AP Calculus AB · Topic 5.5

Curve Sketching Practice

Part of Analytical Applications of Differentiation.(FUN-4.E)

Practice questions

4

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Sample questions

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  1. Sample 1difficulty 2/5

    For f(x)=x3f(x) = x^3, the function is

    • A

      Always increasing, with inflection at x=0x = 0

      check_circle
    • B

      Concave up for x<0x < 0, concave down for x>0x > 0

    • C

      Always decreasing

    • D

      Both A and not C

    Why

    f(x)=3x20f'(x) = 3x^2 \ge 0 — increasing (strictly except at x=0x = 0). f(x)=6xf''(x) = 6x changes sign at 00 — inflection.

  2. Sample 2difficulty 3/5

    Given f>0f' > 0 on (,2)(-\infty, 2), f<0f' < 0 on (2,)(2, \infty), and ff continuous, ff has

    • A

      A local maximum at x=2x = 2

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    • B

      A local minimum at x=2x = 2

    • C

      An inflection point at x=2x = 2

    • D

      No critical point

    Why

    ff' goes from ++ to - at x=2x = 2 → local max.

  3. Sample 3difficulty 3/5

    f<0f' < 0 and f>0f'' > 0 on (a,b)(a, b) describes a graph that is

    • A

      Going up and curving up

    • B

      Going down and curving up (like the right half of a U)

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    • C

      Going down and curving down

    • D

      Going up and curving down

    Why

    Decreasing + concave up — slope is negative but slope is increasing (becoming less negative).

  4. Sample 4difficulty 3/5

    A function with f(0)=0f(0) = 0, ff increasing on (0,)(0, \infty), and concave down everywhere on (0,)(0, \infty) — could be

    • A

      f(x)=xf(x) = -x

    • B

      f(x)=xf(x) = \sqrt x

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    • C

      f(x)=xf(x) = x

    • D

      f(x)=x2f(x) = x^2

    Why

    x\sqrt x: increases, concave down (since f=14x3/2<0f'' = -\tfrac{1}{4}x^{-3/2} < 0).