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

Question

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

Answer 1

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.