# How do you find the local extremas for  f(x)= (x-3)^3?

May 1, 2017

No local extrema.

#### Explanation:

Take the derivative of $f \left(x\right)$. Use chain rule (although the derivative of $x - 3$ is just $1$ so it doesn't really matter.:

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

Find when $f ' \left(x\right) = 0$

$0 = 3 {\left(x - 3\right)}^{2}$

This happens when $x = 3$.

Now check both sides of the zero.

However, check when you check both sides of $f ' \left(x\right)$ they're both positive. So there's no local extremas.

You can check this with the graph.

graph{(x-3)^3 [-10, 10, -5, 5]}

As you can see, there's never a change in slope based on this graph. There are however, absolute maximums and minimums at $x = \infty$ and $x = - \infty$ respectively.