1
$\begingroup$

Given the function $f(x) = x^2$ with the domain $[0, \infty)$ and $g(x) = \sin(x)$ with domain $(- \infty, \infty)$. What are the domain and range of $f(g(x))$ and $(g(f(x))$?

I start the problem by finding both $f(g(x))$ and $g(f(x))$. It appears that:

$g(f(x)) = \sin(x^2)$ and $f(g(x)) = (\sin(x))^2$

First, I consider the composition function $f(g(x))$:

We know that $Dom(g)$ is given by $(- \infty, \infty)$, and the $Ran(g)$ is $ [-1,1].$ We also know that $Dom(f)$ is given by $[0,\infty)$, and $Ran(f) = [0,\infty)$.

I realize that $Ran(g)$ is not a subset of $Dom(f)$. So, this does not exist (?)

For the second one, I realize that $Ran(f) \subset Dom(g),$ hence, the domain for $g(f(x)) = [0,\infty)$. Since $\sin(x)$ oscillates, then the range for $g(f(x)) = [-1,1]$.

Am i right? Thank you in advance.

$\endgroup$
0
$\begingroup$

This answer, unlike my previous answer, assumes that $f(x)=x^2$ is the restricted function with domain $[0,\infty)$.

You are right about $g(f(x))$: the domain is $\mathbb R=(-\infty,\infty)$ and its range is $[-1,1]$.

As for $f(g(x))$, it is defined whenever $g(x)\ge 0$, so in contrast with what you wrote it does sometimes exist. Since $g(x)=\sin(x)$ the condition holds when $x$ is in the first or second quadrant, i.e. for $x\in[2k\pi,(2k+1)\pi]$ for some $k\in\mathbb Z$, or we could say for $\operatorname{frac}\left(\frac x{2\pi}\right)\le\frac 12$.

This restricted domain does not change the range of $f(g(x))$ since for every $x$ where $\sin(x)<0$ we can use $-x$ where $f(g(x))$ is defined and the values are equal. So the range of $f(g(x))$ is $[0,1]$, just as for the unrestricted $f$. enter image description here

$\endgroup$
1
$\begingroup$

You are wrong about the domain of $f$.

$f(x)=x^2$ has the domain $(-\infty,\infty)$. Therefore $Ran(g)$ is actually a subset of $Dom(f)$.

The domains of $f(g(x))$ and $g(f(x))$ are both $(-\infty,\infty)$: i.e. $\mathbb R$, all real numbers.

The range of $f(g(x))=(\sin(x))^2$ is $[0,1]$. The zero is attained at $x=0$ and the one at $x=\frac{\pi}2$. The continuity of the functions guarantees that all intermediate values are also attained.

The range of $g(f(x))=\sin(x^2)$ is $[-1,1]$. The minus-one is attained at $x=\sqrt{\frac{3\pi}2}$ and the one at $x=\sqrt{\frac\pi 2}$. The continuity of the functions guarantees that all intermediate values are also attained.

enter image description here

$\endgroup$
  • $\begingroup$ actually not that the domain is wrong, but the question may be referring to "restricted domain" $\endgroup$ – Joshua Feb 23 '15 at 1:13
  • $\begingroup$ If so, the question was badly worded. I have added another answer for that definition of $f(x)$. $\endgroup$ – Rory Daulton Feb 23 '15 at 1:43

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.