# Find the image set of function $\frac{\sin(\pi x)}{x(1-x)}$

The function $\displaystyle \frac{\sin(\pi x)}{x(1-x)}$ and $x \in (0, 1)$. What is the image set for this one?

I thinking about finding the min and the max for this function. But it is not easy.

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Note that the function is invariant under the transformation $x\to1-x$. Since the numerator and denominator both tend to $0$ for $x\to0$ and $x\to1$, you'll need to apply L'Hôpital's rule at one of these points. –  joriki Feb 23 '12 at 3:29

Your idea of finding the minimum and maximum values is correct. To find the minimum and maximum values, we set the derivative equal to $0$: $$0=\frac{d}{dx}\frac{\sin(\pi x)}{x(1-x)}=\frac{\pi x(1-x)\cos(\pi x)-(1-2x)\sin(\pi x)}{x^2(1-x)^2}$$ which simplifies to $\pi x(1-x)\cos(\pi x)=(1-2x)\sin(\pi x)$. By inspection, we see that $\cos(\pi\frac{1}{2})=0$ and $(1-2\frac{1}{2})=0$, so this holds when $x=\frac{1}{2}$. Can you show that this is the only point at which these are equal? If so, you get three critical points for the function: $0,\frac{1}{2},1$. All points in the image will lie between the values $\frac{\sin(\pi x)}{x(1-x)}$ takes at these points, and the image will be the set of values between these three, minus the images of $0$ and $1$.

EDIT: The function is not defined at $0,1$, but the limit as it goes to these points is (use L'Hôpital's rule to evaluate it) and this should be used instead.

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In a similar vein, it's slightly inaccurate to speak of the minimum and maximum values, since the function as stated has no minimum value, only an infimum. Again, not to nitpick, just to avoid confusion in case the OP isn't already secure in using these concepts. –  joriki Feb 23 '12 at 3:34
@joriki True, and thanks for pointing that out. –  Alex Becker Feb 23 '12 at 3:48

The function is symmetric about $x=1/2$, since $\sin z$ is symmetric about $z=\pi/2$, and $x(1-x)$ is symmetric about $x=1/2$. By looking at the derivative, you can fairly easily show that the function is increasing on $(0,1/2)$ and decreasing on $(1/2,1)$. The maximum value is therefore reached at $x=1/2$. There, the value is $4$.

Now you need to find the number that our function approaches as $x$ approaches $0$ from the right, or equivalently by symmetry as $x$ approaches $1$ from the left. For this, we can use L'Hospital's Rule, and find that this limit is $\pi$. Or else we can use the fact that as $x$ approaches $0$, $\dfrac{\sin(\pi x)}{\pi x}$ approaches $1$, and that $1-x$ approaches $1$, to conclude that the limit is $\pi$.

We conclude that as $x$ ranges over the interval $(0,1)$, our function ranges over the interval $(\pi,4)$.

Remark: It is lucky that the interval we are asked about is so small, since finding a closed form expression for the absolute minimum is probably not possible. It is not hard to see that the function function reaches an absolute absolute maximum at $x=1/2$.

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