When a Composite Function Exists

Where fg can fail

fg(x) means f(g(x)): x goes into g first, and the output of g goes into f. So every output g can give must be a number f accepts.

Take $f\left(x\right) = \sqrt{x}$, whose domain is $x \ge 0$, and g(x) = x − 3, which accepts every number. At x = 1, g gives 1 − 3 = −2, and $f\left(-2\right) = \sqrt{-2}$ is not a real number. As written, fg does not exist.

The condition

fg exists exactly when the range of g lies inside the domain of f: range of $g \subseteq$ domain of f.

Only the range of g matters, not its domain: f never sees the x values, only what g makes of them.

To make fg exist here, restrict g. On $x \ge 3$, g(x) = x − 3 gives every value $y \ge 0$, and every one of those is in the domain of f.

The range of g on $x \ge 3$ is $y \ge 0$. It starts at the same point as the domain of f and runs the same way, so it lies inside it.

The domain and range of fg

With g restricted, fg$\left(x\right) = \sqrt{x - 3}$. Its domain is the domain of g, $x \ge 3$, because x goes into g first.

Its range comes out of f: as x runs from 3 upward, x − 3 runs from 0 upward, and so does its square root. The range of fg is $y \ge 0$.

The other order

gf$\left(x\right) = g\left(f\left(x\right)\right) = \sqrt{x} - 3$. Now the outputs of f go into g. The range of f is $y \ge 0$, and g accepts every number, so gf exists on the whole domain of f, $x \ge 0$.

Its range is $y \ge -3$: $\sqrt{x}$ starts at 0, and g takes 3 away. One composite can exist while the other does not, so check each order separately.

A composite that always exists

Take $f\left(x\right) = \sqrt{x}$ again and $g\left(x\right) = x^{2} + 1$. The range of g is $y \ge 1$, which lies inside $x \ge 0$, so fg$\left(x\right) = \sqrt{x^{2} + 1}$ exists for every x.

With $g\left(x\right) = x^{2} + k$, the range of g is $y \ge k$, and fg exists for $f\left(x\right) = \sqrt{x - a}$ when $k \ge a$.

The usual slips

Reading the domain of $\sqrt{x - a}$ as $x \ge -a$. The expression under the root must be at least 0, so $x - a \ge 0$ and $x \ge a$.

Checking the domain of g instead of its range. What reaches f is the outputs of g.

Mixing up the order. fg feeds g into f and needs the range of g inside the domain of f; gf needs the range of f inside the domain of g.