# Thread: Range of Composite Functions

1. ## Range of Composite Functions

Hi.

If:
$\displaystyle f:R\rightarrow R,\;f(x)=x^2-4$
and
$\displaystyle g:R+\cup\left\{0\right\}\rightarrow R,\;g(x)=\sqrt x$

Then what is the range of $\displaystyle f\circ g$ and why?

And what's the latex command for the little + symbol that normally goes next to R in the domain?

2. $\displaystyle fog=(g(x))^2-4=(\sqrt x)^2-4\geq 0-4\geq -4$

Thus range$\displaystyle =(-4,\infty]$

3. Thanks heaps for your reply, but I don't get that, and my text says the answer is $\displaystyle [-4,\infty)$

4. Originally Posted by Stroodle
Hi.

If:
$\displaystyle f:R\rightarrow R,\;f(x)=x^2-4$
and
$\displaystyle g:R+\cup\left\{0\right\}\rightarrow R,\;g(x)=\sqrt x$

Then what is the range of $\displaystyle f\circ g$ and why?
g(x) maps the non-negative real numbers to the set of non-negative real numbers: given any $\displaystyle y\ge 0$, take $\displaystyle x= y^2$ so y= g(x).
Now apply f(x) to such a number. If $\displaystyle x\ge 0$, then $\displaystyle x^2\ge 0$ so $\displaystyle x^2- 4\ge -4$. The range of $\displaystyle f\circ g(x)$ is all $\displaystyle x\ge -4$.

And what's the latex command for the little + symbol that normally goes next to R in the domain?
$\displaystyle R^+$ is the set of positive real numbers. Adding 0 to that set gives the set of "non-negative" real numbers. You need that for the domain of g because you cannot take the square root of a negative number.

Thanks heaps for your reply, but I don't get that, and my text says the answer is $\displaystyle [-4,\infty)$