Given #f'(x) = 8x^3-16x# and assuming that domain of #f# includes the zeros of #f'#, we proceed:

**Finding Critical Numbers for #f#**

A critical number for #f# is a number in the domain of #f# at which either #f(x)=0# or #f'(x)# does not exist.

In this problem, #f'# is a polynomial, so it exists for all #x#. The critical numbers for this #f# are the zeros of #f'#.

#f'(x) = 8x^3-16x =8x(x^2-2) = 0# at #-sqrt2#, #0#, and #sqrt2#

These are our critical numbers.

**Note** To use the first derivative test for local extrema, we check the sign of #f'# on both sides of each critical number.

**Testing the Critical Numbers**

To use the second derivative test for local extrema, we check the sign of #f''# **at** each critical number.

#f''(x) = 24x^2-16#

At #-sqrt2#, we get #f''(-sqrt2) = 24(-sqrt2)^2-6# which is clearly positive.

#f(-sqrt2)# is a local minimum.

At #0#, we get #f''(0) = 24(0)^2-16# which is negative.

#f(0)# is a local maximum.

At #sqrt2#, we get #f''(sqrt2) = 24(sqrt2)^2-6# which is clearly positive.

#f(sqrt2)# is a local minimum.