Domain of composite function $\left( f \circ g \right)\left( x \right).$

Question

Question:

Given that $f\left( x \right) = \sqrt{x - 3}$ and $g\left( x \right) = x + 1$, find the domain of $\left( f \circ g \right)\left( x \right)$.

My attempt:

$\left( f \circ g \right)\left( x \right) = \sqrt{x - 2}$, hence the domain of $\left( f \circ g \right)\left( x \right)$ is $\{x\mid x \geq 2\}$.

However, I also aware that a composite function can only exist over a domain where both component functions exist. This implies that the domain of $\left( f \circ g \right)\left( x \right)$ is $\{x\mid x \geq 3\}$.

Could someone please explain the second solution to me because it is perplexing.

Answer 1

The natural domain of $g(x) = x + 1$ is the set of all real numbers. Hence, the only restriction domain of the composite function $$(f \circ g)(x) = f(g(x)) = f(x + 1) = \sqrt{x + 1 - 3} = \sqrt{x - 2}$$ is that $x + 1 \geq 3$. Thus, the domain of $f \circ g$ is $[2, \infty) = \{x \in \mathbb{R} \mid x \geq 2\}$. Note that we require that $x + 1 \geq 3$ since we have replaced $x$ by $g(x) = x + 1$ when we formed the composite function $f \circ g$.

Answer 2

Your solution is correct. However your statement in the second part that composition function always exist over a common domain is not correct. That is the case when we define sum, difference or product of two real functions.