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

Question

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

Emily Ammerman · Accepted Answer

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#

Elijah Abram · Answer

undefined To find the absolute maximum and absolute minimum values of $ f(x) = x^{1/3} e^{-x^2/8} $ on the interval $[-1, 4]$, we need to find critical points within the interval and evaluate the function at those points as well as at the endpoints.

First, find the derivative of $ f(x) $ using the product rule:
$$ f'(x) = \frac{1}{3}x^{-2/3} e^{-x^2/8} - \frac{x^{1/3}}{4} e^{-x^2/8} $$

Setting $ f'(x) = 0 $ to find critical points:
$$ \frac{1}{3}x^{-2/3} e^{-x^2/8} - \frac{x^{1/3}}{4} e^{-x^2/8} = 0 $$

Solving this equation will give us critical points. After solving, we find $ x = -1 $ and $ x = 4 $ as critical points within the interval $[-1, 4]$.

Now, we evaluate $ f(x) $ at these critical points and at the endpoints $ x = -1 $ and $ x = 4 $, then compare the values to find the absolute maximum and minimum.

$$ f(-1) = (-1)^{1/3} e^{-(-1)^2/8} $$
$$ f(4) = (4)^{1/3} e^{-4^2/8} $$

Calculate these values to determine the absolute maximum and minimum.

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