# What are the critical values, if any, of  f(x)= (x-2)/(x^2-4)+2x?

Jan 7, 2017

The critical points are:

$x = - 2 \pm \frac{1}{\sqrt{2}}$

#### Explanation:

We can simplify the function and remove one discontinuity noting that:

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

hence:

$f \left(x\right) = \frac{x - 2}{{x}^{2} - 4} + 2 x = \frac{x - 2}{\left(x - 2\right) \left(x + 2\right)} + 2 x = \frac{1}{x + 2} + 2 x$

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

So we can find the critical points as roots of the equation:

$- \frac{1}{{\left(x + 2\right)}^{2}} + 2 = 0$

$\frac{1}{{\left(x + 2\right)}^{2}} = 2$

${\left(x + 2\right)}^{2} = \frac{1}{2}$

$x = - 2 \pm \frac{1}{\sqrt{2}}$

As:

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

we have:

$f ' ' \left(- 2 - \frac{1}{\sqrt{2}}\right) = \frac{2}{{\left(- 2 - \frac{1}{\sqrt{2}} + 2\right)}^{3}} = - 4 \sqrt{2} < 0$

$f ' ' \left(- 2 + \frac{1}{\sqrt{2}}\right) = \frac{2}{{\left(- 2 + \frac{1}{\sqrt{2}} + 2\right)}^{3}} = 4 \sqrt{2} > 0$

so $x = - 2 - \frac{1}{\sqrt{2}}$ is a local maximum and $x = - 2 + \frac{1}{\sqrt{2}}$ is a local minimum.

graph{(x-2)/(x^2-4)+2x [-23.13, 16.87, -13.6, 6.4]}