Would I be correct in assuming that the domain of the composite function $(f∘g)(x)$ is the intersection of the range of $g$ and the domain of $x$? If not, how do I find the domain of a composite function?
Would I find the range of $(f∘g)$ as I would find the range of a non-composite function, or do I need to use the components $g$ and $f$?
P.S.: Are $D_f$ and $R_f$ accepted notations for the domain and range of $f$ respectively?
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$\begingroup$NOTE: I'm going to use the more precise term "image" instead of "range" because I've heard people use range to mean either image OR codomain and I'd prefer to be unambiguous.
Given two functions $f: X \to Y$ and $g: Y \to Z$, the composition $g\circ f$ has domain $X$, codomain $Z$, and the image (range) is given by the image of the restriction of $g$ to the image of $f$.
Example: Let $f: \Bbb R \to \Bbb R$ be given by $f(x)=x^2$ and let $g: \Bbb R \to \Bbb R$ be given by $g(x) = 2x+3$. Then the composition $g\circ f$ is given by $(g\circ f)(x)=g(f(x)) = g(x^2) = 2x^2+3$. Because the domain of $f$ is $\Bbb R$, the domain of $g\circ f$ is $\Bbb R$. Because the codomain of $g$ is $\Bbb R$, the codomain of $g\circ f$ is $\Bbb R$. With that we can write down the signature of the function as $g\circ f: \Bbb R \to \Bbb R$.
But the image of the function takes a little thought to figure out. We need to find the image of the restriction of $g$ to the image of $f$. So first, what's the image of $f$? We can see that it is the nonnegative real numbers $\Bbb R_{\ge 0}$. So then what is the image of the function $\left.g\right|_{\operatorname{Im}(f)}: \Bbb R_{\ge 0} \to \Bbb R$? I'm not sure how to show it without some calculus, but hopefully it's pretty clear that it's the interval $[3, \infty)$. So that is the image of $g\circ f$.
So, to sum up: $$\begin{align}\operatorname{Dom}(g\circ f) &= \Bbb R \\ \operatorname{Codom}(g\circ f) &= \Bbb R \\ \operatorname{Im}(g\circ f) &= [3,\infty)\end{align}$$
As to your question about notation, I haven't seen that notation before (for whatever that's worth), but as long as you define it somewhere on whatever paper/ homework you're using it in, it should be fine.
$\endgroup$ 5 $\begingroup$The domain of f(g(x)) consists of all x in the domain of g such that g(x) is in the domain of f.
( See Calculus by Briggs and Cochran 1st Edition page 3)
I think this means that the domain of f(g(x) is the intersection of the domain of g and the inverse image of the range of g that is also in the domain of f.
Consider f(g(x)= $\sqrt{sin x}$.
If g=sin x, the domain of g is all reals and the range is [-1,1]. If f(x)=$\sqrt{x}$, the domain of f is all non-negative reals. The elements of the range of g that are also in the domain of f is [0,1]. The inverse image of [0,1] under the sine function that intersects the domain of g is the inverse image of [0,1] itself under the sine function and that is the domain of f(g(x), i.e. the union of the intervals
2(n-1)n$\pi\le$x$\le$(2n-1)$\pi$, -2n$\pi\le$x$\le$ -(2n-1)$\pi$
n=1,2,3...
The domain of a composite function $f(g(x))$ is the set of those inputs $x$ in the domain of $g$ for which $ g(x)$ is in the domain of $f$.
Think of it this way, In order to find domain of any function the value inside it must be permissible i.e for a $y=g(x)$ to be a function; for every $x$, $g$ should yield 1 value of $y$.
for a $y=f(g(x))$ to be a function; for every $g(x)$ , $f$ should yield a $y$. And for a $g(x)$ to be a function; for every $x$, $g$ should yield a $y'$(a value). so, $x$, $g(x)$ will restrict domains.
for a $ y=h(f(g(x)))$ to be a function; for every $f(g(x))$, $h$ should yield a $y$. And if $f(g(x))$ is to be a function; for every $g(x)$, $f$ should yield a $y'$(a value). And for a $g(x)$ to be a function; for every $x$, $g$ should yield a $y''$(a value). so, $x$, $g(x)$, $f(x)$ will all restrict domains.
This is how I understand it.
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