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How can I find the critical points of a square root function?

For example: What are the critical points of $f(x) = \sqrt{7x^2 + 2x - 2}$?

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The standard way of finding where the derivative vanishes doesn't work? – J. M. Oct 25 '10 at 5:41
...or doesn't exist? – Vandermonde Oct 25 '10 at 5:42
  1. Take the derivative. This is straightforward here using the chain rule: $$\frac{14x + 2}{2\sqrt{7x^2 + 2x - 2}} = \frac{7x+1}{\sqrt{7x^2 + 2x - 2}}.$$

  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.

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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


$$g'(x) = \frac{h'(x)}{2\sqrt{h(x)}}$$

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:


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.

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