AP Calculus AB · Lesson 1

The Mean Value Theorem and extrema

Know Know it

Mean Value Theorem (MVT). If ff is continuous on [a,b][a, b] and differentiable on (a,b)(a, b), then for at least one cc with a<c<ba < c < b, f′(c)=f(b)−f(a)b−af'(c) = \frac{f(b) - f(a)}{b - a}. Somewhere inside, the instantaneous rate equals the average rate. When f(a)=f(b)f(a) = f(b), this says f′(c)=0f'(c) = 0 for some cc.

Extreme Value Theorem (EVT). If ff is continuous on a closed interval [a,b][a, b], then ff has an absolute maximum and an absolute minimum on [a,b][a, b].

A critical point of ff is an input cc inside the domain where f′(c)=0f'(c) = 0 or f′(c)f'(c) does not exist. Relative extrema of ff can happen only at critical points.

First derivative test. At a critical point, if f′f' changes from positive to negative, ff has a relative maximum; from negative to positive, a relative minimum. If f′f' does not change sign, there is no extremum.

Candidates test (absolute extrema on a closed interval): list the critical points inside the interval and both endpoints, evaluate ff at each, and compare. The largest value is the absolute maximum; the smallest is the absolute minimum.

Trap Name both conditions before you use the theorem

The MVT needs continuity on the closed interval and differentiability on the open interval. A function with a corner or a jump can have an average rate that f′f' never equals. On free-response answers, state both conditions, with the reason they hold (for example, "ff is a polynomial").

Worked example Easy

Let f(x)=x3f(x) = x^3. Which value of cc in the interval (0,3)(0, 3) satisfies the conclusion of the Mean Value Theorem on [0,3][0, 3]?

  1. c=1.5c = 1.5
  2. c=3c = \sqrt{3}Answer
  3. c=3c = 3
  4. c=9c = 9

How to solve it

  1. ff is a polynomial, so it is continuous on [0,3][0, 3] and differentiable on (0,3)(0, 3).
  2. Average rate: f(3)−f(0)3−0=273=9\frac{f(3) - f(0)}{3 - 0} = \frac{27}{3} = 9.
  3. Solve f′(c)=3c2=9f'(c) = 3c^2 = 9: c=3≈1.73c = \sqrt{3} \approx 1.73, which lies in (0,3)(0, 3).

Why each choice is right or wrong

  • A. Incorrect. c=1.5c = 1.5 is the midpoint of the interval, but f′(1.5)=6.75f'(1.5) = 6.75, not the average rate 9.
  • B. Correct. The average rate is 27−03=9\frac{27 - 0}{3} = 9, and f′(c)=3c2=9f'(c) = 3c^2 = 9 gives c=3c = \sqrt{3}, which is inside (0,3)(0, 3).
  • C. Incorrect. c=3c = 3 is the right endpoint; the theorem promises a value strictly inside the interval, and f′(3)=27f'(3) = 27.
  • D. Incorrect. c=9c = 9 is the average rate of change itself, not the input where f′f' equals it.

Explain Explain it

Task verbs: justify, determine

To justify an MVT conclusion, write three things: the function is continuous on the closed interval and differentiable on the open one (and why), the average rate of change as a number, and the conclusion that f′(c)f'(c) equals that number for some cc in the open interval.

To determine an absolute extremum, show the candidates in a short list, such as f(−1)=3f(-1) = 3, f(2)=−5f(2) = -5, f(4)=7f(4) = 7, and then name the winner. A sentence like "the absolute minimum is -5, at x=2x = 2" finishes the answer.

Worked example Medium

Let f(x)=x3−3x2−9x+2f(x) = x^3 - 3x^2 - 9x + 2. What is the absolute minimum value of ff on the closed interval [−2,4][-2, 4]?

  1. -25Answer
  2. -18
  3. 0
  4. 7

How to solve it

  1. f′(x)=3x2−6x−9=3(x−3)(x+1)f'(x) = 3x^2 - 6x - 9 = 3(x - 3)(x + 1), which is 0 at x=−1x = -1 and x=3x = 3, both inside [−2,4][-2, 4].
  2. Evaluate: f(−2)=0f(-2) = 0, f(−1)=7f(-1) = 7, f(3)=−25f(3) = -25, f(4)=−18f(4) = -18.
  3. The smallest value is -25, at x = 3, so that is the absolute minimum.

Why each choice is right or wrong

  • A. Correct. f′(x)=3(x−3)(x+1)f'(x) = 3(x - 3)(x + 1), so the critical points are x=−1x = -1 and x=3x = 3. The candidates are f(−2)=0f(-2) = 0, f(−1)=7f(-1) = 7, f(3)=−25f(3) = -25 and f(4)=−18f(4) = -18; the smallest is -25.
  • B. Incorrect. -18 is f(4)f(4), the smaller endpoint value; checking only the endpoints misses the critical point x=3x = 3, where f(3)=−25f(3) = -25.
  • C. Incorrect. 0 is f(−2)f(-2), the value at the left endpoint, which is not the smallest of the candidates.
  • D. Incorrect. 7 is f(−1)f(-1), the relative maximum and also the absolute maximum on the interval, not the minimum.

Justify Your turn

A function hh is differentiable for all real numbers, with h(1)=4h(1) = 4 and h(5)=−8h(5) = -8. Explain why there must be a number cc with 1<c<51 < c < 5 and h′(c)=−3h'(c) = -3.

Write it, then compare. Nothing you type here is saved or graded.

Compare with a model answer

A full answer does this

  • Says hh is continuous on [1,5][1, 5] and differentiable on (1,5)(1, 5), because it is differentiable for all real numbers
  • Computes the average rate of change h(5)−h(1)5−1=−3\frac{h(5) - h(1)}{5 - 1} = -3
  • Concludes, by the Mean Value Theorem, that h′(c)=−3h'(c) = -3 for some cc in (1,5)(1, 5)

Model answer

Since hh is differentiable for all real numbers, it is differentiable on (1,5)(1, 5) and also continuous on [1,5][1, 5], because a differentiable function is continuous. Its average rate of change on [1,5][1, 5] is h(5)−h(1)5−1=−8−44=−3\frac{h(5) - h(1)}{5 - 1} = \frac{-8 - 4}{4} = -3. So by the Mean Value Theorem, h′(c)=−3h'(c) = -3 for some cc with 1<c<51 < c < 5.

Apply Apply it

Practice these topics. AP Practice questions on exactly what this lesson covers, with an explanation for every answer.

Practice these topics