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
Alan P.
Apr 16, 2016

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]}

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