TITLE: Find the image set of function $\frac{\sin(\pi x)}{x(1-x)}$
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: 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 $x=1/2$.