Critical Points of a Square Root Function How can I find the critical points of a square root function?
For example: What are the critical points of $f(x) = \sqrt{7x^2 + 2x - 2}$?
 A: *

*Take the derivative. This is straightforward here using the chain rule:
$$\frac{14x + 2}{2\sqrt{7x^2 + 2x - 2}} = \frac{7x+1}{\sqrt{7x^2 + 2x - 2}}.$$

*Find the points in the domain of $f(x)$ where the derivative does not exist or is equal to zero.
Hint the first: There are two points that are in the domain of $f(x)$ where the derivative does not exist.
Hint the second: A fraction is equal to zero if and only if the numerator is $0$ and the denominator is not zero.
A: to find the critical points of this function, you can either do f'(x) = 0 directly using the chain rule or you can note the following:
Please consider the following:
$h(x)>0$ and exists everywhere
$g(x) = \sqrt{h(x)}$ --> definition for us
then,
$$g'(x) = \frac{h'(x)}{2\sqrt{h(x)}}$$
thus since $h(x)>0$ at any point $x$, then $2\sqrt{h(x)}>0$ at any point $x$.
thus the points at which $g'(x)=0$ are the same as where $f'(x)=0$ being that $f'(x)$ is the numerator of $g'(x)$.
The preceding works as well for $g(x)$ and $h(x)=g(x)^2$; which is just squaring both sides of the previous argument.
This helps you in the following way:
$f(x)=\sqrt{7x^2+2x-2}$
the critical points are now found by using a function $g(x)$ where:
$g(x)=f(x)^2 = 7x^2+2x-2$
$g'(x) = 14x+2$
critical points are then located at $x$ such that $14x+2=0$ which is that same as the numerator of the previous post.
