Find the absolute maximum and absolute minimum values of $f(x) = x^(1/3) e^(−x^2/8)$ on the interval [−1, 4]?

Question

4497 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-17T15:31:18

The maximum is $root(6)(4/3)e^(-1/6) ~~ 0.888$ (it occurs at $x=sqrt(4/3)$ ) and the minimum is $-e^(-1/8) ~~ -0.882$ (at $x=-1)$ .

$f'(x) = (4-3x^2)/(12x^(2/3)e^(x^2/8))$

$f'$ is undefined at $x=0$ and $f'(x) = 0$ for $x = +- sqrt(4/3)$

The critical numbers in $[-1,4]$ are $0$ , $sqrt(4/3)$ .

Evaluating, we find that

$f(-1) ~~ - 0.8825$

$f(0) = 0$

$f(sqrt(4/3)) ~~ 0.888$

$f(3) ~~ 0.468$

Find the absolute maximum and absolute minimum values of f(x) = x^(1/3) e^(−x^2/8)f(x) = x^(1/3) e^(−x^2/8) on the interval [−1, 4]?