Q.If functions f : R → R and g : R → R are defined by f(x) = x/2 and g(x) = 2x, then gof(x) is -

Domain of a Composite Function

Picture a building with two doors: the first opens with a blue pass, the second with a red pass. Walking through $f(g(x))$ means passing the inner door $g$ first, then the outer door $f$. The domain of the composite is simply: which inputs make it through both doors?

The intuition

Take $f(x) = \sqrt{x}$ and $g(x) = x - 5$, so $f(g(x)) = \sqrt{x-5}$.

• The inner $g(x) = x - 5$ accepts every real number.
• The outer $f$ accepts only non-negative inputs.

So the real question is: which $x$ make $g(x)$ land inside the domain of $f$? The domain of the composite is not just the domain of $g$, nor just the domain of $f$ — it is the overlap seen through $g$.

The precise statement

Two steps, in order:

1. Keep only the $x$ that $g$ can handle.
2. Among those, keep only the $x$ for which $g(x)$ is something $f$ can handle.

A worked check

For $f(x) = \sqrt{x}$, $g(x) = \dfrac{1}{x-1}$:

• Domain of $g$: $x \ne 1$.
• Outer condition: $g(x) \ge 0 \Rightarrow \dfrac{1}{x-1} \ge 0 \Rightarrow x - 1 > 0 \Rightarrow x > 1$.

So $\text{Domain}(f\circ g) = (1, \infty)$ — the condition from $f$ already excludes $x = 1$.

Order matters …