Finding first and second derivative of an function with an absolute value

Question

Given the equation $f(x)= |x^2-9|$ where $-4\le x\le 5$, I must find the extremes, as well as the concavities.

This I know how to do.

The issue is I'm unfamiliar on how to find the first and second derivative of the given function.

Thanks in advance.

Answer 1

The function $f(x)$ is

$$f(x)= \begin{cases} x^2-9,&\text{for}\,\,|x|\ge 3\\\\ 9-x^2,&\text{for}\,\,|x|\le 3& \end{cases}$$

Thus, the derivative $f'(x)$ is

$$f'(x)= \begin{cases} 2x,&\text{for}\,\,|x|\ge 3\\\\ -2x,&\text{for}\,\,|x|\le 3 \end{cases}$$

Both the derivative from the left and right are zero at $x=0$ and so this is a local extremum. To check to see if this is a local maximum or a local minimum we can either take a second derivative (which is $-2<0\implies \text{a local maximum}$) or observe that the derivative decreases to the right of $x=0$ and increases to the left (which implies a local maximum).

Thus, the other possible critical points are at the endpoints of the interval $[-4,5]$ or at $x=\pm 3$. The task then is to test these points, compare and determine the absolute extrema.

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NOTE:

Aside, it might be of interest to note that inasmuch as

$$\frac{d|x|}{dx}=\frac{x}{|x|}$$

for $x\ne0$, then

$$\frac{d|f(x)|}{dx}=f'(x)\,\left(\frac{f(x)}{|f(x)|}\right)$$

for $f(x)\ne 0$

Answer 2

For $x>0$,

$$|x|=x,\ |x|'=1,\ |x|''=0.$$

For $x<0$,

$$|x|=-x,\ |x|'=-1,\ |x|''=0.$$

For $x=0$, neither $|x|'$ nor $|x|''$ are defined.

You can summarize with

$$\text{for }x\ne0,\ |x|'=\text{sign}(x),\ |x|''=0.$$

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Let us solve the exercise:

For $x^2-9\ne0$,

$$f'(x)=2x\text{ sign}(x^2-9)$$

has a zero at $x=0$, and changes sign at $x=\pm3$.

$$-\infty\searrow-3\nearrow0\searrow3\nearrow\infty$$

As $$f''(x)=2\text{ sign}(x^2-9)$$ the concavities go like

$$-\infty\smile-3\frown3\smile\infty.$$

$x=0$ is a maximum, and due to the discontinuity of the derivative, $x=\pm3$ are angular points.