A function has $f'(x) = (x + 1)(x - 2)^2(x - 4)$ . Where does it have a local minimum?

x = 4

onlyCorrect answerYour answerAt

x = -1

onlyCorrect answerYour answerAt

x = 2

and

x = 4

Correct answerYour answerAt

x = -1

and

x = 4

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer A

Correct answer: A

Signs of $f'$ : positive for $x < -1$ , negative on $(-1, 2)$ and on $(2, 4)$ , positive for $x > 4$ .
Negative to positive only at $x = 4$ : the local minimum. At $x = -1$ it goes positive to negative (a local maximum); at $x = 2$ it does not change.

Worth noting

The squared factor $(x - 2)^2$ makes $f'(2) = 0$ without a change of sign, so $x = 2$ is not an extremum.

Ten questions from Maxima & Minima, at your own pace.

A function has f′(x)=(x+1)(x−2)2(x−4)f'(x) = (x + 1)(x - 2)^2(x - 4)f′(x)=(x+1)(x−2)2(x−4). Where does it have a local minimum?