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

Dec 8, 2017

$x = - 7 , x = 1$ 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 $f ' \left(x\right)$. I will not expand the parenthesis (because I'll have to do less factoring later) and instead use the chain rule. If we let $u = {x}^{2} + 6 x - 8$, we get:
$\frac{d}{\mathrm{dx}} \left({\left({x}^{2} + 6 x - 7\right)}^{2}\right) = \frac{d}{\mathrm{du}} \left({u}^{2}\right) \frac{d}{\mathrm{dx}} \left({x}^{2} + 6 x - 7\right)$

$2 u \left(2 x + 6\right) = 2 \left({x}^{2} + 6 x - 7\right) \left(2 x + 6\right)$

Now we set this expression equal to $0$.
$2 \left({x}^{2} + 6 x - 7\right) \left(2 x + 6\right) = 0$

Factoring ${x}^{2} + 6 x - 7$ gives:
$2 \left(x + 7\right) \left(x - 1\right) \left(2 x + 6\right) = 0$

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.