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

Dec 3, 2015

They are $x = - 3 \pm \sqrt{6}$, which are approximately $- 0.5505$ and $- 5.4495$.

#### Explanation:

By the Quotient Rule, the derivative is

$f ' \left(x\right) = \frac{\left(x + 3\right) \cdot 2 x - \left({x}^{2} - 3\right) \cdot 1}{{\left(x + 3\right)}^{2}} = \frac{{x}^{2} + 6 x + 3}{{\left(x + 3\right)}^{2}}$.

This equals zero when ${x}^{2} + 6 x + 3 = 0$. By the quadratic formula, the roots of this are

$x = \frac{- 6 \pm \sqrt{36 - 12}}{2} = - 3 \pm \sqrt{4} \frac{\sqrt{6}}{2} = - 3 \pm \sqrt{6}$.

These are the critical values of $f$.

You can check, with the First or Second Derivative Test, that $f$ has a local maximum at $- 3 - \sqrt{6}$ and a local minimum at $- 3 + \sqrt{6}$.

Here's the graph of $f$:

graph{(x^2-3)/(x+3) [-80, 80, -40, 40]}