So I am asked to find the maximum and minimum values of $f(x)=6(x-1)^\frac{1}{3}-15(x-1)^\frac{2}{3}+12|x|$ on the interval $[-7,2]$
This was easy enough. I took the derivative which was
$f'(x)=2(x-1)^\frac{-2}{3}-10(x-1)^\frac{-1}{3}+\frac{12x}{|x|}$
I'm having trouble finding the critical points though.
I can get $x=1,0$ and of course the endpoints $x=-7,2$ but the others are $x=\frac{28}{27},\frac{9}{8}$ and I have no idea how to get those values.
I am trying to solve for $x$ when $f'(x)$ is 0 but I am having a lot of trouble with the algebra. Any help?
May be, you could consider the two cases
If $x >0$, $$f(x)=6(x-1)^\frac{1}{3}-15(x-1)^\frac{2}{3}+12x\implies f'(x)=\frac{2}{(x-1)^{2/3}}-\frac{10}{(x-1)^{1/3}}+12$$ Setting $y=\frac{1}{(x-1)^{1/3}}$ the derivative is zero when $2y^2-10y+12=0$ that is to to say $y=2$ and $y=3$ corresponding to $x=\frac {9}{8}$ and to $x=\frac {28}{27}$.
If $x <0$, $$f(x)=6(x-1)^\frac{1}{3}-15(x-1)^\frac{2}{3}-12x\implies f'(x)=\frac{2}{(x-1)^{2/3}}-\frac{10}{(x-1)^{1/3}}-12$$ Setting $y=\frac{1}{(x-1)^{1/3}}$ the derivative is zero when $2y^2-10y-12=0$ that is to to say $y=-1$ or $y=6$ corresponding to an undefined value of $x$ and to $x=\frac {217}{216}$ which is a contradiction.
To summarize $f'(x)$ only cancels for $x=\frac {9}{8}$ and $x=\frac {28}{27}$.