MTH 132, Calculus I (Michigan State University) | Chapter 3, Sections 5, 7, 9 Source: Chapter 3 lecture notes, Applications of Differentiation
Tags: curve sketching, even function, odd function, symmetry, domain, intercepts, asymptotes, optimisation, optimization, constraint equation, objective function, antiderivative, initial value problem, IVP, general antiderivative, constant of integration, Calculus I, MTH 132, MSU
Difficulty: Intermediate to Advanced Prerequisites: All earlier sections of Chapter 3 (critical numbers, EVT, MVT, first/second derivative tests, concavity, limits at infinity, horizontal and slant asymptotes). Solid algebra skills, including factoring, polynomial long division, and inequality solving.
This is where Chapter 3 comes together. Section 5 combines every tool from Sections 1–4 into a systematic curve-sketching procedure. Section 7 applies maxima and minima to real-world word problems (optimisation), the kind of problem that turns up on nearly every Calculus I exam. Section 9 reverses the differentiation process by introducing antiderivatives, setting the stage for integral calculus in Chapter 4.
Curve sketching follows a checklist: domain, intercepts, symmetry, asymptotes, intervals of increase/decrease, concavity, inflection points, then sketch. Optimisation problems require translating a word problem into a function to maximise or minimise, often using a constraint to reduce to one variable. Antiderivatives are functions whose derivative equals a given function; the most general antiderivative includes an arbitrary constant C, and initial conditions pin that constant down.
Even function
A function satisfying f(-x) = f(x) for all x. Its graph is symmetric about the y-axis. Example: f(x) = x^2.
Odd function
A function satisfying f(-x) = -f(x) for all x. Its graph is symmetric about the origin. Example: f(x) = x^3.
Objective function
The function you are trying to maximise or minimise in an optimisation problem. Think of it as the quantity you care about (area, cost, distance, revenue).
Constraint equation
An equation that relates the variables in an optimisation problem, reflecting a physical or geometric restriction (fixed perimeter, fixed volume, fixed budget). You use it to eliminate a variable so the objective function depends on only one variable.
Antiderivative
A function F is an antiderivative of f on an interval I if F'(x) = f(x) for all x in I. In simple terms, it is the function you get by "undoing" differentiation.
General antiderivative
F(x) + C, where F is any one antiderivative of f and C is an arbitrary constant. The +C accounts for the fact that the derivative of a constant is zero, so infinitely many functions share the same derivative.
Initial value problem (IVP)
A differential equation paired with an initial condition such as f(c) = a. The initial condition pins down the value of C, giving a unique solution.
Differential equation
An equation involving the derivative(s) of an unknown function. In this section, the simplest kind: f'(x) = g(x), solved by antidifferentiation.
For a given function f(x), work through these steps in order:
Domain. Where is f defined? Look for division by zero, even roots of negatives, and logarithms of non-positives.
Intercepts. Set x = 0 for the y-intercept. Set f(x) = 0 for x-intercepts.
Symmetry. Check if f is even (symmetric about y-axis), odd (symmetric about the origin), or neither. This can halve your work.
Asymptotes. Vertical: where the denominator is zero and the numerator is not. Horizontal: evaluate limits as x -> infinity and x -> negative infinity. Slant: if deg(numerator) = deg(denominator) + 1, do polynomial long division.
Intervals of increase/decrease. Find f'(x), locate critical numbers, build a sign chart.
Local extrema. Apply the First or Second Derivative Test at each critical number.
Concavity and inflection points. Find f''(x), locate where f'' = 0 or is undefined, build a sign chart for f''.
Sketch. Plot key points (intercepts, extrema, inflection points), draw asymptotes, then connect with a curve that respects the sign charts.
To check algebraically: compute f(-x). If f(-x) = f(x), the function is even. If f(-x) = -f(x), it is odd. If neither, it is neither.
Examples: p(x) = x^3 + x + 1 is neither (the constant 1 spoils both). f(x) = 5 is even (constant functions are even). g(x) = (4x^3 - x)/(x^2 + 1) is odd. h(x) = sin(4x) - x is odd.
Domain: all real numbers. Intercepts: x = 0 and x = 4 (x-intercepts); y = 0 (y-intercept). No asymptotes (polynomial). f'(x) = (x - 4)^3 + 3x(x - 4)^2 = (x - 4)^2(4x - 4) = 4(x - 4)^2(x - 1). Critical numbers: x = 4 and x = 1. Sign of f': positive on (-infinity, 1), negative on (1, 4), positive on (4, infinity). So x = 1 is a local max and x = 4 is neither (no sign change). For concavity, f''(x) leads to inflection points that can be found by differentiating again and testing sign changes.
Domain: all x except x = 2. No x-intercepts (discriminant of x^2 - 2x + 4 is negative). y-intercept: f(0) = 4/(-2) = -2. Vertical asymptote: x = 2. Long division: f(x) = x + 4/(x - 2), so slant asymptote y = x. Given f'(x) = x(x - 4)/(x - 2)^2: critical numbers x = 0 and x = 4. f' > 0 on (-infinity, 0) and (4, infinity), f' < 0 on (0, 2) and (2, 4). Local max at x = 0 with f(0) = -2; local min at x = 4 with f(4) = 6. Given f''(x) = 8/(x - 2)^3: f'' < 0 when x < 2 (concave down), f'' > 0 when x > 2 (concave up). No inflection points (f is not continuous at x = 2).
Understand the problem. Read it fully. Identify what is given and what is unknown.
Draw a diagram. Label all relevant quantities.
Introduce notation. Assign a variable to the quantity being maximised or minimised (the objective), and assign variables to other quantities.
Write the objective function. Express the quantity to optimise in terms of your variables.
Use the constraint. Use any restricting equation to eliminate variables, reducing the objective to a function of one variable. Identify the domain of that variable.
Find the absolute max or min. Use the Closed Interval Method (if the domain is a closed interval) or the First/Second Derivative Test (if the domain is open or unbounded).
Find the maximum area of a rectangle inscribed in an equilateral triangle of side length 6, with one side on the base.
Let x = half the width of the rectangle and h = height of the rectangle. The triangle's height is 3sqrt(3). By similar triangles, h = 3sqrt(3)(1 - x/3) = 3sqrt(3) - sqrt(3)x. Area A = 2x * h = 2x(3sqrt(3) - sqrt(3)x) = 2sqrt(3)x(3 - x). Domain: 0 <= x <= 3. A'(x) = 2sqrt(3)(3 - 2x). Setting A'(x) = 0 gives x = 3/2. A(0) = 0, A(3) = 0, A(3/2) = 2sqrt(3)(3/2)(3/2) = (9sqrt(3))/2. Maximum area = 9sqrt(3)/2.
A fish tank with a square base (side x) and height h has glass sides ($3/sq ft) and a plastic base ($2/sq ft), no top. Volume constraint: x^2 * h = 5. Cost = 2x^2 (base) + 4xh * 3 (four sides) = 2x^2 + 12xh. Substitute h = 5/x^2: Cost(x) = 2x^2 + 60/x. Domain: x > 0. Cost'(x) = 4x - 60/x^2. Setting this to zero: 4x^3 = 60, so x^3 = 15, giving x = (15)^(1/3). Since Cost''(x) = 4 + 120/x^3 > 0, this is a minimum.
The sum of two positive numbers is 16. Minimise the sum of their squares. Let x be one number; the other is 16 - x. Objective: S(x) = x^2 + (16 - x)^2 = 2x^2 - 32x + 256. Domain: 0 < x < 16. S'(x) = 4x - 32 = 0 gives x = 8. S''(8) = 4 > 0, confirming a minimum. Both numbers are 8, and the minimum sum of squares is 128.
A farmer has 300 ft of fencing to enclose a rectangular area divided into two pens (a divider parallel to one side). Let x = length of each of the two sides perpendicular to the divider, and y = length of the three parallel fences (two sides + divider). Constraint: 2x + 3y = 300. Area A = xy. Substitute y = (300 - 2x)/3: A(x) = x(300 - 2x)/3 = (300x - 2x^2)/3. A'(x) = (300 - 4x)/3 = 0 gives x = 75. A(75) = (30075 - 25625)/3 = (22500 - 11250)/3 = 3750 sq ft. This is the maximum (A'' = -4/3 < 0).
A function F is an antiderivative of f if F'(x) = f(x). In other words, you reverse the differentiation process. Example: an antiderivative of f(x) = 2x + 3 is F(x) = x^2 + 3x, because F'(x) = 2x + 3.
If F is one antiderivative of f, then every antiderivative of f has the form F(x) + C, where C is an arbitrary constant. This follows from the MVT corollary: two functions with the same derivative differ by a constant.
Antiderivative of x^n (n not equal to -1): x^(n+1)/(n+1) + C.
Antiderivative of cos(x): sin(x) + C.
Antiderivative of sin(x): -cos(x) + C.
Antiderivative of sec^2(x): tan(x) + C.
Antiderivative of 1/x^2 = x^(-2): -1/x + C.
These rules work by recognising the derivative in reverse. Sums and constant multiples carry through just as they do for derivatives.
Find the general antiderivative of f(x) = cos(x) + 1/x^2.
Rewrite 1/x^2 as x^(-2). Antiderivative: sin(x) + x^(-1)/(-1) + C = sin(x) - 1/x + C.
Find the general antiderivative of f(x) = sqrt(x)(6 + 7x).
Expand: 6sqrt(x) + 7xsqrt(x) = 6x^(1/2) + 7x^(3/2). Antiderivative: 6*(2/3)x^(3/2) + 7*(2/5)x^(5/2) + C = 4x^(3/2) + (14/5)x^(5/2) + C.
Find the general antiderivative of g(x) = 8x - 3sec^2(x).
Antiderivative: 4x^2 - 3tan(x) + C.
An initial value problem gives you f'(x) and a condition f(c) = a. First find the general antiderivative, then plug in the condition to solve for C.
Example: f'(x) = 1 + 3sqrt(x), f(4) = 25. General antiderivative: f(x) = x + 3(2/3)x^(3/2) + C = x + 2x^(3/2) + C. Plug in f(4) = 25: 4 + 2*(8) + C = 25, so 20 + C = 25, giving C = 5. Answer: f(x) = x + 2x^(3/2) + 5.
Example: f'(x) = 2/sqrt(x), f(4) = 20. Rewrite as 2x^(-1/2). Antiderivative: 2*(2)x^(1/2) + C = 4sqrt(x) + C. f(4) = 42 + C = 8 + C = 20, so C = 12. Answer: f(x) = 4*sqrt(x) + 12.
If velocity v(t) is given, position s(t) is the antiderivative of v(t). If acceleration a(t) is given, velocity is the antiderivative of a(t), and then position is the antiderivative of velocity. Each step introduces a constant C that requires an initial condition to determine.
Example: v(t) = sin(t) - cos(t), s(0) = 0. s(t) = -cos(t) - sin(t) + C. s(0) = -1 - 0 + C = 0, so C = 1. Answer: s(t) = -cos(t) - sin(t) + 1.
Example: a(t) = t^2 - 4t + 6, s(1) = 20, s(0) = 0. First find v(t) = t^3/3 - 2t^2 + 6t + C1. Then s(t) = t^4/12 - 2t^3/3 + 3t^2 + C1*t + C2. Use s(0) = 0: C2 = 0. Use s(1) = 20: 1/12 - 2/3 + 3 + C1 = 20, so C1 = 20 - 1/12 + 2/3 - 3 = 17 + 7/12 = 211/12. Answer: s(t) = t^4/12 - 2t^3/3 + 3t^2 + (211/12)t.
Optimisation is one of the most practical skills in calculus. Engineers use it to minimise material cost for containers, architects use it to maximise usable floor area, and economists use it to set prices that maximise revenue. The farmer's fencing problem and the fish-tank cost problem are simplified versions of real design constraints. Antiderivatives connect directly to physics: given an object's acceleration (from a force model), you integrate once to get velocity and again to get position.
Students often forget to identify the domain of the objective function in optimisation problems. If lengths, areas, or counts must be positive, the domain is restricted. Without this, you may find a critical number that is physically meaningless.
A common error is trying to optimise a function of two variables without using the constraint to eliminate one. The derivative test only works on functions of a single variable at this stage.
Students sometimes write the general antiderivative without +C. Without the constant, you have one particular antiderivative, not the general one. On exams, missing the +C typically costs marks.
When solving initial value problems, students sometimes forget to substitute the initial condition to solve for C. The final answer should contain no unknown constants.
⚠️ Curve-sketching problems are comprehensive: they test domain, intercepts, asymptotes, derivatives, and concavity all at once. Work through the checklist systematically.
⚠️ Optimisation word problems appear on nearly every Calculus I exam. The hardest part is setting up the equations; once you have a single-variable function, it is a standard max/min problem.
⚠️ Know the common antiderivative rules by heart. The power rule in reverse (x^n -> x^(n+1)/(n+1)) covers most problems, but trig antiderivatives (sin, cos, sec^2) are tested too.
⚠️ Initial value problems are straightforward if you do not skip steps: antidifferentiate, add C, plug in the condition, solve for C, write the answer.
True or False: Every odd function passes through the origin. Answer: False. An odd function satisfies f(-x) = -f(x), which means f(0) = 0 only if 0 is in the domain. But a function like f(x) = 1/x is odd and undefined at 0.
Fill in the blank: The most general antiderivative of x^n (for n not equal to -1) is ____. Answer: x^(n+1)/(n+1) + C.
True or False: In an optimisation problem, you always use the Closed Interval Method. Answer: False. The Closed Interval Method requires a closed interval domain. If the domain is open or unbounded, you need the First or Second Derivative Test instead.
Fill in the blank: If F'(x) = f(x) and G'(x) = f(x), then F(x) - G(x) = ____. Answer: a constant C.
Q: Find the general antiderivative of f(x) = sqrt(x^3) + cube root of x^2.
A: Rewrite: x^(3/2) + x^(2/3). Antiderivative: x^(5/2)/(5/2) + x^(5/3)/(5/3) + C = (2/5)x^(5/2) + (3/5)x^(5/3) + C.
Q: A poster has total area 180 in^2 with 1-inch margins on the bottom and sides and a 2-inch margin on top. What dimensions maximise the printed area?
A: Let w = width, h = height. Total area: wh = 180. Printed area: (w - 2)(h - 3). Substitute h = 180/w: P(w) = (w - 2)(180/w - 3) = 180 - 3w - 360/w + 6. P'(w) = -3 + 360/w^2. Set P'(w) = 0: w^2 = 120, so w = 2sqrt(30). Then h = 180/(2sqrt(30)) = 90/sqrt(30) = 3*sqrt(30). These dimensions maximise the printed area.
Q: Solve the initial value problem f'(x) = 2/sqrt(x), f(4) = 20.
A: f'(x) = 2x^(-1/2). Antiderivative: f(x) = 4x^(1/2) + C = 4sqrt(x) + C. f(4) = 42 + C = 8 + C = 20, so C = 12. Answer: f(x) = 4*sqrt(x) + 12.
Q: True or False: If F(x) is an antiderivative of f(x), then F(x) + 3 is also an antiderivative of f(x).
A: True. The derivative of F(x) + 3 is F'(x) + 0 = f(x).
Q: True or False: The antiderivative of a nonzero constant function is itself a constant function.
A: False. The antiderivative of the constant k is kx + C, which is a linear function.
Curve sketching synthesises everything from Chapters 1–3: limits, continuity, derivative rules, critical numbers, the derivative tests, and asymptotes. In Calculus II, these skills return when you analyse functions defined by integrals. Optimisation is the practical payoff of the entire derivatives unit; the same framework (objective function, constraint, single-variable reduction, derivative test) reappears in multivariable calculus with Lagrange multipliers. Antiderivatives are the entry point for integral calculus: the Fundamental Theorem of Calculus (Chapter 4) connects definite integrals to antiderivatives, making everything in Section 9 essential preparation.
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