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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

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

2 Answers 2

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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.

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  • $\begingroup$ 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. $\endgroup$ Jun 5, 2013 at 11:50
  • $\begingroup$ You mean to say: "...in such way that *f'(x)*=0 in both endpoints..." ?? $\endgroup$
    – dajoker
    Jun 5, 2013 at 11:54
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    $\begingroup$ I didn't see uniqueness, so I proposed another method. $\endgroup$ Jun 5, 2013 at 12:04
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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.

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