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Mean Value Theorem and Extreme Value Theorem

AP Calculus AB· difficulty 3/5

For f(x)=x3−12xf(x) = x^3 - 12x on [−3,3][-3, 3], the absolute maximum is

  • A

    −9-9

  • B

    1616

    check_circle
  • C

    99

  • D

    00

Explanation

f′(x)=3x2−12=0⇒x=±2f'(x) = 3x^2 - 12 = 0 \Rightarrow x = \pm 2. Values: f(−3)=−27+36=9f(-3) = -27 + 36 = 9, f(−2)=−8+24=16f(-2) = -8 + 24 = 16, f(2)=−16f(2) = -16, f(3)=−9f(3) = -9. Max =16= 16.

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