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Extrema: First Derivative Test

AP Calculus AB· difficulty 3/5

For f(x)=∣x∣f(x) = |x| on [−1,1][-1, 1], the absolute minimum occurs at

  • A

    x=−1x = -1

  • B

    x=1x = 1

  • C

    x=0x = 0

    check_circle
  • D

    Both endpoints

Explanation

f(0)=0f(0) = 0 is the minimum (endpoints both give 11). x=0x = 0 is a critical point because f′f' is undefined there.

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