f is differentiable at the point g(x), then the composition f(g(x)) is differentiable at the point x. Moreover, if y = f(g(x)) and u = g(x), then y = f(u) and dy/dx = dy/du · du/dx.

📌 Example: Find dy/dx if y = 4 cos(x³). Let u = x³ so that y = 4 cos(u). By the Chain Rule: dy/dx = dy/du · du/dx = d/du[4 cos(u)] · d/dx[x³] = (-4 sin(u))(3x²) = (-4 sin(x³))(3x²) = -12x² sin(x³).

💡 Why this matters: The formula is easy to remember if you think of canceling the du on the top and bottom resulting in dy/dx, but this is only a memory technique.

Generalized Derivative formula

The chain rule gives dy/dx = dy/du · du/dx. Since y = f(u) gives dy/du = f'(u) upon differentiation with respect to u, substituting gives a powerful and simple formula: d/dx[f(u)] = f'(u) · du/dx.

🔑 Definition — Generalized Derivative Formula: d/dx[f(u)] = f'(u) · du/dx, where u is a function of x.

📌 Example: f(x) = (x² - x + 1)²³ Let u = x² - x + 1, so f(u) = u²³. Applying the formula: d/dx[(x² - x + 1)²³] = d/dx[u²³] = 23u²² · du/dx = 23(x² - x + 1)²² · d/dx(x² - x + 1) = 23(x² - x + 1)²² · (2x - 1).

💡 Why this matters: When u = x, the formula simplifies to the standard power rule: d/dx[x²³] = 23x²².

More Generalized Derivative formula

The formula can be applied to a wide range of functions, including trigonometric and radical functions.

📌 Example: d/dx[sin(2x)] Let u = 2x. Then d/dx[sin(2x)] = d/dx[sin(u)] = cos(u) · du/dx = cos(2x) · 2 = 2 cos(2x).

📌 Example: d/dx[tan(x² + 1)] Let u = x² + 1. Then d/dx[tan(x² + 1)] = d/dx[tan(u)] = sec²(u) · du/dx = sec²(x² + 1) · 2x = 2x · sec²(x² + 1).

📌 Example: d/dx[√(x³ + cosec(x))] Let u = x³ + cosec(x). Then: d/dx[√(x³ + cosec(x))] = d/dx[√u] = (1/(2√u)) · du/dx = (1/(2√(x³ + cosec(x)))) · d/dx[x³ + cosec(x)] = (1/(2√(x³ + cosec(x)))) · (3x² - cosec(x) cot(x)).

💡 Why this matters: To choose what u equals, make the substitution so the result is a function you already know how to differentiate.

An alternative approach to using Chain Rule

If we don’t label g(x) as u but keep it in the original notation, we get: d/dx[f(g(x))] = f'(g(x)) · g'(x). Informally, the Chain Rule says: “Derivative of the OUTER function f, then Derivative of the INNER function g, and multiply the two together.”

🔑 Definition — Alternative Chain Rule: d/dx[f(g(x))] = f'(g(x)) · g'(x).

📌 Example: d/dx[cos(3x + 1)] Here f(x) = cos(x), g(x) = 3x + 1. So: d/dx[cos(3x + 1)] = [cos(3x + 1)]' · (3x + 1)' = -sin(3x + 1) · 3 = -3 sin(3x + 1).

⭐ Key Takeaways

The Chain Rule is essential for differentiating any composition of functions, expressed either as dy/dx = dy/du · du/dx or d/dx[f(g(x))] = f'(g(x)) · g'(x). The generalized formula d/dx[f(u)] = f'(u) · du/dx streamlines this process by directly multiplying the derivative of the outer function (evaluated at u) by the derivative of the inner function. You must identify the outer and inner functions correctly, and the rule applies to all function types—power, trigonometric, and radical—as demonstrated with examples like (x² - x + 1)²³, sin(2x), tan(x² + 1), and √(x³ + cosec(x)). The informal "derivative of outer times derivative of inner" approach provides an efficient alternative to substitution.

🧠 Quick Revision Questions

1. State the Chain Rule formula in terms of y, u, and x, and explain what each part represents.
2. Find the derivative of y = cos(x⁵) using the Chain Rule with substitution u = x⁵.
3. Using the generalized formula, find d/dx[sin(4x + 3)].
4. Use the alternative approach to find d/dx[(2x² - 5x)⁶] without explicit substitution.
5. Find d/dx[√(sin(x) + eˣ)] using the Chain Rule.

📘 Lecture 19 — Implicit Differentiation

📖 Overview: This lecture introduces the method of implicit differentiation, which allows us to find derivatives when it is inconvenient or impossible to solve for ( y ) explicitly. It also covers the derivative rule for rational powers of ( x ) and discusses the differentiability of implicit functions. This is essential for handling equations where ( y ) and ( x ) are intertwined.

🗂️ Topics Covered

The lecture covers the method of implicit differentiation as an alternative to solving for ( y ) first, with examples showing how to find ( dy/dx ) and even second derivatives from implicit equations. It then extends the power rule to rational exponents. Finally, it touches on the differentiability of implicit functions.

📝 Lecture Summary

The method of Implicit Differentiation

Consider the equation ( xy = 1 ). To find ( dy/dx ), one method is to solve for ( y ) first: ( y = 1/x ), then differentiate to get ( dy/dx = -1/x^2 ). However, when solving for ( y ) is difficult or impossible, we can use implicit differentiation.

We treat ( y ) as an unknown function of ( x ) and differentiate both sides of the equation with respect to ( x ). For ( xy = 1 ), we use the product rule: [ \frac{d}{dx}(xy) = \frac{d}{dx}(1) ] [ x\frac{d}{dx}(y) + y\frac{d}{dx}(x) = 0 \quad \Rightarrow \quad x\frac{dy}{dx} + y(1) = 0 ] [ x\frac{dy}{dx} = -y \quad \Rightarrow \quad \frac{dy}{dx} = -\frac{y}{x} ] Since ( xy = 1 ) implies ( y = 1/x ), substituting gives ( dy/dx = -1/x^2 ), which matches the explicit result. This demonstrates the core idea of implicit differentiation: differentiate without isolating ( y ) first.

Example: Finding ( dy/dx )

Find ( dy/dx ) if ( 5y^2 + \sin y = x^2 ). It is hard to separate ( y ), so use implicit differentiation: [ \frac{d}{dx}(5y^2 + \sin y) = \frac{d}{dx}(x^2) ] [ 5\frac{d}{dx}(y^2) + \frac{d}{dx}(\sin y) = 2x ] Using the chain rule (since ( y ) is a function of ( x )): [ 5\left(2y \frac{dy}{dx}\right) + \cos(y)\frac{dy}{dx} = 2x ] [ (10y + \cos y)\frac{dy}{dx} = 2x ] [ \frac{dy}{dx} = \frac{2x}{10y + \cos y} ] The derivative formula involves both ( x ) and ( y ), and since the original equation cannot be solved for ( y ), the formula is left in this form.

Example: Finding the Slope of a Tangent Line

Find the slope of the tangent line at the point ( (4,0) ) on the graph of ( 7y^4 + x^3 y + x = 4 ). Use implicit differentiation: [ \frac{d}{dx}(7y^4 + x^3 y + x) = \frac{d}{dx}(4) ] [ 28y^3\frac{dy}{dx} + \left(x^3\frac{dy}{dx} + y\frac{d}{dx}(x^3)\right) + 1 = 0 \quad \text{(Product Rule and Chain Rule)} ] [ 28y^3\frac{dy}{dx} + x^3\frac{dy}{dx} + 3yx^2 + 1 = 0 ] [ \frac{dy}{dx}(28y^3 + x^3) = -3yx^2 - 1 ] [ \frac{dy}{dx} = -\frac{3yx^2 + 1}{28y^3 + x^3} ] At the point ( (4, 0) ), substitute ( x = 4 ) and ( y = 0 ): [ m_{\text{tan}} = -\frac{3(0)(4^2) + 1}{28(0)^3 + 4^3} = -\frac{1}{64} ]

Example: Finding the Second Derivative

Find ( \frac{d^2 y}{dx^2} ) if ( 4x^2 - 2y^2 = 9 ). First, differentiate implicitly to find the first derivative: [ 8x - 4y\frac{dy}{dx} = 0 \quad \Rightarrow \quad \frac{dy}{dx} = \frac{2x}{y} ] Now differentiate ( dy/dx ) with respect to ( x ): [ \frac{d^2 y}{dx^2} = \frac{d}{dx}\left(\frac{2x}{y}\right) ] Use the quotient rule and remember that ( y ) is a function of ( x ): [ \frac{d^2 y}{dx^2} = \frac{y(2) - 2x\frac{dy}{dx}}{y^2} = \frac{2y - 2x\left(\frac{2x}{y}\right)}{y^2} = \frac{2y - \frac{4x^2}{y}}{y^2} = \frac{2y^2 - 4x^2}{y^3} ] From the original equation ( 4x^2 - 2y^2 = 9 ) we have ( 2y^2 - 4x^2 = -9 ), so: [ \frac{d^2 y}{dx^2} = -\frac{9}{y^3} ] 💡 Why this matters: This shows how to compute second derivatives implicitly, which is often required for analyzing function behavior (e.g., concavity).

Derivatives of Rational Powers of ( x )

We know the power rule holds for all integers: ( \frac{d}{dx}[x^n] = nx^{n-1} ). We now extend it to rational numbers ( r ): [ \frac{d}{dx}[x^r] = rx^{r-1} ] Where ( r ) is a rational number. This rule allows us to differentiate functions like ( x^{1/2} ) or ( x^{3/4} ) directly.

⭐ Key Takeaways

• Use implicit differentiation when it is inconvenient or impossible to solve for ( y ) explicitly.
• When differentiating implicitly, treat ( y ) as a function of ( x ) and apply the chain rule to any term containing ( y ).
• The derivative of an implicit function often involves both ( x ) and ( y ) and cannot always be simplified to an expression in ( x ) alone.
• Implicit differentiation can be extended to find higher-order derivatives by differentiating the expression for ( dy/dx ).
• The power rule for derivatives holds for all rational exponents: ( \frac{d}{dx}[x^r] = rx^{r-1} ).

🧠 Quick Revision Questions

1. Why would you choose implicit differentiation over explicit differentiation? (What problem does it solve?)
2. In implicit differentiation, when you differentiate ( y^2 ), why do you get ( 2y \frac{dy}{dx} )?
3. Find ( dy/dx ) if ( x^2 + y^2 = 1 ), using implicit differentiation.
4. If ( x^3 y = 6 ), find ( \frac{dy}{dx} ).
5. What is the derivative of ( x^{2/3} ) according to the rational powers rule?

📘 Lecture 20 — Derivatives of Logarithmic and Exponential Functions and Inverse functions and their derivatives

📖 Overview: This lecture focuses on deriving the formulas for differentiating logarithmic and exponential functions, including the natural logarithm and base e. It also introduces logarithmic differentiation as a powerful technique for simplifying complex derivatives, extends the power rule to all real numbers, and defines inverse functions along with a formula for their derivatives.

🗂️ Topics Covered

The lecture covers the derivative of the general logarithmic function leading to the natural log, the derivative of the natural log function with chain rule applications, the technique of logarithmic differentiation for simplifying messy functions, and using this method to prove the power rule for irrational exponents. Finally, it derives the derivative of exponential functions and introduces inverse functions and their derivative formula.

📝 Lecture Summary

Derivative of the logarithmic function

The derivative of the general logarithmic function ( f(x) = \log_b(x) ) is found using the limit definition of the derivative. By applying properties of logarithms and the limit definition of the number ( e ), we derive the general formula. This formula is then simplified using a change of base to give a more practical derivative expression.

🔑 Definition — logarithmic function ( y = \log_b(x) ): The value ( y ) such that ( b^y = x ), where ( b ) is the base. 📐 Formula: ( \frac{d}{dx} [\log_b(x)] = \frac{1}{x} \log_b(e), \ x > 0 ) → The derivative of a log function is 1/x times the log of e to the given base. 📐 Formula (Simplified): ( \frac{d}{dx} [\log_b(x)] = \frac{1}{x \ln(b)}, \ x > 0 ) → This is a more practical formula using the natural logarithm of the base.

When the base is ( e ), we get the derivative of the natural logarithm. 📐 Formula: ( \frac{d}{dx} [\ln(x)] = \frac{1}{x}, \ x > 0 ) → The derivative of the natural log of x is simply 1/x. 💡 Why this matters: The base ( e ) is the only base for which the derivative of ( \log_b(x) ) is so simple, making it the fundamental logarithm for calculus.

Derivative of the Natural log functions

The derivative formula for ( \ln(x) ) can be generalized using the chain rule for a composition of functions where the outer function is ( \ln(u) ).

📐 Formula: ( \frac{d}{dx} [\ln(u)] = \frac{1}{u} \cdot \frac{du}{dx} ) → The derivative of the natural log of a function u(x) is 1/u times the derivative of u.

📌 Example: Find ( \frac{d}{dx} [\ln(x^2 + 1)] ) Here, ( u = x^2 + 1 ), so ( \frac{du}{dx} = 2x ). ( \frac{d}{dx} [\ln(x^2 + 1)] = \frac{1}{x^2+1} \cdot 2x = \frac{2x}{x^2+1} ).

📌 Example: Find ( \frac{d}{dx} \left[ \ln \left( \frac{x^2 \sin(x)}{\sqrt{1+x}} \right) \right] ) First, use log properties to expand: ( \ln(x^2) + \ln(\sin x) - \frac{1}{2}\ln(1+x) = 2\ln(x) + \ln(\sin x) - \frac{1}{2}\ln(1+x) ). Then, differentiate term by term: ( \frac{2}{x} + \frac{\cos x}{\sin x} - \frac{1}{2(1+x)} ).

Logarithmic Differentiation

Logarithmic differentiation is a technique where you take the natural log of both sides of an expression for y, simplify using log properties, and then differentiate implicitly. This is useful for simplifying derivatives of products, quotients, and powers of functions.

📌 Example: Find ( y' ) for ( y = \frac{x^2 \sqrt[3]{7x-14}}{(1+x^2)^4} ) Step 1: Take (\ln) of both sides and simplify: ( \ln y = 2\ln x + \frac{1}{3}\ln(7x-14) - 4\ln(1+x^2) ) Step 2: Differentiate implicitly with respect to (x): ( \frac{1}{y} y' = \frac{2}{x} + \frac{7}{3(7x-14)} - \frac{8x}{1+x^2} ) Step 3: Solve for ( y' ) by multiplying by ( y ): ( y' = \left( \frac{2}{x} + \frac{7}{3(7x-14)} - \frac{8x}{1+x^2} \right) \cdot \frac{x^2 \sqrt[3]{7x-14}}{(1+x^2)^4} )

Derivatives of Irrational powers of x

The power rule ( \frac{d}{dx} [x^r] = r x^{r-1} ) is proven to hold for all real numbers ( r ), including irrationals, by using logarithmic differentiation.

Proof: Let ( y = x^r ). Then ( \ln y = \ln(x^r) = r \ln x ). Differentiate: ( \frac{dy}{dx} / y = \frac{r}{x} ). Multiply by ( y ): ( \frac{dy}{dx} = \frac{r}{x} \cdot x^r = r x^{r-1} ).

Derivatives of Exponential functions

The derivative of an exponential function ( f(x) = b^x ) is found by using logarithmic differentiation.

Solve: Let ( y = b^x ). Then ( \ln y = x \ln b ). Differentiate: ( \frac{dy}{dx} / y = \ln b ). Multiply by y: ( \frac{dy}{dx} = b^x \ln b ).

📐 Formula: ( \frac{d}{dx} [b^x] = b^x \cdot \ln b ) → The derivative of a general exponential function is the function itself times the natural log of its base. 📐 Formula (Chain Rule): ( \frac{d}{dx} [b^u] = b^u \cdot \ln b \cdot \frac{du}{dx} ) → A general version for a composite function.

For the special case where ( b = e ): 📐 Formula: ( \frac{d}{dx} [e^x] = e^x ) → The derivative of ( e^x ) is simply ( e^x ). 📐 Formula (Chain Rule): ( \frac{d}{dx} [e^u] = e^u \cdot \frac{du}{dx} ) → For a composite function.

Inverse Functions

Two functions ( f ) and ( g ) are inverse functions if applying one after the other returns the original input. A function must be one-to-one (no two x-values map to the same y-value) to have an inverse. The inverse of a function ( f ) is denoted by ( f^{-1} ).

🔑 Definition — Inverse Functions (Definition 7.4.1): Two functions ( f ) and ( g ) are inverses if ( f(g(x)) = x ) for all ( x ) in the domain of ( g ), and ( g(f(x)) = x ) for all ( x ) in the domain of ( f ). 📌 Example: ( f(x) = 2x ) and ( g(x) = \frac{1}{2}x ) are inverses because ( f(g(x)) = 2(\frac{1}{2}x) = x ) and ( g(f(x)) = \frac{1}{2}(2x) = x ).

Derivatives of Inverse Functions

There is a theorem (Theorem 7.4.7) that relates the derivative of a function to the derivative of its inverse. This can be written in a simpler, more practical form.

📐 Formula: If ( y = f^{-1}(x) ) (so ( x = f(y) )), then ( \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} ). → This means the derivative of the inverse function at a point ( x ) is the reciprocal of the derivative of the original function at the point ( y = f^{-1}(x) ).

⭐ Key Takeaways

The most critical concepts are the derivative formulas for ( \ln(x) ) and ( e^x ), which are ( 1/x ) and ( e^x ) respectively, as these are the simplest and most fundamental for calculus. Logarithmic differentiation is a powerful technique taught in this lecture that simplifies the differentiation of complex products, quotients, and functions to a power by using log properties. The lecture also completes the proof of the power rule for all real exponents and connects exponential and logarithmic functions as inverses of each other, culminating in the reciprocal formula for the derivative of an inverse function.

🧠 Quick Revision Questions

1. What is the formula for the derivative of ( \log_b(x) ) using the natural log of the base?
2. What is the derivative of ( \ln(x^3 + 2x) ) and what rule is used?
3. How does logarithmic differentiation simplify finding the derivative of ( y = x^{\sin x} )?
4. Prove the power rule ( \frac{d}{dx}[x^r] = r x^{r-1} ) for any real number ( r ).
5. If ( f(3) = 5 ) and ( f'(3) = 7 ), what is the derivative of the inverse function ( (f^{-1}) ) at ( x = 5 )?

📘 Lecture 21 — Applications of Differentiation

📖 Overview: This lecture explores how derivatives are applied to solve real-world problems involving related rates and to analyze the behavior of functions through increasing/decreasing intervals and concavity. Understanding these applications is crucial for modeling dynamic systems and sketching accurate function graphs.

🗂️ Topics Covered

This lecture covers three main applications of differentiation: Related Rates, which involve finding how fast one quantity changes relative to another using the chain rule; Increasing and Decreasing Functions, where the sign of the first derivative determines whether a function is rising or falling on an interval; and Concavity of Functions, which uses the second derivative to describe the curvature of a graph.

📝 Lecture Summary

Related Rates

Related Rates are real-life problems that involve finding the rate at which one quantity changes with respect to another quantity. For example, we may be interested in finding out how fast the polar ice caps are melting with respect to changes in temperature, or how fast a satellite is changing altitude with respect to changes in time or gravity. To solve these problems, we use the idea of derivatives, which measure the rate of change.

Example: Assume that oil spilled from a ruptured tanker spreads in a circular pattern whose radius increases at a constant rate of 2 ft/sec. How fast is the area of the spill increasing when the radius of the spill is 60 ft?

Let:

• ( t ) = number of seconds elapsed from the time of the spill
• ( r ) = radius of the spill in feet after ( t ) seconds
• ( A ) = area of the spill in square feet after ( t ) seconds

We want to find (\frac{dA}{dt} \big|_{r=60}) given that (\frac{dr}{dt} = 2 \text{ ft/sec}).

The spill is circular, so ( A = \pi r^2 ). Differentiating with respect to ( t ): (\frac{dA}{dt} = 2\pi r \frac{dr}{dt}).

At ( r = 60 ): (\frac{dA}{dt} = 2\pi (60)(2) = 240\pi \text{ ft}^2/\text{sec}).

💡 Why this matters: The rate of change of area depends on both the current radius and the rate at which the radius is changing.

The following steps are helpful in solving related rate problems:

1. Draw a figure and label the quantities that change.
2. Identify the rates of change that are known and those that are to be found.
3. Find an equation that relates the quantity whose rate of change is to be found to those quantities whose rates of change are known.
4. Differentiate the equation with respect to the variable that quantities are changing in respect to (usually time).
5. Evaluate the derivative at appropriate points.

Example: A five-foot ladder is leaning against a wall. It slips in such a way that its base is moving away from the wall at a rate of 2 ft/sec at the instant when the base is 4 ft from the wall. How fast is the top of the ladder moving down the wall at that instant?

Let:

• ( t ) = number of seconds after the ladder starts to slip
• ( x ) = distance in feet from the base of the ladder to the wall
• ( y ) = distance in feet from the top of the ladder to the floor

We want (\frac{dy}{dt} \big|{x=4}) given that (\frac{dx}{dt} \big|{x=4} = 2 \text{ ft/sec}).

Using the Pythagorean theorem: ( x^2 + y^2 = 25 ). Differentiating with respect to ( t ) and using the chain rule: ( 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 ). Solving for (\frac{dy}{dt}): (\frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt}).

When ( x = 4 ), from the Pythagorean theorem: ( 4^2 + y^2 = 25 ), so ( y = 3 ). Thus: (\frac{dy}{dt} \big|_{x=4} = -\frac{4}{3} (2) = -\frac{8}{3} \text{ ft/sec}).

🔑 Definition — Related Rates: Problems that involve finding the rate of change of one quantity given the rate of change of another quantity, using implicit differentiation with respect to time. 📐 Formula: (\frac{dA}{dt} = 2\pi r \frac{dr}{dt}) → The rate of change of the area of a circle is the circumference times the rate of change of the radius. 📌 Example: As the ladder base moves right at 2 ft/sec when x=4 ft, the top slides down the wall at 8/3 ft/sec. The negative sign indicates the top is moving downward.

Increasing and Decreasing Functions

We can use derivatives to get accurate information about the behavior of a function's graph on an interval as we move from left to right. An increasing function on an interval means that as we move from left to right in the x-direction, the y-values increase in magnitude. A decreasing function on an interval means that as we move from left to right in the x-direction, the y-values decrease in magnitude.

Definition 4.2.1: A function ( f ) is said to be increasing on an interval ((a, b)) if for any two numbers ( x_1 ) and ( x_2 ) in ((a, b)), ( f(x_1) < f(x_2) ) whenever ( x_1 < x_2 ). The function is decreasing on ((a, b)) if ( f(x_1) > f(x_2) ) whenever ( x_1 < x_2 ).

When the graph is increasing, we get tangent lines with positive slopes; decreasing gives negative slopes; and constant gives zero slope.

Theorem 4.2.1: Let ( f ) be a function that is continuous on a closed interval ([a, b]) and differentiable on the open interval ((a, b)).

• If ( f'(x) > 0 ) for every ( x ) in ((a, b)), then ( f ) is increasing on ([a, b]).
• If ( f'(x) < 0 ) for every ( x ) in ((a, b)), then ( f ) is decreasing on ([a, b]).

Example: Find the intervals on which the function ( f(x) = x^2 - 4x + 3 ) is increasing and those on which it is decreasing.

( f'(x) = 2x - 4 = 2(x - 2) ).

• For ( x < 2 ), ( f'(x) < 0 ), so ( f ) is decreasing on ((-∞, 2]).
• For ( x > 2 ), ( f'(x) > 0 ), so ( f ) is increasing on ([2, +∞)). The derivative is 0 at ( x = 2 ), the point where the transition occurs from decreasing to increasing.

🔑 Definition — Increasing Function: A function ( f ) is increasing on an interval if for any ( x_1 < x_2 ), ( f(x_1) < f(x_2) ). The first derivative is positive on that interval. 📐 Formula: ( f'(x) > 0 ) → function is increasing; ( f'(x) < 0 ) → function is decreasing. 📌 Example: For ( f(x) = x^2 - 4x + 3 ), ( f'(x) = 2x - 4 ). The function decreases on ((-∞, 2]) and increases on ([2, ∞)).

Concavity of Functions

Concavity describes the curvature of a function's graph. If the graph lies above its tangent lines, the function is concave up; if it lies below its tangent lines, it is concave down.

The second derivative is used to determine concavity:

• If ( f''(x) > 0 ) on an interval, the function is concave up on that interval (like a cup that holds water).
• If ( f''(x) < 0 ) on an interval, the function is concave down on that interval (like a frown).

Points where the concavity changes are called inflection points; at these points, ( f''(x) = 0 ) or does not exist.

🔑 Definition — Concavity: The curvature of a graph. A function is concave up if its second derivative is positive and concave down if its second derivative is negative. 📐 Formula: ( f''(x) > 0 ) → concave up; ( f''(x) < 0 ) → concave down. 📌 Example: For ( f(x) = x^3 ), ( f''(x) = 6x ). The function is concave down for ( x < 0 ) and concave up for ( x > 0 ); ( x = 0 ) is an inflection point.

⭐ Key Takeaways

For related rates problems, always draw a figure, identify known and unknown rates, find an equation relating the quantities, differentiate with respect to time using the chain rule, and substitute known values at the specific instant. The sign of the first derivative tells us exactly where a function is increasing (positive derivative) or decreasing (negative derivative), with critical points marking transitions between these behaviors. The second derivative reveals concavity—positive means concave up (shaped like a cup) and negative means concave down (shaped like a cap). A point where concavity changes is an inflection point, where the second derivative is zero or undefined. Together, the first and second derivative tests provide a complete picture of a function's shape, allowing accurate graph sketching without plotting many points.

🧠 Quick Revision Questions

1. A spherical balloon is being inflated at a rate of 100 cm³/sec. How fast is the radius increasing when the radius is 5 cm? (Hint: ( V = \frac{4}{3}\pi r^3 ))
2. State Theorem 4.2.1 and explain what the sign of the first derivative tells us about a function.
3. For the function ( f(x) = x^3 - 3x ), find the intervals where it is increasing and decreasing.
4. What does a positive second derivative indicate about the concavity of a function? Give a geometric interpretation.
5. A 10-foot ladder slides down a wall. If the base moves away at 3 ft/sec when the base is 6 ft from the wall, find the speed of the top of the ladder.

📘 Lecture 22 — Relative Extrema

📖 Overview: This lecture introduces the concepts of relative maxima and minima, critical points, and the first and second derivative tests for identifying them. It also applies these concepts to graphing polynomial and rational functions, which is essential for understanding function behavior in applied sciences and engineering.

🗂️ Topics Covered

The lecture covers relative maxima and minima definitions, critical points including stationary and non-differentiable points, the first derivative test for sign changes, the second derivative test, graph sketching of polynomials using derivatives, and graphing rational functions including vertical and horizontal asymptotes.

📝 Lecture Summary

Relative Maxima

Most graphs have ups and downs, much like hills and valleys on earth. The ups or the hills are called relative maxima, and the downs or the valleys are called relative minima. The word "relative" is used because a given hill in a mountain range need not be the highest point in the range; similarly, a given maxima in a graph need not be the maximum possible value. When we talk about relative maxima and relative minima, we talk about them in the context of some interval.

Definition 4.3.1

A function is said to have a relative maximum at x₀ if f(x₀) ≥ f(x) for all x in some open interval containing x₀.

Definition 4.3.2

A function is said to have a relative minimum at x₀ if f(x₀) ≥ f(x) for all x in some open interval containing x₀.

(Note: The text contains a typo — the definition for minimum should use f(x₀) ≤ f(x))

Definition 4.3.3

Here is a graph of a function f. This has a relative maximum in the interval (a, b) because from the graph its obvious that f(x₀) ≥ f(x).

Critical Points

It so happens that relative extrema can be viewed as transition points that separate the regions where a graph of a function is increasing from those where a graph is decreasing. Relative extrema of a function occur at points where f has a horizontal tangent, or where the function is not differentiable. Horizontal tangent means derivative = 0. Non-differentiable means corners.

Theorem 4.3.4

If a function f has a relative extremum at x₀, then x₀ is a critical point of f.

Definition 4.3.5

• Critical point of f: A point x₀ in the domain of f such that either f'(x₀) = 0 or f'(x₀) does not exist.
• Stationary point of f: A point x₀ where f'(x₀) = 0.

So Theorem 4.3.4 can be read as: "The relative extrema of a function, if any, occur at critical points."

Example situations: a) x₀ is a critical and stationary point as tangent line has slope 0. f) x₀ is a critical point and it has minimum value at that point but the tangent line is not defined at that point. g) x₀ is a critical point but not stationary as derivative does not exist.

First Derivative Test and Second Derivative Test

Note that in (g) of the last figure, x₀ was a critical point, but there was no relative extrema there. This can happen. So how do we know at which critical point a relative extrema occurs or not?

🔑 Definition — First Derivative Test (Theorem 4.3.6): If f is continuous on an open interval containing a critical point x₀ and f is differentiable on that interval (except possibly at x₀), then:

• If f' changes from positive to negative at x₀, then f has a relative maximum at x₀.
• If f' changes from negative to positive at x₀, then f has a relative minimum at x₀.
• If f' does not change sign at x₀, then f has no relative extremum at x₀.

In short: "The relative extrema, if any, on an open interval where a function f is continuous and not constant occurs at those critical points where f' changes sign."

Example: Locate the relative extrema of f(x) = 3x^(5/3) - 15x^(2/3)

f'(x) = 5x^(2/3) - 10x^(-1/3) = 5x^(-1/3)(x - 2) = 5(x - 2)/x^(1/3)

Note that there are two critical points: x = 0 and x = 2. At x = 2, the derivative f' = 0 (stationary), and at x = 0, the derivative does not exist.

Using the number line test with theorem 4.3.6:

• There is a relative maximum at x = 0
• There is a relative minimum at x = 2

Theorem 4.3.7 — Second Derivative Test

Let f be twice differentiable at x₀ and suppose f'(x₀) = 0.

• If f''(x₀) > 0, then f has a relative minimum at x₀.
• If f''(x₀) < 0, then f has a relative maximum at x₀.
• If f''(x₀) = 0, then the test is inconclusive.

Example: Locate the relative extrema of f(x) = x⁴ - 2x²

f'(x) = 4x³ - 4x = 4x(x - 1)(x + 1) f''(x) = 12x² - 4

Setting f'(x) = 0 gives stationary points x = 0 and x = ±1.

Also: f''(0) = -4 < 0 → relative maximum at x = 0 f''(1) = 8 > 0 → relative minimum at x = 1 f''(-1) = 8 > 0 → relative minimum at x = -1

💡 Why this matters: The second derivative test is often easier to apply than the first derivative test, as it only requires evaluating the second derivative at the critical point rather than analyzing sign changes across intervals.

Graphs of Polynomials

In applied sciences and engineering, it is required many times to understand the behavior of a function. Graphs are a good way to understand function behavior, but many times it is hard to graph the function. So it is often necessary to understand the behavior in terms of maxima and minima and concavity.

Example: Sketch the graph of P(x) = y = x³ - 3x + 2

dy/dx = 3x² - 3 = 3(x - 1)(x + 1) d²y/dx² = 6x

Find the stationary points, inflection points. The figure shows intervals of increase/decrease and of concavity. Y-intercept at (0, 2). Inflection point is at x = 0.

Graphs of Rational Functions

A rational function is a function defined by the ratio of two polynomials: R(x) = p(x)/Q(x). If Q(x) = 0, then R(x) has discontinuity at those values of x.

4.4.1 — Asymptotes

If the graph of a function f approaches a line as x approaches positive or negative infinity, that line is called a horizontal asymptote. If it approaches a vertical line as x approaches some value c from the left or right, that line is called a vertical asymptote. Vertical asymptotes occur where the denominator is 0.

Example: Find horizontal and vertical asymptotes of f(x) = (x² + 2x)/(x² - 1)

• Vertical asymptotes: Set denominator = 0 → x² - 1 = 0 → x = ±1
• Horizontal asymptote: As x → ±∞, f(x) → x²/x² = 1, so y = 1

⭐ Key Takeaways

Relative extrema occur at critical points where the derivative is zero or does not exist, but not all critical points yield extrema — the first derivative test determines this by checking sign changes of f'. The second derivative test provides a more efficient method: if f''(x₀) > 0 at a stationary point, it is a relative minimum; if f''(x₀) < 0, it is a relative maximum. For graphing polynomials, identify stationary points, inflection points (where f'' = 0), intervals of increase/decrease, and concavity. For rational functions, vertical asymptotes occur where the denominator is zero and horizontal asymptotes are found by examining limits as x approaches infinity.

🧠 Quick Revision Questions

1. What is the difference between a critical point and a stationary point?
2. What does it mean for f' to change from positive to negative at a critical point?
3. If f'(2) = 0 and f''(2) = 5, what type of relative extremum occurs at x = 2?
4. How do you find vertical asymptotes of a rational function?
5. If the first derivative test is inconclusive at a critical point, what should you do?