How do you use the first derivative test to find the local max and local min values of #f(x)= -x^3 + 12x#?

1 Answer
Apr 16, 2016

Answer:

For local minimum and maximum values the derivative is equal to zero.

Explanation:

Given #f(x)=-x^3+12x#

The first derivative is
#color(white)("XXX")f'(x)=-3x^2+12#

For local minimum and maximum values
#color(white)("XXX")f'(x) = 0#

So
#color(white)("XXX")-3x^2+12=0#

#color(white)("XXX")rarr x^2-4=0#

#color(white)("XXX")rarr x=-2 or x=+2#

If #x=-2#
#color(white)("XXX")f(x=-2)= -(-2)^3+12(-2)#
#color(white)("XXXXXXXXXX")=8-24#
#color(white)("XXXXXXXXXX")=-16#

If #x=+2#
#color(white)("XXX")f(x=+2) = -(+2)^3+12(+2)#
#color(white)("XXXXXXXXXX")=-8+24#
#color(white)("XXXXXXXXXX")=+16#

So #x=-2# gives a local minimum
and #x=+2# give a local maximum

This can be verified by checking the graph of the original function:
graph{-x^3+12x [-41.1, 41.07, -20.53, 20.55]}