# What are the local extrema, if any, of f (x) =(x^3-3)/(x+6)?

Nov 12, 2015

The one real number critical point of this function is $x \approx - 9.01844$. A local minimum occurs at this point.

#### Explanation:

By the Quotient Rule, the derivative of this function is

$f ' \left(x\right) = \frac{\left(x + 6\right) \cdot 3 {x}^{2} - \left({x}^{3} - 3\right) \cdot 1}{{\left(x + 6\right)}^{2}} = \frac{2 {x}^{3} + 18 {x}^{2} + 3}{{\left(x + 6\right)}^{2}}$

This function equals zero if and only if $2 {x}^{3} + 18 {x}^{2} + 3 = 0$. The roots of this cubic include on negative irrational (real) number and two complex numbers.

The real root is $x \approx - 9.01844$. If you plug in a number just less than this into $f '$, you'll get a negative output and if you plug a number just greater than this into $f '$, you'll get a positive output. Therefore, this critical point gives a local minimum value of $f$ (and $f \left(- 9.01844\right) \approx 244$ is the local minimum value (output).