Composite Functions

One function after another

Take two functions: f(x) = x + 3, which adds 3, and g(x) = 2x, which doubles. The output of one can be used as the input of the other. A function made this way, by applying one function and then another to its output, is called a composite function.

gf(x) means g(f(x)): first apply f to x, then apply g to the result. The function written nearest to the x acts first, just as the inside bracket of g(f(x)) is worked out first. So a composite is read from right to left.

gf means f first

Work out gf(4). The f is nearer to the 4, so f acts first: f(4) = 4 + 3 = 7. Then g acts on that output: g(7) = 2 × 7 = 14. So gf(4) = g(f(4)) = g(7) = 14.

For gf(4), the 4 goes through f first, which adds 3 to make 7, and then through g, which doubles 7 to make 14.

The other order

fg(4) means f(g(4)), so g acts first this time: g(4) = 2 × 4 = 8, and then f(8) = 8 + 3 = 11. The same two functions, applied in the other order, give 11 instead of 14.

The two orders do different things to the 3 that f adds. In gf, the 3 is added first and then doubled along with everything else, so it becomes 6. In fg, the doubling is already over when the 3 is added, so it stays 3.

For fg(4), the 4 goes through g first, which doubles it to 8, and then through f, which adds 3 to make 11.

The composite as a formula

To write gf(x) as a formula, put the whole of f(x) in as the input of g. Since g doubles its input, gf(x) = g(x + 3) = 2(x + 3) = 2x + 6. Check: gf(4) = 2 × 4 + 6 = 14.

For fg(x), put g(x) in as the input of f. Since f adds 3 to its input, fg(x) = f(2x) = 2x + 3. Check: fg(4) = 2 × 4 + 3 = 11.

So gf(x) = 2x + 6 and fg(x) = 2x + 3. For every x, gf(x) is 3 more than fg(x), so the two composites never agree: fg and gf are different functions, not just different at x = 4.

The order can change more than a number. Take f(x) = x + 3 and $h\left(x\right)=x^2$h(x) = x². Then hf$\left(x\right)=h\left(x+3\right)={\left(x+3\right)}^2$(x) = h(x + 3) = (x + 3)², which adds 3 and then squares, while fh$\left(x\right)=f\left(x^2\right)=x^2+3$(x) = f(x²) = x² + 3, which squares and then adds 3. At x = 1, hf$\left(1\right)=4^2=16$(1) = 4² = 16 but fh(1) = 1 + 3 = 4.

Each output must be an input the next function accepts

In a composite, the outputs of the first function become the inputs of the second, so they must lie in the second function's domain. Take $f\left(x\right)=\sqrt{x}$f(x) = √x, whose domain is $x\ge 0$x ≥ 0, and g(x) = x − 5. Then fg$\left(x\right)=f\left(x-5\right)=\sqrt{x-5}$(x) = f(x − 5) = √(x − 5).

Try fg(3). First g(3) = 3 − 5 = −2. That is a perfectly good output of g, but it is not an input f accepts, because −2 has no real square root. So fg(3) has no value. fg(x) has a value only when g(x) = x − 5 is 0 or more, which means $x\ge 5$x ≥ 5: the domain of fg is $x\ge 5$x ≥ 5, even though g on its own accepts every number.

In the other order, gf$\left(x\right)=g\left(\sqrt{x}\right)=\sqrt{x}-5$(x) = g(√x) = √x − 5. Here f acts first, so the input must be 0 or more, and then any output of f can go into g. The domain of gf is $x\ge 0$x ≥ 0.

For $f\left(x\right)=\sqrt{x}$f(x) = √x and g(x) = x − 5, the input 3 goes through g to give −2, and −2 is not in the domain of f, so fg(3) has no value.

A function applied twice

A function can be composed with itself. For f(x) = x + 3, ff(x) = f(x + 3) = (x + 3) + 3 = x + 6: adding 3 twice adds 6. For g(x) = 2x, gg(x) = g(2x) = 2 × 2x = 4x: doubling twice multiplies by 4.

Some books write ff(x) as $f^2\left(x\right)$f²(x). There the 2 means "apply f twice", not "square the output": for g(x) = 2x, $g^2\left(3\right)=$g²(3) = gg(3) = 12, while ${\left(g\left(3\right)\right)}^2=6^2=36$(g(3))² = 6² = 36.

Solving an equation in a composite

Which input x gives gf(x) = 20, for f(x) = x + 3 and g(x) = 2x? Write the composite as a formula first, gf(x) = 2x + 6, then solve 2x + 6 = 20: subtract 6 from both sides to get 2x = 14, so x = 7. Check through the machines: f(7) = 10, and g(10) = 20.

The usual mistakes

Reading the composite from left to right. gf(4) is not "g first": the f is next to the 4, so f acts first, and gf(4) = g(7) = 14, not f(8) = 11.

Using only one of the functions. 2 × 4 = 8 is only g; for gf(4) the adding of 3 has to happen as well, and before the doubling.

Multiplying the two rules together. fg(x) is one function put inside the other, not the product f(x) × g(x). For f(x) = x + 3 and g(x) = 2x, fg(x) = 2x + 3, while the product is 2x(x + 3).