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

1 Answer
Dec 8, 2017

Answer:

#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.