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

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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#