# 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!

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.
• $log{\frac{x+3}{x-3}}-1$ is negative for $x<\frac{3+3\exp{1}}{\exp{1}-1}$. – Jan Feb 2 '17 at 18:49