Consider the real function $f(x)=\frac{5}{4} x^{\frac{4}{5}} - |x-2|$.
$f$ is a continuous function and it is differentiable on $\mathbb{R}$ except at $\{0,2\}$.
Indeed, $f$ is as follows:(i) If $x \geq 2$, then $f(x)=\frac{5}{4} x^{\frac{4}{5}}-x+2$.(ii) If $x < 2$, then $f(x)=\frac{5}{4} x^{\frac{4}{5}}+x-2$.
Hence,(i) If $x > 2$, then $f'(x)= x^{-\frac{1}{5}}-1$.(ii) If $x < 2$, then $f'(x)= x^{-\frac{1}{5}}+1$. Here we must exclude $x=0$ where $f'$ is not defined.(iii) If $x=2$, then $f'$ does not exist (the one-sided limits of $f'$ are different).
Solving $f'=0$ we obtain $x=-1$.
Therefore, we have three critical points $\{-1,0,2\}$.
By checking the sign of the derivative $f'$, we see that $x=-1$ is maximum, $x=0$ is minimum and $x=2$ is maximum.
My question is: Why in Wolframalpha the graph of $f$ looks different from what I have found? Only $x=2$ looks like a maximum point. $\{-1,0\}$ do not look like extreme points.
Thank you very much!