P3 1.2 - Composite and inverse functions

Composite function

The composite $fg$ means “apply $g$ first, then $f$”:

An input $x$ is allowed only if it belongs to the domain of $g$ and its intermediate output $g(x)$ belongs to the domain of $f$.

Worked example 1: a restriction at the second stage

Question. Let $f(x)=\sqrt{x}$, with domain $x\ge0$, and $g(x)=2x-6$, with domain $\mathbb R$. Find $fg(x)$ and its domain.

Plan. Apply the linear rule first and require its output to be an allowed square-root input.

The first stage accepts every real input. The second requires

Answer. $fg(x)=\sqrt{2x-6}$, with domain $[3,\infty)$.

Check. At $x=5$, $g$ gives $4$, then $f$ gives $2$. The composite formula also gives $\sqrt4=2$. At $x=2$, the intermediate value is $-2$, so the composition fails.

Worked example 2: reversing the order

Question. For the same functions, find $gf(x)$ and its domain.

Plan. Apply the square root first, then the linear rule.

The input must satisfy $x\ge0$. The output $\sqrt{x}$ is always an allowed input for $g$, which accepts every real number.

Answer. $gf(x)=2\sqrt{x}-6$, with domain $[0,\infty)$.

Check. At $x=4$, $f$ gives $2$, then $g$ gives $-2$. Negative final outputs are allowed: the square root has already been taken.

Watch out

$fg(x)$ is neither $f(x)g(x)$ nor $g(f(x))$. Write the nested form $f(g(x))$ before substituting to keep the order clear.

Inverse function

For a one-one function $f$, its inverse $f^{-1}$ reverses every input-output pair. The domain of $f^{-1}$ is the range of $f$, and its range is the domain of $f$. On the appropriate domains,

The superscript $-1$ denotes an inverse function, not a reciprocal.

Worked example 3: an inverse with an excluded input

Question. Find $f^{-1}$ for $f(x)=\dfrac{3x+1}{x-2}$, with domain $x\ne2$.

Plan. Collect the terms containing the original input before dividing, then determine the range of $f$.

Output $y=3$ is impossible: it would require $3x-6=3x+1$. Every $y\ne3$ gives an input from the final rule, and that input is not $2$, since $2y+1=2(y-3)$ is impossible.

Answer. $f^{-1}(x)=\dfrac{2x+1}{x-3}$, with domain $x\ne3$. The unique recovered input also establishes that $f$ is one-one.

Check. $f(0)=-1/2$, and $f^{-1}(-1/2)=0$. Also $f^{-1}(1)=-3/2$, and $f(-3/2)=1$: the rules undo one another in both directions.

Worked example 4: retaining the inverse domain

Question. Find $g^{-1}$ for $g(x)=\dfrac{3x-4}{x-2}$, with domain $x\ne2$.

Plan. Reverse the mapping by solving $y=g(x)$ for $x$.

Again $y=3$ would give a contradiction, now $-6=-4$. For any $y\ne3$, the recovered input cannot equal $2$.

Answer. $g^{-1}(x)=\dfrac{2x-4}{x-3}$, with domain $x\ne3$.

Check. $g(0)=2$ and $g^{-1}(2)=0$. Conversely $g^{-1}(4)=4$, and $g(4)=4$.

Watch out

For $f(x)=2x+3$, the inverse is $(x-3)/2$, while the reciprocal is $1/(2x+3)$. Only the inverse reverses the input-output pair $4\mapsto11$.