Difficulty: Intermediate | Prerequisites: Chapters 1 and 2 (limits, continuity, basic differentiation rules, the definition of the derivative).
Chapter 3 is where differentiation stops being abstract and starts doing useful work. You already know how to take derivatives; now you use them to find the highest and lowest values a function reaches, to guarantee that the derivative hits a particular slope somewhere on an interval, and to prove uniqueness of solutions. These ideas underpin optimisation problems in engineering, economics and the sciences, and they set the stage for curve sketching and antiderivatives later in the chapter. If you are behind, make sure you are comfortable with continuity on closed intervals and the basic derivative rules before diving in.
The Extreme Value Theorem guarantees that a continuous function on a closed interval has an absolute max and min; you find them by checking critical numbers and endpoints. Rolle's Theorem is the special case of the Mean Value Theorem where the endpoint values match, guaranteeing a horizontal tangent inside. The MVT itself says there is always a point where the instantaneous rate of change equals the average rate of change over the interval.
Absolute (global) maximum
The value f(c) is the absolute maximum of f on a domain D if f(c) >= f(x) for every x in D. In simple terms, it is the single highest output the function reaches anywhere on the domain.
Absolute (global) minimum
The value f(c) is the absolute minimum of f on D if f(c) <= f(x) for every x in D. Think of it as the lowest point on the entire curve over the domain.
Local (relative) maximum
f(c) is a local maximum if f(c) >= f(x) for all x near c (in some open interval around c). In simple terms, it is a hilltop on the graph, even if there is a higher hill somewhere else.
Local (relative) minimum
f(c) is a local minimum if f(c) <= f(x) for all x near c. Think of it as a valley on the graph.
Extreme values (extrema)
A catch-all term for maximums and minimums, whether local or absolute.
Critical number
A number c in the domain of f where either f'(c) = 0 or f'(c) does not exist. In simple terms, it is any x-value where the derivative is zero (a flat tangent) or undefined (a cusp or corner). Absolute extrema on an interval can only occur at critical numbers or endpoints.
Extreme Value Theorem (EVT)
If f is continuous on a closed interval [a, b], then f attains both an absolute maximum and an absolute minimum somewhere in [a, b]. Think of it as a guarantee that a continuous curve on a closed interval must have a highest and a lowest point.
Rolle's Theorem
If f is continuous on [a, b], differentiable on (a, b), and f(a) = f(b), then there is at least one c in (a, b) with f'(c) = 0. In simple terms, if the function starts and ends at the same height, there must be a horizontal tangent somewhere in between.
Mean Value Theorem (MVT)
If f is continuous on [a, b] and differentiable on (a, b), then there exists a c in (a, b) such that f'(c) = (f(b) - f(a)) / (b - a). Think of it as saying there is always a moment where the instantaneous speed equals the average speed over the trip.
Corollary (zero derivative implies constant)
If f'(x) = 0 for all x in (a, b), then f is constant on that interval.
Corollary (equal derivatives differ by a constant)
If f'(x) = g'(x) for all x in (a, b), then f(x) = g(x) + C for some constant C. This is the reason antiderivatives always carry a "+ C."
A function must be continuous on the interval. f(x) = 1/x on [-2, 2] is not continuous at x = 0, so the EVT does not apply, and the function has no absolute minimum on that interval.
The interval must be closed. f(x) = x on (0, 2] has no absolute minimum because the open left endpoint lets values get arbitrarily close to 0 without reaching it.
Both conditions (continuous AND closed interval) are required for the EVT to guarantee extrema exist.
To find the absolute max and min of a continuous function f on [a, b]:
Find all critical numbers of f in (a, b). These are x-values where f'(x) = 0 or f'(x) does not exist.
Evaluate f at each critical number.
Evaluate f at the endpoints a and b.
The largest value from steps 2 and 3 is the absolute maximum. The smallest is the absolute minimum.
f'(x) = 3x^2 - 12 = 3(x^2 - 4) = 3(x - 2)(x + 2)
Critical numbers: x = 2 and x = -2. Only x = 2 lies in (1, 4).
f(1) = 1 - 12 + 1 = -10
f(2) = 8 - 24 + 1 = -15
f(4) = 64 - 48 + 1 = 17
Absolute max = 17 at x = 4. Absolute min = -15 at x = 2.
g'(t) = 1 - 2t^(-1/3) = 1 - 2 / t^(1/3)
Set g'(t) = 0: 1 = 2 / t^(1/3), so t^(1/3) = 2, hence t = 8. Also, g'(t) is undefined at t = 0. Both t = 0 and t = 8 lie in (-1, 27).
g(-1) = -1 - 3(1) = -4
g(0) = 0
g(8) = 8 - 3(4) = -4
g(27) = 27 - 3(9) = 0
Absolute max = 0 at t = 0 and t = 27. Absolute min = -4 at t = -1 and t = 8.
f'(t) = (14/3)t^(-1/3) - 15t^(2/3) = t^(-1/3)[(14/3) - 15t]
f'(t) = 0 when 14/3 = 15t, so t = 14/45.
f'(t) is undefined at t = 0, and 0 is in the domain.
Critical numbers: t = 0 and t = 14/45.
Rolle's Theorem is a stepping stone to the MVT. It requires three conditions:
f is continuous on [a, b]
f is differentiable on (a, b)
f(a) = f(b)
Conclusion: there exists at least one c in (a, b) where f'(c) = 0. Geometrically, if the function starts and ends at the same height, the curve must flatten out (horizontal tangent) somewhere in between.
Requires two conditions:
f is continuous on [a, b]
f is differentiable on (a, b)
Conclusion: there exists at least one c in (a, b) such that f'(c) = (f(b) - f(a)) / (b - a).
Geometrically, there is a point where the tangent line is parallel to the secant line connecting (a, f(a)) and (b, f(b)).
f(x) = |x| on [-2, 2]: continuous on [-2, 2] but not differentiable at x = 0, so the hypotheses are not satisfied. The MVT does not apply.
Any function with a discontinuity in [a, b] or a non-differentiable point in (a, b) fails the hypotheses.
f is a polynomial, so it is continuous and differentiable everywhere. MVT applies.
f(0) = 5, f(1) = 9.
(f(1) - f(0)) / (1 - 0) = 4.
f'(x) = 2x + 3. Set 2c + 3 = 4, so c = 1/2. This lies in (0, 1).
f is continuous and differentiable on [1, 3] (no issues since 0 is outside the interval). MVT applies.
f(1) = 1, f(3) = 1/3.
(f(3) - f(1)) / (3 - 1) = (1/3 - 1) / 2 = -1/3.
f'(x) = -1/x^2. Set -1/c^2 = -1/3, so c^2 = 3, c = sqrt(3). This lies in (1, 3).
If f'(x) = 0 for all x in (a, b), then f is constant on (a, b). The derivative being zero everywhere means the function never climbs or falls.
If f'(x) = g'(x) for all x in (a, b), then f(x) = g(x) + C for some constant. Two functions with the same derivative differ only by a vertical shift. This is why the general antiderivative includes "+ C."
To show 3x + 2cos(x) + 5 = 0 has at most one solution, let f(x) = 3x + 2cos(x) + 5. Suppose there were two solutions a and b. Then f(a) = f(b) = 0, so by Rolle's Theorem there would be a c with f'(c) = 0. But f'(x) = 3 - 2sin(x), and since -1 <= sin(x) <= 1, we have f'(x) >= 3 - 2 = 1 > 0 for all x. This contradiction means there cannot be two solutions.
\text{EVT: } f \text{ continuous on } [a,b] \Rightarrow f \text{ attains an absolute max and min on } [a,b]\text{Critical number: } c \in \text{dom}(f) \text{ where } f'(c)=0 \text{ or } f'(c) \text{ DNE}\text{Rolle: } f(a)=f(b),\; f \text{ cont. on } [a,b],\; f \text{ diff. on } (a,b) \Rightarrow \exists\, c\in(a,b):\; f'(c)=0\text{MVT: } f \text{ cont. on } [a,b],\; f \text{ diff. on } (a,b) \Rightarrow \exists\, c\in(a,b):\; f'(c)=\frac{f(b)-f(a)}{b-a}The closed interval method is the backbone of optimisation in engineering: choose a design parameter (radius of a pipe, thickness of insulation), write a cost or performance function, and find the parameter value that maximises or minimises it. The MVT is the mathematical reason a speed camera can prove you were speeding between two checkpoints: if the average speed over the interval exceeds the limit, there must have been at least one instant where your actual speed exceeded it too.
Students often think every critical number is a local extremum. It is not. A critical number is merely a candidate; the derivative might not change sign there (e.g. f(x) = x^3 at x = 0).
Students forget to check endpoints when finding absolute extrema on a closed interval. The absolute max or min very often occurs at an endpoint, not at a critical number.
Students assume the MVT tells you the value of c. It does not. It guarantees c exists but you have to solve for it yourself.
Students try to apply the MVT to functions that are not differentiable on the open interval (a, b). If f has a corner, cusp, or vertical tangent inside the interval, the theorem does not apply. Check the hypotheses first.
The closed interval method is one of the most commonly tested procedures in this chapter. Expect at least one problem asking you to find absolute extrema on [a, b].
You will almost certainly be asked to verify the hypotheses of the MVT and then find the value of c.
Rolle's Theorem problems often appear as "show there exists a c" questions, requiring you to set up the hypotheses carefully.
Proof-by-contradiction questions using the MVT (e.g. "show at most one root") appear regularly on exams.
True or false: A continuous function on an open interval (a, b) always has an absolute maximum.
Answer: False. The EVT requires a closed interval.
Fill in the blank: A critical number of f is a value c in the domain of f where f'(c) = ______ or f'(c) ______.
Answer: 0; does not exist.
True or false: If f'(c) = 0, then f has a local extremum at c.
Answer: False. The derivative being zero is necessary but not sufficient; f'(x) may not change sign.
True or false: The MVT guarantees a unique value of c.
Answer: False. It guarantees at least one; there can be several.
Fill in the blank: Rolle's Theorem is a special case of the MVT where f(a) = ______.
Answer: f(b).
Q: Find the absolute maximum and minimum of f(x) = (x - 1)(x - 5)^3 + 3 on [0, 6].
A: Expand or use the product rule to find f'(x). Set f'(x) = 0 to get the critical numbers. Evaluate f at each critical number in (0, 6) and at the endpoints x = 0 and x = 6. Compare all values. The largest is the absolute max, the smallest is the absolute min.
Q: Let g(s) = 1/(s - 5) on [-5, -4]. Find the absolute maximum and minimum.
A: g'(s) = -1/(s - 5)^2, which is never zero. No critical numbers in [-5, -4]. g(-5) = 1/(-10) = -1/10, g(-4) = 1/(-9) = -1/9. Absolute max = -1/10 at s = -5. Absolute min = -1/9 at s = -4.
Q: Consider f(x) = cube root of x on [0, 1]. Verify the MVT hypotheses and find all c satisfying the conclusion.
A: f is continuous on [0, 1] (cube root is continuous everywhere) and differentiable on (0, 1) (f'(x) = (1/3)x^(-2/3) exists for x > 0). The MVT applies. (f(1) - f(0))/(1 - 0) = 1. Set f'(c) = 1: (1/3)c^(-2/3) = 1, so c^(2/3) = 1/3, c = (1/3)^(3/2) = 1/(3 sqrt(3)), which is in (0, 1).
Q: Show that the equation 3x + 2cos(x) + 5 = 0 has at most one solution.
A: Let f(x) = 3x + 2cos(x) + 5. Then f'(x) = 3 - 2sin(x) >= 1 > 0 for all x, so f is strictly increasing. A strictly increasing function can cross zero at most once, so the equation has at most one solution.
Q: Find all c satisfying the MVT for f(x) = 2sin(x)cos(x) on [0, pi].
A: f(x) = sin(2x), so f'(x) = 2cos(2x). f(0) = 0, f(pi) = 0. (f(pi) - f(0)) / (pi - 0) = 0. Set 2cos(2c) = 0, so cos(2c) = 0, 2c = pi/2 or 3pi/2, c = pi/4 or 3pi/4. Both lie in (0, pi).
The critical number concept introduced here feeds directly into the First and Second Derivative Tests in Section 3 of this chapter, where you classify critical numbers as local maxes, local mins, or neither. The MVT corollary about equal derivatives differing by a constant is the theoretical foundation for Section 9 on antiderivatives and the "+ C" that appears in every indefinite integral. Optimisation problems in Section 7 are applied versions of the closed interval method.
absolute maximum, absolute minimum, global max, global min, local max, local min, relative extrema, extreme values, critical point, critical number, stationary point, closed interval method, Extreme Value Theorem, EVT, Rolle's Theorem, Mean Value Theorem, MVT, secant line, tangent line parallel to secant, proof by contradiction, constant function, antiderivative + C, Calculus I, MTH 132, MSU, Michigan State, applications of differentiation