(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Show that the function [tex]f(x) = x\sin\left( \frac{\pi}{x} \right)[/tex] has an infinite number of relative extrema.

2. Relevant equations

3. The attempt at a solution

Using the product rule and chain rule to find the derivative f'(x):

[tex] f'(x) = \sin\left( \frac{\pi}{x} \right) - \frac{x\pi}{x^2} \cos\left( \frac{\pi}{x} \right) [/tex]

[tex] f'(x) = \sin\left( \frac{\pi}{x} \right) - \frac{\pi}{x} \cos\left( \frac{\pi}{x} \right) [/tex]

To try to find the relative extrema, I will determine for which values f'(x) = 0.

[tex] \sin\left( \frac{\pi}{x} \right) - \frac{\pi}{x} \cos\left( \frac{\pi}{x} \right) = 0 [/tex]

[tex] \sin\left( \frac{\pi}{x} \right) = \frac{\pi}{x} \cos\left( \frac{\pi}{x} \right) [/tex]

[tex] \frac{\sin\left( \frac{\pi}{x} \right)}{\cos\left( \frac{\pi}{x} \right)} = \frac{\pi}{x} [/tex]

[tex] \frac{\pi}{x}=\tan\left( \frac{\pi}{x} \right ) [/tex]

Now I can replace π/x with k, if k ≠ 0:

[tex] k=\tan(k) [/tex]

The solution to the equation above are the points of interception between the line y = k and the curve y = tan(k). Since there is an infinite number of interceptions, the equation above has an infinite number of solutions.

There is an infinite number of interceptions because, in the function y = tan(k), there are infinitely many vertical asymptotes, so, the line y = k will intercept the curve y = tan(k) in infinitely many points.

So, f(x) = x sin(π/x) has an infinite number of relative extrema.

Is this correct? Is there some important step missing in the middle?

Thank you in advance.

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# Calculus question involving sine and relative extrema

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