Determine the the points where the function $\sin^n x \sin nx$ is maximum or minimum.
Differentiating,we get that the critical points turn out to be $\frac{k\pi}{n+1}$. But from this how can the points at minima or maxima be differentiated when we derived only one form as critical points? Does that mean the function will always be either at maxima or either at minima at these points?
You're overlooking an important point. (I have assumed n is positive and non zero)
For $$ x= \frac{kπ}{n+1} $$ The value of k ranges as 1,2,3,...
When k is odd, it generates a local maxima and when k is even, it generates a local minima.
It can be proven by finding out the second order derivative:- $$f''(x)= cos (x+nx)(1+n) $$
Plugging the value of x in it gives $cos(kπ)(1+n)$.
If k is odd, $cos(kπ)=-1$
If k is even, $cos(kπ)=1$
So if k is odd, $f''(x)<0 $ which denotes $f(\frac{kπ}{n+1})$= value of local maxima
And if k is even, $f''(x)>0 $ which denotes $f(\frac{kπ}{n+1})$= value of local minima