# finding the domain

Suppose that a function $f$ has domain $(-2,2)$ and range $(-3,5)$. What is the domain of the function $f(\sqrt{x})$?

I'm having a bit of trouble trying to find the new function's domain. I set up $-2 < \sqrt{x} < 2$ and got $(-\inf,4)U(4, \inf)$, but that isn't right. Could someone give me a hint?

• You should first find the domain of function $\sqrt\cdot$. After that you must find for which values $x$ in that domain $\sqrt x$ is an element of the domain of $f$ (so that composition $f\circ\sqrt{}$ is defined). – drhab Dec 25 '17 at 14:01

The domain of $g=f\left(\sqrt{x}\right)$ is the domain of $\sqrt{x}$, which is $E=[0,\;+\infty)$ intersection the interval where $\sqrt{x}\in(-2,2)$ that is $D=[0,4)$

So $\text{dom }g=D\cap E=[0,\;4)$

Hope this can be useful

If $-2 < \sqrt{x} < 2$ we get that $x \geq 0$ and $x < 4$.

The first follows from the definition of the square root (otherwise it isn't defined in $\mathbb{R}$

The second property you should try to verify for yourself.

You should first find the domain of function $\sqrt\cdot$ which is $[0,\infty)$.

However not for every $x\in[0,\infty)$ we have $\sqrt x\in(-2,2)=\mathsf{dom} f$ so the composition $f\circ\sqrt{}$ is not properly defined.

To repair that you must work with a restriction of the function $\sqrt:[0,\infty)\to\mathbb R$.

$\sqrt x\in(-2,2)$ demands that $x\in[0,4)$ and the set $[0,4)$ can be labeled as the domain of $f\circ g$ where $g$ is the function $g:[0,4)\to(-2,2)$ prescribed by $x\mapsto\sqrt x$.