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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?

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  • $\begingroup$ Check that the function is not differentiable at 0. $\endgroup$ – Sachchidanand Prasad Nov 7 '16 at 3:46
  • $\begingroup$ Yeah I know. the derivative of the absolute value does not exist at 0. What about the other values though? $\endgroup$ – Future Math person Nov 7 '16 at 3:58
  • $\begingroup$ Wait I'm not so sure any of the x-values you listed are correct. When $x = 1$, for instance, $f'(x) = 12$, not $0$; when $x = 0, f'(x) = -12$. $\endgroup$ – Horse Nov 7 '16 at 4:01
  • $\begingroup$ 1 and 0 come from the fact the derivative is undefined at that point. $\endgroup$ – Future Math person Nov 7 '16 at 4:06
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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 $x=\frac {28}{27}$.

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  • $\begingroup$ Thanks. My mistake was I let y equal the cubic instead of 1/cubic. $\endgroup$ – Future Math person Nov 7 '16 at 17:07
  • $\begingroup$ @SubhashisChakraborty. You are welcome. Good to know that you found the mistake. Cheers. $\endgroup$ – Claude Leibovici Nov 7 '16 at 18:42

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