For , at which positive value of does have a local minimum?

Answer and explanation

Correct answer: C. 4

  1. Step 1: f'(x) = x^2 - 16, which is zero at x = 4 or x = -4.
  2. Step 2: f' changes from negative to positive at x = 4, so that is a local minimum.
Answer
4.
Why not the others: At x = -4 the derivative changes from positive to negative, giving a local maximum; 16 and 0 do not make the derivative zero.

Practise more Calculus.

Free questions with an explanation for every one, and full timed Tawjihiya Bahrain (Secondary School Certificate) mock tests.

All Calculus questionsTake a mock test

More Calculus questions