<s>[INST] Max and min of a function with absolute values I have this function : 
$f(x) =
\begin{cases}
\left\lvert\frac{x-3}{x+3}\right\rvert\log \left\lvert\frac{x+3}{x-3}\right\rvert -3,  & \text{if $x\neq \pm3$} \\[2ex]
-3, & \text{if $x= \pm3$}
\end{cases}$
How do you find the derivative?
I usually split the functions when there are absolute values but this time I've found some irregularities about the min and max points.
I would split in this way: 
$f(x)_1 =
\begin{cases}
\left(\frac{x-3}{x+3}\right)\log \left(\frac{x+3}{x-3}\right)-3,  & \text{if $x\lt -3 \land x \gt +3$} \\[2ex]
-3, & \text{if $x= \pm3$}
\end{cases}$
$f(x)_2 =
\begin{cases}
\left(\frac{3-x}{x+3}\right)\log \left(\frac{-x-3}{x-3}\right)-3,  & \text{if  $ -3 \lt x \lt +3$} \\[2ex]
-3, & \text{if $x= \pm3$}
\end{cases}$
And so the derivatives are:
$f(x)_1' =
\begin{cases}
\left(\frac{6}{(x+3)^2}\right)\left(\log \left(\frac{x+3}{x-3}\right) -1\right),  & \text{if $x\lt -3 \land x \gt +3$} \\[2ex]
0, & \text{if $x= \pm3$}
\end{cases}$
$f(x)_2' =
\begin{cases}
\left(-\frac{6}{(x+3)^2}\right)\left(\log \left(\frac{-x-3}{x-3}\right) -1\right),  & \text{if  $ -3 \lt x \lt +3$} \\[2ex]
0, & \text{if $x= \pm3$}
\end{cases}$
I don't know what I got wrong.. I know for sure that it's something related the absolute value.I should get two max and a minimum.
Can you help me please?
Thanks in advance! 
 [/INST] I think what you have is correct. Namely, you can prove using your derivatives on different intervals that the function is strictly decreasing of $x<-3$. On $(-3,3)$ and on $(3,\infty)$ the function is probably concave and hence has one local maximum on each interval. Comparing values at those local maxima with $-3$, values at $x=\pm3$, you get your solution.Plotting the function usually helps.

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