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Mean Value Theorem and Extreme Value Theorem

AP Calculus AB· difficulty 3/5

For f(x)=x2f(x) = x^2 on [1,4][1, 4], the value of cc guaranteed by MVT satisfies f′(c)=(f(4)−f(1))/(4−1)f'(c) = (f(4) - f(1))/(4-1). So c=c =

  • A

    1.51.5

  • B

    3.53.5

  • C

    2.52.5

    check_circle
  • D

    33

Explanation

(16−1)/3=5(16 - 1)/3 = 5. f′(c)=2c=5⇒c=2.5f'(c) = 2c = 5 \Rightarrow c = 2.5.

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