# How do use the first derivative test to determine the local extrema f(x) = x³+3x²-9x+15?

Aug 9, 2015

$x = - 3$ is a local maximum and $x = 1$ a local minimum.

#### Explanation:

First we look for points that might be the extrema by solving $f ' \left(x\right) = 0$:
$f ' \left(x\right) = 3 {x}^{2} + 6 x - 9 = 3 \left(x + 3\right) \left(x - 1\right)$
$3 \left(x + 3\right) \left(x - 1\right) = 0 \implies x = - 3 \vee x = 1$
$x = - 3$ and $x = 1$ are two potential extrema.

Now, we can use the graph of the first derivative to determine whether those points are extrema and which type of extrema:

if when passing through point $x$ the first derivative changes it's sign from positive to negative then $x$ is a local maximum
and if from negative to positive - a locac minimum.

graph{3(x+3)(x-1) [-10, 10, -5.21, 5.21]}

In this case $x = - 3$ is a local maximum and $x = 1$ a local minimum.