# What are the local extrema of f(x)= x^3-6x^2+15, if any?

Dec 3, 2015

$\left(0 , 15\right) , \left(4 , - 17\right)$

#### Explanation:

A local extremum, or a relative minimum or maximum, will occur when the derivative of a function is $0$.

So, if we find $f ' \left(x\right)$, we can set it equal to $0$.

$f ' \left(x\right) = 3 {x}^{2} - 12 x$

Set it equal to $0$.

$3 {x}^{2} - 12 x = 0$

$x \left(3 x - 12\right) = 0$

Set each part equal to $0$.

$\left\{\begin{matrix}x = 0 \\ 3 x - 12 = 0 \rightarrow x = 4\end{matrix}\right.$

The extrema occur at $\left(0 , 15\right)$ and $\left(4 , - 17\right)$.

Look at them on a graph:

graph{x^3-6x^2+15 [-42.66, 49.75, -21.7, 24.54]}

The extrema, or changes in direction, are at $\left(0 , 15\right)$ and $\left(4 , - 17\right)$.