What are the local extrema, if any, of #f(x)= 120x^5 - 200x^3#?

1 Answer
Mar 11, 2016

Answer:

Local maximum of #80# (at #x=-1#) and local minimum of #-80# (at #x=1#.

Explanation:

#f(x)= 120x^5 - 200x^3#

#f'(x)= 600x^4 - 600x^2 = 600x^2(x^2 - 1)#

Critical numbers are: #-1#, #0#, and #1#

The sign of #f'# changes from + to - as we pass #x=-1#, so #f(-1) = 80# is a local maximum.

(Since #f# is odd, we can immediately conclude that #f(1)=-80# is a relative minimum and #f(0)# is not a local extremum.)

The sign of #f'# does not change as we pass #x= 0#, so #f(0)# is not a local extremum.

The sign of #f'# changes from - to + as we pass #x=1#, so #f(1) = -80# is a local minimum.