# Question #cbfb8

Feb 27, 2017

There is one critical number, at $x = 1$, which is neither a maximum nor a minimum.

#### Explanation:

Start by finding the derivative by the power rule.

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

The critical points will occur when the derivative equals $0$ or is undefined. This is a quadratic function, so it is defined on all of $x$.

$0 = 3 {x}^{2} - 6 x + 3$

$0 = 3 \left({x}^{2} - 2 x + 1\right)$

$0 = \left(x - 1\right) \left(x - 1\right)$

$x = 1$

Next, we must verify whether this is a local minimum or a local maximum. Note that this function will never have an absolute maximum/minimum, as shows the table below.

We can verify whether this point is a local min or a local max by finding the intervals of increase/decrease.

Test point 1: $x = 0$

$f ' \left(0\right) = 3 {\left(0\right)}^{2} - 6 \left(0\right) + 3 = 3$

Test point 2: $x = 3$

$f ' \left(3\right) = 3 {\left(3\right)}^{2} - 6 \left(3\right) + 3 = 27 - 18 + 3 = 12$

Since $f ' \left(x\right) > 0$ on both sides of $x = 1$, our single critical point is neither a maximum nor a minimum.

Hopefully this helps!