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.
 A: 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.
A: 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 $x=1/2$.
