AP Calculus AB · Free-response skill

Justify extrema and the Mean Value Theorem

The move. Find and classify the extrema of a function from its derivative, giving the sign-change or candidates reason in words, and show why the Mean Value Theorem applies on an interval.

How to do it

  1. Start every part from the derivative. Write it, factor it, and list the critical points that lie inside the interval.
  2. For a relative extremum, give the reason as a sign change of the derivative in words: "changes from positive to negative" means a maximum, "from negative to positive" a minimum. A derivative of 0 alone is not a reason.
  3. For an absolute extremum on a closed interval, make a short list of the function’s values at the critical points and at both endpoints, then name the largest or smallest. A relative maximum can lose to an endpoint.
  4. For the Mean Value Theorem, say why the function is continuous on the closed interval and differentiable on the open one (for example, it is a polynomial), compute the average rate of change, and state the conclusion with the theorem’s name.
  5. This kind of question can be on either part of the free-response section, so practice it by hand.

Practice prompt Try it

Let g(x)=x3−6x2+9x+2g(x) = x^3 - 6x^2 + 9x + 2 on the closed interval [−1,5][-1, 5].

Answer each part. Show the work that leads to each answer.

  1. Find the x-coordinate of each critical point of gg in the interval (−1,5)(-1, 5).
  2. For each critical point from part (a), say whether gg has a relative maximum, a relative minimum or neither there, and give a reason based on g′g'.
  3. Find the absolute minimum value of gg on [−1,5][-1, 5] by comparing the values of gg at the endpoints and at the critical points.
  4. Use the Mean Value Theorem to explain why there is a number cc with 1<c<31 < c < 3 such that g′(c)=−2g'(c) = -2. Say why the theorem applies.

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Checklist

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0 of 6 points

Model answer, annotated

(a) g′(x)=3x2−12x+9=3(x−1)(x−3)g'(x) = 3x^2 - 12x + 9 = 3(x - 1)(x - 3), so g′(x)=0g'(x) = 0 at x=1x = 1 and x=3x = 3. Both are in (−1,5)(-1, 5), and g′g' is defined everywhere, so these are the critical points.

  • Point 1 Differentiates, factors and lists both zeros of the derivative inside the interval.

(b) Test values in g′(x)=3(x−1)(x−3)g'(x) = 3(x - 1)(x - 3): g′(0)=9>0g'(0) = 9 > 0, g′(2)=−3<0g'(2) = -3 < 0 and g′(4)=9>0g'(4) = 9 > 0. So for −1<x<1-1 < x < 1, g′(x)>0g'(x) > 0; for 1<x<31 < x < 3, g′(x)<0g'(x) < 0; for 3<x<53 < x < 5, g′(x)>0g'(x) > 0. At x=1x = 1, g′g' changes from positive to negative, so gg has a relative maximum there. At x=3x = 3, g′g' changes from negative to positive, so gg has a relative minimum there.

  • Point 2 Classifies x = 1 as a maximum with the positive-to-negative sign change of g prime.
  • Point 3 Classifies x = 3 as a minimum with the negative-to-positive sign change of g prime.

(c) The candidates are the endpoints and the critical points: g(−1)=−14g(-1) = -14, g(1)=6g(1) = 6, g(3)=2g(3) = 2 and g(5)=22g(5) = 22. The least of these is -14, so the absolute minimum value of gg on [−1,5][-1, 5] is -14, at x=−1x = -1.

  • Point 4 Lists all four candidate values and names the smallest, which is at an endpoint, not at the relative minimum.

(d) gg is a polynomial, so it is continuous on [1,3][1, 3] and differentiable on (1,3)(1, 3). Its average rate of change on [1,3][1, 3] is g(3)−g(1)3−1=2−62=−2\frac{g(3) - g(1)}{3 - 1} = \frac{2 - 6}{2} = -2. By the Mean Value Theorem, g′(c)=−2g'(c) = -2 for some cc with 1<c<31 < c < 3.

  • Point 5 States both conditions of the theorem with the reason they hold.
  • Point 6 Computes the average rate with the values of g and names the theorem for the conclusion.