Determine the maxima and minima of $(x-a)^{\frac{1}{3}}(2x-a)^{\frac{2}{3}}$ where $a>0$.
Doing the first derivative,we get the points $x=a,\frac{a}{2}$ where the first derivative is undefined and a defined critical point $x=\frac{5}{6}a$. But the calculation of 2nd derivative seemed to me a daunting task,so instead of differentiating again,I tried to look for the intervals where $f(x)$ is increasing or decreasing. But doing that I found that at $f'(x)$ is positive in both left and right neighbourhood of $a$,meaning it is neither maxima nor minima. But since $a$ is a critical point,it should give a sign change. Now,I am unsure of the approach. It will be very helpful if lights are shed on it.
The value of x which you get after solving the first derivative are $\frac{a}{2},\frac{5a}{6}$, for which you're supposed to get minima and maxima.
If we take $$y=\sqrt[3]{(x-a)(2x-a)^2}$$
Regarding critical points
We notice that y is 0 at $\frac{a}{2}$ and $a$. We can see that the factor $(2x-a)$ has an even power. It denotes that sign of y will not change on either sides of $\frac{a}{2}$ until it reaches $a$ on its right, although it's a critical point.
Reason for this:-
Let $x_0$ be any point before $a$ Then if $$y=\sqrt[3]{(x_0-a)(2x_0-a)^2}$$ then y is always negative because $(2x-a)^2$ is always positive, irrespective of whether $x_0>\frac{a}{2}$ or $x_0<\frac{a}{2}$. It will only become positive when $x_0$ becomes more than $a$.
As $\frac{a}{2}$ is a root, y=0 when $$x=\frac{a}{2}$$
It's a constantly increasing function after x exceeds the local minima. So there exists a local maxima at $x=\frac{a}{2}$ where $y=0$ and local minima at $x=\frac{5a}{6}$ where $y<0$ ( because $\frac{a}{2}<\frac{5a}{6}<a$)
The presence of local minima and maxima can also be determined by the signs of $$f''(x)=24x-16a$$ after plugging $x=\frac{a}{2}$ and $\frac{5a}{6}$ in it. ( Negative value for local maxima and positive for local minima)
Hence value of local maxima is 0 and local minima is $$\sqrt[3]{(\frac{5a}{6}-a)(2.\frac{5a}{6}-a)^2}= \frac {-a\sqrt[3]{16}}{6}$$