Question

@phi please helpp

Answer 1

How do you find a critical value ?

Answer 2

critical values are where f' = 0
see
http://tutorial.math.lamar.edu/classes/calcI/criticalpoints.aspx

Answer 3

btw, a critical value is also where f' is not defined (e.g. we divided by 0). x=2 would be a critical value, except they ruled out x=2.

so all you have to do is solve for when the "top" equals 0

Answer 4

x^2-16 =0

you can solve by factoring (a difference of squares)
(x-4)(x+4)=0
and solve each factor for when it's 0:
x-4=0 and x+4=0
you get x=4 and x=-4

or you could write the problem as
x^2=16
take the square rt of each side
\[ x= \pm \sqrt{16} = \pm 4\]