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 $f\left(x\right)=\sqrt{x}$f(x) = √x, whose domain is $x\ge 0$x ≥ 0, and g(x) = x − 3, which accepts every number. At x = 1, g gives 1 − 3 = −2, and $f\left(-2\right)=\sqrt{-2}$f(−2) = √(−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 $g\subseteq$g ⊆ 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 $x\ge 3$x ≥ 3, g(x) = x − 3 gives every value $y\ge 0$y ≥ 0, and every one of those is in the domain of f.

The range of g on $x\ge 3$x ≥ 3 is $y\ge 0$y ≥ 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$\left(x\right)=\sqrt{x-3}$(x) = √(x − 3). Its domain is the domain of g, $x\ge 3$x ≥ 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 $y\ge 0$y ≥ 0.

The other order

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

Its range is $y\ge -3$y ≥ −3: $\sqrt{x}$√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 $f\left(x\right)=\sqrt{x}$f(x) = √x again and $g\left(x\right)=x^2+1$g(x) = x² + 1. The range of g is $y\ge 1$y ≥ 1, which lies inside $x\ge 0$x ≥ 0, so fg$\left(x\right)=\sqrt{x^2+1}$(x) = √(x² + 1) exists for every x.

With $g\left(x\right)=x^2+k$g(x) = x² + k, the range of g is $y\ge k$y ≥ k, and fg exists for $f\left(x\right)=\sqrt{x-a}$f(x) = √(x − a) when $k\ge a$k ≥ a.

The usual slips

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