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given function:

$$y = x(x-1)^{\frac{1}{3}}$$

steps:

$$y'=\frac{(x^2-x)^{\frac{-2}{3}}2x-1}{3}$$

After simplifying:

$$ y' = \frac{2x-1}{(3x^2-3x)^{\frac{2}{3}}}$$

therefore

$$ x = \frac{1}{2}$$ $$x \ne 0$$

Am I right with calculus?

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    I think you have applied the product rule incorrectly. I get the stationary point as $x = 1/4$. Why don't you edit your post to include your working to find $y'$. – 2017-02-24
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    @M.Mass: Assuming the problem statement is correctly written, your first step is wrong. The equation is $$y = (x) \left((x-1)^{\frac{1}{3}}\right)$$ not $$y = \left(x(x-1)\right)^{\frac{1}{3}}$$ – 2017-02-24
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    Also, for any critical point, you need to _test_ it to see whether of not it corresponds to a local extreme point. – 2017-02-24
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    And recall: A critical point is a value of $x$ in the domain of the function such that _either_ the derivative is 0, or the derivative doesn't exist. Thus, for this function, you will get critical points at $x = 3/4%$ and $x = 1$. – 2017-02-24

2 Answers 2

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The function is $$ f(x)=x\sqrt[3]{1-x} $$ not $$ \sqrt[3]{x(1-x)} $$ at least according to common notation. Thus the derivative is $$ f'(x)=(1-x)^{1/3}+x\cdot\frac{1}{3}(1-x)^{-2/3}\cdot(-1)= \frac{1}{3(1-x)^{2/3}}(3-3x-x)=\frac{3-4x}{3(1-x)^{2/3}} $$ which vanishes at $x=3/4$ and is undefined at $x=1$.

Note that the function is increasing over $(-\infty,3/4]$ and decreasing over $[3/4,\infty)$ (the point where the function is not differentiable doesn't harm).

Alternatively, use the fact that $x\mapsto x^3$ is a continuous bijection $\mathbb{R}\to\mathbb{R}$, so the function $f$ has the same extremal points as $$ g(x)=(f(x))^3=x^3(1-x)=x^3-x^4 $$ Since $g'(x)=3x^2-4x^3=x^2(3-4x)$, we get the same conclusions as before.

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    samll, but logical mistake in my calculus, thats why I wanted someone to check me, thanks much! – 2017-02-24
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for the first derivative we get by the product rule $$f'(x)=(x-1)^{1/3}+\frac{x}{3\sqrt[3]{(x-1)^2}}$$