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.
$\endgroup$ 62 Answers
$\begingroup$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.
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 $f(x)\ne 0$
$\endgroup$ 2 $\begingroup$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
$$\text{for }x\ne0,\ |x|'=\text{sign}(x),\ |x|''=0.$$
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.
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