# What are the critical values, if any, of  f(x)= |x^3 - 3 x^2 + 2| ?

Jul 29, 2016

The critical numbers are $1 - \sqrt{3}$, $\text{ }$ $1$, $\text{ }$ $1 + \sqrt{3}$ $\text{ }$ and $x = 0$, $\text{ }$ $2$ $\text{ }$

#### Explanation:

$\left\mid {x}^{3} - 3 {x}^{2} + 2 \right\mid = \left\{\begin{matrix}{x}^{3} - 3 {x}^{2} + 2 & \text{ if " & x^3-3x^2+2 > 0 \\ -(x^3-3x^2+2) & " if } & {x}^{3} - 3 {x}^{2} + 2 > 0\end{matrix}\right.$

We need to solve the inequalities.
First we solve:

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

Using the rational zeros theorem or obsrving that the sum of the coefficients is $0$, we learn that $1$ is a solution.

Therefore $x - 1$ is a factor. Use division or trial and error to factor and get

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

The quadratic can be solved using the formula or completing the square. We get:

$x = 1 - \sqrt{3}$, $\text{ }$ $1$, $\text{ }$ $1 + \sqrt{3}$

Analyzing the sign, we get

{: (bb"Interval:",(-oo,1-sqrt3),(1-sqrt3,1),(1,1+sqrt3),(1+sqrt3,oo)), (darrbb"Factors"darr,"========","======","=====","======"), (x-2, bb" -",bb" -",bb" +",bb" +"), (x^2-2x-2,bb" +",bb" -",bb" -",bb" +"), ("==========","========","======","=====","======"), (x^3-3x^2+2,bb" -",bb" +",bb" -",bb" +") :}

So we can write

$f \left(x\right) = \left\mid {x}^{3} - 3 {x}^{2} + 2 \right\mid = \left\{\begin{matrix}{x}^{3} - 3 {x}^{2} + 2 & \text{ if " & x < 1-sqrt3 \\ -(x^3-3x^2+2) & " if " & 1-sqrt3 < x < 1 \\ x^3-3x^2+2 & " if " & 1 < x < 1+sqrt3 \\ -(x^3-3x^2+2) & " if } & 1 + \sqrt{3} < x\end{matrix}\right.$

Differentiating each piece yields

$f ' \left(x\right) = \left\{\begin{matrix}3 {x}^{2} - 6 x & \text{ if " & x < 1-sqrt3 \\ -3x^2+6x & " if " & 1-sqrt3 < x < 1 \\ 3x^2-6x & " if " & 1 < x < 1+sqrt3 \\ -3x^2+6x & " if } & 1 + \sqrt{3} < x\end{matrix}\right.$

We can quickly see that $f ' \left(x\right) = 0$ at $0$ and at $2$.

It takes a moment to realize that $f$ is not differentiable at the three cut points. $x = 1 - \sqrt{3}$, $\text{ }$ $1$, $\text{ }$ $1 + \sqrt{3}$.

The derivative on the left and right of the cuts have opposite signs because they are not $0$ at the cuts.

(We can actually calculate if you like, the left derivative at $1 - \sqrt{3}$ is $6$ and the right derivative is $- 6$. So the (two-sided) derivative does not exist at this point.

The critical numbers are

$x = 1 - \sqrt{3}$, $\text{ }$ $1$, $\text{ }$ $1 + \sqrt{3}$ $\text{ }$ (where $f '$ does not exist)

and
$x = 0$, $\text{ }$ $2$ $\text{ }$ (where $f '$ is $0$)