# Prove: if $f(x) =x\sin (\pi x)$ then $f'(x)$ vanishes at a unique point in $( n + 1/2, n)$

Let $f(x) = x\sin (\pi x), x > 0$. Then prove that for all natural numbers n, $f'(x)$ vanishes at a unique point in $( n + 1/2, n)$

The given solution shows a graph, but is there any algebraic method? Hints please :D

• Perhaps you mean $x\sin\pi x$? Or the interval should by from $(n+1/2)\pi$ to $n\pi$? Commented Jun 5, 2013 at 11:41
• Sorry. Just editing. Commented Jun 5, 2013 at 11:43

There is an error on your interval, but the idea is you should have $f(x)=0$ on your interval boundaries and f is differentiable, so by Rolle's theorem $f'(x) =0$ somewhere on the interval.

Another way would be to solve $f'(x) = 0$, this is equivalent to solve $u=-\tan(u)$ on each interval, where $u=\pi x$. The existence of a solution is ensured by the fact that on each interval $\tan(u)$ go from $-\infty$ to $+\infty$ and u is finite. The uniqueness is given by the strict monotony of $u+\tan(u)$. An interesting and easy exercise would be to draw both u and $-\tan(u)$ to observe the solution and their asymptotical comportement.

• Even if the interval in the question is changed in such way that $f(x)=0$ in both endpoints, this still does not guarantee uniqueness. Commented Jun 5, 2013 at 11:50
• You mean to say: "...in such way that *f'(x)*=0 in both endpoints..." ?? Commented Jun 5, 2013 at 11:54
• I didn't see uniqueness, so I proposed another method. Commented Jun 5, 2013 at 12:04

Solving by Rolle's theorem you would have to put $f(n+1/2)=f(n+1)$ and see if they are equal or not. If they are equal then there exists a unique point in the interval where $f'(x)= 0$.

After solving you wold notice $f(n+1/2)=f(n+1)=0$.

And if imorin is active i would like to know which method.