Calculus question involving sine and relative extrema

[pc2-brazil]

Homework Statement

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

Homework Equations

The Attempt at a Solution

Using the product rule and chain rule to find the derivative f'(x):
f'(x) = \sin\left( \frac{\pi}{x} \right) - \frac{x\pi}{x^2} \cos\left( \frac{\pi}{x} \right)
f'(x) = \sin\left( \frac{\pi}{x} \right) - \frac{\pi}{x} \cos\left( \frac{\pi}{x} \right)

To try to find the relative extrema, I will determine for which values f'(x) = 0.
\sin\left( \frac{\pi}{x} \right) - \frac{\pi}{x} \cos\left( \frac{\pi}{x} \right) = 0
\sin\left( \frac{\pi}{x} \right) = \frac{\pi}{x} \cos\left( \frac{\pi}{x} \right)
\frac{\sin\left( \frac{\pi}{x} \right)}{\cos\left( \frac{\pi}{x} \right)} = \frac{\pi}{x}
\frac{\pi}{x}=\tan\left( \frac{\pi}{x} \right )
Now I can replace π/x with k, if k ≠ 0:
k=\tan(k)
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.

 

[micromass]

That is ok, but I have one annoying point.

That f^\prime(x)=0 does NOT mean that there is a relative extremum at x. Indeed, look at the function f(x)=x^3. This has derivative 0 at 0, but it is no extremum there.

In order to be a relative extremum, you additionaly need f^{\prime\prime}(x)\neq 0.

 

[abr_pr90]

To know that, try plotting k=tan(k) and f(x)=x sin(x/pi) in the same graph.

the sin function gives you the maximum value when sin(pi/x)=1--> pi/x=(2n+1)*pi/2 for n= 0,1,...

So 1/x= n+1/2--> x=2/(2n+1)

(2n+1)pi/2= tan(pi*(2n+1)/2)For small k the equation tank=k is an identity

 

[micromass]

For small k the equation tank=k is an identity

No it is never an identity for small k (only for k=0). That it is a good approximation does NOT mean that it is equal.

 

[pc2-brazil]

That is ok, but I have one annoying point.

That f^\prime(x)=0 does NOT mean that there is a relative extremum at x. Indeed, look at the function f(x)=x^3. This has derivative 0 at 0, but it is no extremum there.

In order to be a relative extremum, you additionaly need f^{\prime\prime}(x)\neq 0.

That's true, the points where f''(x) = 0 cannot be relative extrema. In order to be a relative extremum f''(x) has to be nonzero. If f''(a) > 0, then the function is concave upwards in f(a), and, if f''(a) < 0, it's concave downwards in f(a). In f(x) = x³, f(0) is an inflection point, not a relative extremum, since f''(0) = 0 and f''(x) changes sign in x = 0.
Calculating f''(x) using product rule and chain rule:
f&#039;&#039;(x)=-\frac{\pi\cos\left(\frac{\pi}{x}\right)}{x^2}-\pi \left( -\frac{\cos\left(\frac{\pi}{x}\right)}{x^2}+\frac{\pi\sin\left(\frac{\pi}{x}\right)}{x^3} \right)
f&#039;&#039;(x)=-\frac{\pi^2\sin\left(\frac{\pi}{x}\right)}{x^3}
The expression above will be zero if and only if \sin\left(\frac{\pi}{x}\right) = 0. This will be true if and only if \frac{\pi}{x} = n\pi, where n is any nonzero integer (nonzero because \frac{\pi}{x} \neq 0).
Since if \sin\left(\frac{\pi}{x}\right) = 0, then \tan\left(\frac{\pi}{x}\right) = 0, the values for which f''(x) = 0 can't be solutions to k = tan(k) where k = π/x. Therefore, all solutions to k = tan(k) are points in which f&#039;&#039;(x) \neq 0. Thus, they are relative extrema.

 

[micromass]

This is correct! You solved that very neatly!