-1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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). $\endgroup$ – drhab Dec 25 '17 at 14:01
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $x\mapsto\sqrt x$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.