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Intermediate Value Theorem

AP Calculus AB· difficulty 3/5

For f(x)=x3+x−1f(x) = x^3 + x - 1 with f(0)=−1f(0) = -1 and f(1)=1f(1) = 1, the IVT guarantees a c∈(0,1)c \in (0, 1) with

  • A

    f(c)=−1f(c) = -1

  • B

    f(c)=1f(c) = 1

  • C

    f(c)=0f(c) = 0

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  • D

    f′(c)=0f'(c) = 0

Explanation

ff is continuous and changes sign on [0,1][0,1]. By IVT there is a cc with f(c)=0f(c) = 0 between −1-1 and 11.

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