Mean Value Theorem and Extreme Value Theorem

AP Calculus AB· difficulty 4/5

On [1,4][1, 4], the function f(x)=x3f(x) = x^3 satisfies the MVT. The value c(1,4)c \in (1,4) guaranteed by the theorem is

  • A

    c=2c = 2

  • B

    c=7c = \sqrt{7}

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  • C

    c=2.5c = 2.5

  • D

    c=3c = 3

Explanation

MVT: f(c)=(f(4)f(1))/(41)=(641)/3=21f'(c) = (f(4)-f(1))/(4-1) = (64-1)/3 = 21. Solve 3c2=213c^2 = 21 to get c2=7c^2=7, so c=72.65c = \sqrt{7} \approx 2.65.

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