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

Nov 11, 2015

${x}^{3} - 3 x + 6$ has local extrema at $x = - 1$ and $x = 1$

#### Explanation:

The local extrema of a function occur at points where the first derivative of the function is $0$ and the sign of the first derivative changes.
That is, for $x$ where $f ' \left(x\right) = 0$ and either $f ' \left(x - \varepsilon\right) \le 0 \mathmr{and} f ' \left(x + \varepsilon\right) \ge 0$ (local minimum) or
$f ' \left(x - \varepsilon\right) \ge 0 \mathmr{and} f ' \left(x + \varepsilon\right) \le 0$ (local maximum)

To find the local extrema, then, we need to find the points where $f ' \left(x\right) = 0$.

$f ' \left(x\right) = 3 {x}^{2} - 3 = 3 \left({x}^{2} - 1\right) = 3 \left(x + 1\right) \left(x - 1\right)$
so
$f ' \left(x\right) = 0 \iff 3 \left(x + 1\right) \left(x - 1\right) = 0 \iff x = \pm 1$

Looking at the sign of $f '$ we get
$\left\{\begin{matrix}f ' \left(x\right) > 0 \mathmr{if} x < - 1 \\ f ' \left(x\right) < 0 \mathmr{if} - 1 < x < 1 \\ f ' \left(x\right) > 0 \mathmr{if} x > 1\end{matrix}\right.$

So the sign of $f '$ changes at each of $x = - 1$ and $x = 1$ meaning there is a local extremum at both points.

Note: From the change in signs, we can further tell that there is a local maximum at $x = - 1$ and a local minimum at $x = 1$.