What are the absolute extrema of #f(x) =x^4 − 8x^2 − 12 in[-3,-1]#?

1 Answer
Ken C.
May 30, 2016

#-3# (occurring at #x=-3#) and #-28# (occurring at #x=-2#)

Explanation:

Absolute extrema of a closed interval occur at the endpoints of the interval or at #f'(x)=0#.

That means we'll have to set the derivative equal to #0# and see what #x#-values that gets us, and we'll have to use #x=-3# and #x=-1# (because these are the endpoints).

So, starting with taking the derivative:
#f(x)=x^4-8x^2-12#
#f'(x)=4x^3-16x#

Setting it equal to #0# and solving:
#0=4x^3-16x#
#0=x^3-4x#
#0=x(x^2-4)#
#x=0# and #x^2-4=0#
Thus the solutions are #0,2,# and #-2#.

We immediately get rid of #0# and #2# because they are not on the interval #[-3,-1]#, leaving only #x=-3,-2,# and #-1# as the possible places where extrema can occur.

Finally, we evaluate these one by one to see what the absolute min and max are:
#f(-3)=-3#
#f(-2)=-28#
#f(-1)=-19#

Therefore #-3# is the absolute maximum and #-28# is the absolute minimum on the interval #[-3,-1]#.

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