AP Calculus AB · Lesson 2

Increasing, decreasing and concavity

Know Know it

Increasing and decreasing. ff is increasing on an interval where f′(x)>0f'(x) > 0 and decreasing where f′(x)<0f'(x) < 0. To find the intervals, find where f′=0f' = 0 or is undefined, then test the sign of f′f' between those points.

Concavity. The graph of ff is concave up where f′f' is increasing, which happens where f′′>0f'' > 0, and concave down where f′f' is decreasing, where f′′<0f'' < 0.

A point of inflection is where the concavity changes, which is where f′′f'' changes sign (or, on a graph of f′f', where f′f' switches between increasing and decreasing).

Second derivative test. If f′(c)=0f'(c) = 0 and f′′(c)>0f''(c) > 0, ff has a relative minimum at cc. If f′(c)=0f'(c) = 0 and f′′(c)<0f''(c) < 0, it has a relative maximum. If f′′(c)=0f''(c) = 0, the test gives no answer, so use the first derivative test.

What each sign tells you about f
If…then f…
f′>0f' > 0is increasing
f′<0f' < 0is decreasing
f′f' changes from + to −has a relative maximum
f′f' changes from − to +has a relative minimum
f′′>0f'' > 0 (or f′f' increasing)is concave up
f′′<0f'' < 0 (or f′f' decreasing)is concave down
f′′f'' changes signhas a point of inflection

Trap A zero is not the same as a sign change

f′(c)=0f'(c) = 0 does not guarantee an extremum, and f′′(c)=0f''(c) = 0 does not guarantee an inflection point. f(x)=x3f(x) = x^3 has f′(0)=0f'(0) = 0 but keeps increasing, and f(x)=x4f(x) = x^4 has f′′(0)=0f''(0) = 0 but stays concave up. Always check that the sign actually changes.

Worked example Easy

The derivative of a function ff is f′(x)=3x2−12x−36f'(x) = 3x^2 - 12x - 36. On which interval is ff decreasing?

  1. (−∞,−2)(-\infty, -2) and (6,∞)(6, \infty)
  2. (−2,6)(-2, 6)Answer
  3. (2,∞)(2, \infty)
  4. (−∞,2)(-\infty, 2)

How to solve it

  1. Factor: f′(x)=3(x2−4x−12)=3(x−6)(x+2)f'(x) = 3(x^2 - 4x - 12) = 3(x - 6)(x + 2), with zeros at x=−2x = -2 and x=6x = 6.
  2. Test a point between them: f′(0)=−36<0f'(0) = -36 < 0.
  3. So ff is decreasing on (−2,6)(-2, 6) and increasing outside it.

Why each choice is right or wrong

  • A. Incorrect. On (−∞,−2)(-\infty, -2) and (6,∞)(6, \infty), f′(x)=3(x−6)(x+2)f'(x) = 3(x - 6)(x + 2) is positive, so ff is increasing there.
  • B. Correct. f′(x)=3(x−6)(x+2)f'(x) = 3(x - 6)(x + 2) is negative between its zeros, on (−2,6)(-2, 6), so ff is decreasing there.
  • C. Incorrect. (2,∞)(2, \infty) is where f′′(x)=6x−12f''(x) = 6x - 12 is positive, so it describes concave up, not decreasing.
  • D. Incorrect. (−∞,2)(-\infty, 2) is where f′′(x)=6x−12f''(x) = 6x - 12 is negative, so it describes concave down, not decreasing.

Explain Explain it

Task verbs: justify, determine

To justify an extremum, give the sign change of f′f' at that input in words: "ff has a relative maximum at x=2x = 2 because f′f' changes from positive to negative there." Saying only "f′(2)=0f'(2) = 0" is not enough.

To justify concavity or an inflection point, give the behavior of f′′f'', or of f′f' if that is what you were given: "the graph of ff has a point of inflection at x=1x = 1 because f′f' changes from decreasing to increasing there."

Worked example Medium

The derivative of a function ff is f′(x)=x2−3x−10f'(x) = x^2 - 3x - 10. At which value of xx does ff have a relative minimum?

  1. x=−2x = -2
  2. x=1.5x = 1.5
  3. x=5x = 5Answer
  4. x=−2x = -2 and x=5x = 5

How to solve it

  1. f′(x)=(x−5)(x+2)f'(x) = (x - 5)(x + 2) is 0 at x=−2x = -2 and x=5x = 5.
  2. Signs: f′(−3)=8>0f'(-3) = 8 > 0, f′(0)=−10<0f'(0) = -10 < 0, f′(6)=8>0f'(6) = 8 > 0.
  3. At x=5x = 5, f′f' goes from negative to positive: a relative minimum.

Why each choice is right or wrong

  • A. Incorrect. At x=−2x = -2, f′(x)=(x−5)(x+2)f'(x) = (x - 5)(x + 2) changes from positive to negative, so ff has a relative maximum there.
  • B. Incorrect. x=1.5x = 1.5 is where f′′(x)=2x−3f''(x) = 2x - 3 is 0, a point of inflection, not a critical point of ff.
  • C. Correct. f′(x)=(x−5)(x+2)f'(x) = (x - 5)(x + 2) changes from negative to positive at x=5x = 5, so ff has a relative minimum there.
  • D. Incorrect. Both x=−2x = -2 and x=5x = 5 are critical points, but only x=5x = 5 is a minimum; x=−2x = -2 is a maximum.

Justify Your turn

The derivative of a function gg is g′(x)=(x−2)2(x+1)g'(x) = (x - 2)^2(x + 1). Find each value of xx at which gg has a relative extremum, say whether each is a maximum or a minimum, and justify your answer. Also say whether gg has an extremum at x=2x = 2.

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

Compare with a model answer

A full answer does this

  • Says gg has a relative minimum at x=−1x = -1 because g′g' changes from negative to positive there
  • Says gg has no relative extremum at x=2x = 2 because g′g' does not change sign there

Model answer

g′(x)=(x−2)2(x+1)g'(x) = (x - 2)^2(x + 1) is 0 at x=−1x = -1 and x=2x = 2. The factor (x−2)2(x - 2)^2 is never negative, so the sign of g′g' is the sign of x+1x + 1 (except at x=2x = 2). At x=−1x = -1, g′g' changes from negative to positive, so gg has a relative minimum there. At x=2x = 2, g′g' is positive on both sides and does not change sign, so gg has no relative extremum at x=2x = 2.

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