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Extrema: First Derivative Test

AP Calculus AB· difficulty 4/5

On [−2,3][-2, 3], f(x)=x3−3xf(x) = x^3 - 3x has its absolute minimum at

  • A

    x=1x = 1

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  • B

    x=−1x = -1

  • C

    x=3x = 3

  • D

    x=−2x = -2

Explanation

f′=3x2−3=3(x−1)(x+1)f' = 3x^2 - 3 = 3(x-1)(x+1). Critical: ±1\pm 1. f(−2)=−2f(-2) = -2, f(−1)=2f(-1)=2, f(1)=−2f(1)=-2, f(3)=18f(3)=18. Min value −2-2 achieved at both x=−2x=-2 and x=1x=1. Of the choices, x=1x=1 is correct.

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