On $[-2, 2]$ , at which points does $f(x) = 2x^{3} - 6x$ take its least value?

x = 1

onlyCorrect answerYour answerAt

x = -2

and

x = 1

Correct answerYour answerAt

x = -2

onlyCorrect answerYour answerAt

x = -1

and

x = 2

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer B

Correct answer: B

$f'(x) = 6x^{2} - 6 = 0$ at $x = \pm 1$ .
$f(-2) = -4$ , $f(-1) = 4$ , $f(1) = -4$ , $f(2) = 4$ : the least value $-4$ comes at both $x = -2$ and $x = 1$ .

Worth noting

Stopping at the local minimum, $x = 1$ , misses the end $x = -2$ , where $f$ is just as low.

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

On [−2,2][-2, 2][−2,2], at which points does f(x)=2x3−6xf(x) = 2x^{3} - 6xf(x)=2x3−6x take its least value?