Question

How do you find the critical numbers of `f(x)=(x^2+6x-7)^2`?

Answer 1

`x=-7, x=1` and `x=-3`

Explanation:

The critical points of a function are where the function's derivative is either undefined or `0`.

Let's first start by computing `f'(x)`. I will not expand the parenthesis (because I'll have to do less factoring later) and instead use the chain rule. If we let `u=x^2+6x-8`, we get:
`d/dx((x^2+6x-7)^2)=d/(du)(u^2)d/dx(x^2+6x-7)`

`2u(2x+6)=2(x^2+6x-7)(2x+6)`

Now we set this expression equal to `0`.
`2(x^2+6x-7)(2x+6)=0`

Factoring `x^2+6x-7` gives:
`2(x+7)(x-1)(2x+6)=0`

This tells us that `x=-7, x=1` and `x=-3` all are solutions and therefor critical points. This function is never undefined, so these are also the only critical points.