# What are the critical values, if any, of f(x) = (x - 1) / (x + 3) -sqrt(x^2-3)?

Mar 1, 2017

There are critical values at:

$x = \pm \sqrt{3}$
$x = - 4.935$

#### Explanation:

Start by finding the derivative using the quotient and chain rules.

f'(x) = (1(x + 3) - 1(x - 1))/(x + 3)^2 - (2x)/(2sqrt(x^2 - 3)

$f ' \left(x\right) = \frac{x + 3 - x + 1}{x + 3} ^ 2 - \frac{x}{\sqrt{{x}^{2} - 3}}$

$f ' \left(x\right) = \frac{4}{x + 3} ^ 2 - \frac{x}{\sqrt{{x}^{2} - 3}}$

Critical points will occur whenever the derivative equals $0$ (a horizontal tangent or is undefined (vertical tangent).

The derivative will be undefined whenever the denominator equals $0$.

${\left(x + 3\right)}^{2} = 0$

$x = - 3$

AND

$\sqrt{{x}^{2} - 3} = 0$

$x = \pm \sqrt{3}$

However, we disregard the first critical value, $x = - 3$, because it also renders the function undefined.

Now let's see where the derivative equals $0$.

$0 = \frac{4}{x + 3} ^ 2 - \frac{x}{\sqrt{{x}^{2} - 3}}$

$\frac{x}{\sqrt{{x}^{2} - 3}} = \frac{4}{x + 3} ^ 2$

${x}^{2} / \left({x}^{2} - 3\right) = \frac{16}{x + 3} ^ 4$

${x}^{2} {\left(x + 3\right)}^{4} = 16 \left({x}^{2} - 3\right)$

This equation is very difficult, if not impossible to solve algebraically. Use either Newton's Method, a CAS or a graphing calculator to solve.

I'll use the latter. On your equations tab, enter

$\left\{\begin{matrix}{y}_{1} = {x}^{2} {\left(x + 3\right)}^{4} \\ {y}_{2} = 16 \left({x}^{2} - 3\right)\end{matrix}\right.$

Find the intersect-finding function. These will give you your solutions. This will be $x = - 1.839$ and $x = - 4.935$. However, $x = - 1.839$ is extraneous, so it is not a critical value.

So, you're done here!

Hopefully this helps!