Domain of a composite function by using intersections

Question

Can we find the domain of a composite function $f \circ g$ by simplifying $f \circ g$ (using identities or any rewriting rule) and then taking the intersection of the natural domain of it with $g$? Note that some simplifications change the domain of functions. Example: $$f(x)= x^2,\qquad g(x)=\sqrt x,$$ $$\operatorname{Domain}(f)=\mathbb R,\;\operatorname{Domain}(g)=\mathbb R^{\geq0},$$ $$\Rightarrow(f \circ g)(x)=(\sqrt x)^2=x,$$which has a natural domain of $\mathbb R$ after simplification. Therefore, by our method, the domain of the composite is $$\mathbb R \cap \mathbb R^{\geq0}= \mathbb R^{\geq0}$$ which can be verified using the formal definition of domain of composite functions, namely: $$\operatorname{Domain}(f\circ g)=\{x\in \operatorname{Domain}(g)\vert \,g(x)\in \operatorname{Domain}(f)\}$$ The method is usually faster than using the formal definition, and it has worked in every case I tried. But I'm not sure if it always works. Please provide a formal proof.

Answer 1

As you said, it is true that it works in most of the cases but not all. Actually, taking the intersection of domain of both the functions is correct but not sufficient. You should add one more point in that.

$\bullet$ Find those inputs, $x$, in the domain of $g$ for which $g(x)$ is in the domain of $f$. That is, exclude those inputs, $x$, from the domain of gf or which $g(x)$ is not in the domain of $f$. The resulting set is the domain of $f∘g$.

In your example, only taking intersection of domain of both the functions worked. But it will not in case like

Let $f(x) =\sqrt{x}$ and $g(x)=x-2$.

Here, the intersection of domain of both the functions comes out to be $[0,\infty]$. But, if you take into consideration the point which I mentioned then,

$$f\circ g=\sqrt{x-2}$$

The domain for the square root is all inputs that make $x – 2$ non-negative. That is, all $x$ such that $x – 2 \ge 0$. Solving this for $x$ gives the domain as $[2, \infty]$

You can also check this link.