Question

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

Answer 1

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]}