Find all minima and maxima of $f(x) = (x+1)^{1/3}(x^2-2x+1)^{1/5}$ Find all minima and maxima of :
$f(x) = (x+1)^{1/3}(x^2-2x+1)^{1/5}$
I feel kind of ashamed by posting such a question but I am not able to get the right answer. I've tried to do it the Fichtenholtz way - probably the most uncreative one and still it didn't work for me. 


*

*I calculate the derivative $f'(x) = \dfrac{11x^2-10x-1}{15((x-1)^{8/5}(x+1)^{2/3}} $

*I check how all the constituent functions behave in certain intervals. In that case we can suspect $x=\frac{-1}{11}$, $-1$ , $1$. Let's call the function in the numarator $a$ and take $b$, $c$ for two separate "brackets" in the denominator. 
Then we get: 
$(-\infty,-1)$: $a -$ , $b -$ , $c +$ , $d +$, so $f'(x) > 0$ 
$(-1 , -\frac{1}{11})$: $a -$ , $b -$ , $c +$ , $d +$, so $f'(x) > 0$
$(-\frac{1}{11}, 1)$: $a+$ , $b-$ , $c +$ , $d +$, so $f'(x) < 0$
$(1 , + \infty)$: $a,b,c,d$ are positive, so $f'(x) > 0$
Therefore when we look at the signs we can say that there's a minimum at $1$ and a maximum at $-\frac{1}{11}$ which is not true. I realise that's somehow really basic problem but could you tell what I'am doing wrong?   
 A: Let's compute the derivative of
$$
f(x)=\sqrt[3]{x+1}\cdot\sqrt[5]{x^2-2x+1}
$$
with the (formal) method of the logarithmic derivative:
$$
\log f(x)=\frac{1}{3}\log(x+1)+\frac{2}{5}\log(x-1)
$$
and therefore
$$
\frac{f'(x)}{f(x)}=\frac{1}{3(x+1)}+\frac{2}{5(x-1)}=
\frac{5x-5+6x+6}{15(x+1)(x-1)}=\frac{11x+1}{15(x+1)(x-1)}
$$
so that
$$
f'(x)=\frac{11x+1}{15\sqrt[3]{(x+1)^2}\cdot\sqrt[5]{(x-1)^3}}
$$
Note that the derivative is correct for all $x$, even when its value is negative. Note also that the function is not differentiable at $-1$ and $1$.
We can easily see that the derivative has the same signs as $(11x+1)(x-1)$ (where it exists), so the derivative is


*

*positive in $(-\infty,-1/11)$ (excluding $-1$),

*negative in $(-1/11,1)$,

*positive in $(1,\infty)$


Thus $-1/11$ is a point of maximum, while $1$ is a point of minimum.
You're right: there can be no change in sign around $-1$. This is the graph drawn by GeoGebra

A: The Mathematica plot gives roots at $ x=\pm 1 $ with vertical tangents.The former,a multiple "root", the latter is a cusp/multiple "root". These can be confirmed by conventional maxima/minima calculation.
"Root" in the sense for $ y = 1/x $ passing between $ - \infty \; to -\infty, $ derivative not zero, but $\infty$.
Maximum value at $ ( -1/11,\approx 1.00304) $

