Composite and inverse functions

A composite function applies one function and then another. An inverse function reverses the action of a one-to-one function. The two ideas are linked: if $f^{-1}$ exists, composing it with $f$ returns the original input.

This lesson explains how to calculate composites, find inverses, choose valid domains and ranges, and avoid the notation errors that make these questions harder than they need to be.

Before you start

You should be comfortable with:

• substituting into function notation
• rearranging equations
• solving linear and quadratic equations
• identifying the domain and range of a function
• reading graphs of functions

Composite functions

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

$fg(x)=f(g(x)).$

The function nearest to $x$ acts first. You can picture the input passing through two machines:

$x\longmapsto g(x)\longmapsto f(g(x)).$

Similarly,

$gf(x)=g(f(x)),$

so $f$ acts first. In general,

$fg(x)\ne gf(x).$

Example 1: finding two composites

Let

$f(x)=3x-2,\qquad g(x)=x^2+1.$

To find $fg(x)$, substitute the whole expression $g(x)=x^2+1$ wherever $x$ appears in $f$:

$$\begin{aligned} fg(x)&=f(g(x))\\ &=3(x^2+1)-2\\ &=3x^2+1. \end{aligned}$$

For $gf(x)$, substitute $f(x)=3x-2$ into $g$:

$$\begin{aligned} gf(x)&=g(f(x))\\ &=(3x-2)^2+1\\ &=9x^2-12x+5. \end{aligned}$$

These expressions are different because changing the order changes the calculation.

Example 2: evaluating at a particular input

Using the same functions, find $fg(2)$.

One method is to use the composite found above:

$fg(2)=3(2)^2+1=13.$

You can also work from the inside out:

$g(2)=2^2+1=5,$

then

$f(5)=3(5)-2=13.$

The second method is often safer when only one numerical value is required.

Example 3: three functions

Let

$f(x)=x+4,\qquad g(x)=x^2,\qquad h(x)=\frac1x.$

Find $fgh(x)$. Read from right to left:

$$x\xmapsto{h}\frac1x \xmapsto{g}\frac1{x^2} \xmapsto{f}\frac1{x^2}+4.$$

Therefore,

$fgh(x)=\frac1{x^2}+4,\qquad x\ne0.$

The restriction $x\ne0$ comes from $h(x)=1/x$. Simplifying a composite must not make you forget restrictions inherited from the original functions.

Domains of composite functions

For $fg(x)=f(g(x))$ to be defined:

1. $x$ must lie in the domain of $g$.
2. The output $g(x)$ must lie in the domain of $f$.

Example 4: a composite with a square root

Let

$f(x)=\sqrt{x},\qquad g(x)=2x-6.$

Then

$fg(x)=\sqrt{2x-6}.$

For a real square root, its input must be non-negative:

$2x-6\ge0,$

so the domain is

$x\ge3.$

In the reverse order,

$gf(x)=2\sqrt{x}-6,$

whose domain is $x\ge0$. Again, order matters.

Example 5: a hidden excluded value

Let

$f(x)=\frac1{x-2},\qquad g(x)=x+5.$

Then

$fg(x)=\frac1{x+3}.$

The denominator cannot be zero, so $x\ne-3$. Equivalently, $g(x)$ cannot equal $2$, since $2$ is excluded from the domain of $f$.

Inverse functions

An inverse function undoes the original function. If

$f(4)=11,$

then

$f^{-1}(11)=4.$

The symbol $f^{-1}$ does not mean the reciprocal $1/f$. These are different objects:

$f^{-1}(x)\ne\frac1{f(x)}.$

For example, if $f(x)=2x+3$, then its inverse is $(x-3)/2$, while its reciprocal is $1/(2x+3)$.

The inverse relationship

Where both compositions are defined,

$f^{-1}f(x)=x$

and

$ff^{-1}(x)=x.$

The first composition applies $f$ and then undoes it. The second applies the inverse and then reverses that operation with $f$.

How to find an inverse function

Use this reliable procedure:

1. Write $y=f(x)$.
2. Rearrange the equation to make $x$ the subject.
3. Replace $y$ by $x$ and write the result as $f^{-1}(x)$.
4. State the domain of the inverse when it is relevant.

Example 6: inverse of a linear function

Let

$f(x)=5x-7.$

Start with

$y=5x-7.$

Rearrange for $x$:

$y+7=5x,$

$x=\frac{y+7}{5}.$

Therefore,

$f^{-1}(x)=\frac{x+7}{5}.$

Check by composition:

$$f\left(f^{-1}(x)\right) =5\left(\frac{x+7}{5}\right)-7 =x.$$

This check catches most rearrangement errors.

Example 7: inverse of a fractional function

Let

$f(x)=\frac{3x+1}{x-2},\qquad x\ne2.$

Write

$y=\frac{3x+1}{x-2}.$

Multiply both sides by $x-2$:

$y(x-2)=3x+1.$

Collect the terms containing $x$:

$yx-2y=3x+1,$

$yx-3x=2y+1,$

$x(y-3)=2y+1.$

Hence

$x=\frac{2y+1}{y-3},$

so

$f^{-1}(x)=\frac{2x+1}{x-3},\qquad x\ne3.$

The inverse excludes $3$ because $3$ is not in the range of the original function. Notice that the domain and range swap:

$$\operatorname{domain}(f^{-1})=\operatorname{range}(f), \qquad \operatorname{range}(f^{-1})=\operatorname{domain}(f).$$

When does an inverse exist?

A function has an inverse function only if it is one-to-one on its stated domain. This means that every output comes from exactly one input.

The horizontal line test gives a graphical check. If any horizontal line crosses the graph more than once, the function is not one-to-one and has no inverse on that domain.

For example, $f(x)=x^2$ on all real numbers has no inverse function because

$f(2)=f(-2)=4.$

The proposed inverse $\sqrt{x}$ could not know whether to return $2$ or $-2$.

Restricting a domain

We can make $f(x)=x^2$ invertible by restricting its domain.

If

$f(x)=x^2,\qquad x\ge0,$

then

$f^{-1}(x)=\sqrt{x},\qquad x\ge0.$

If instead

$f(x)=x^2,\qquad x\le0,$

then

$f^{-1}(x)=-\sqrt{x},\qquad x\ge0.$

The same formula for $f$ therefore produces different inverses on different restricted domains.

Example 8: inverse of a restricted quadratic

Let

$f(x)=(x-3)^2+5,\qquad x\ge3.$

Find $f^{-1}$.

Begin with

$y=(x-3)^2+5.$

Then

$y-5=(x-3)^2.$

Because the original domain says $x\ge3$, we know $x-3\ge0$. We therefore take the positive square root:

$x-3=\sqrt{y-5},$

$x=3+\sqrt{y-5}.$

Thus

$f^{-1}(x)=3+\sqrt{x-5},\qquad x\ge5.$

The inverse domain $x\ge5$ is the range of $f$.

If the original restriction had been $x\le3$, the inverse would instead be

$f^{-1}(x)=3-\sqrt{x-5}.$

Graphs of inverse functions

The points on a function and its inverse have their coordinates exchanged:

$f(a)=b\quad\Longleftrightarrow\quad f^{-1}(b)=a.$

So if $(a,b)$ lies on $y=f(x)$, then $(b,a)$ lies on $y=f^{-1}(x)$. Exchanging coordinates reflects a graph in the line

$y=x.$

This explains why the domain and range swap. It also provides a visual check: the graphs of a function and its inverse should be mirror images in $y=x$. See transformations of graphs for related graph techniques.

Solving equations using composites and inverses

Example 9: solve a composite equation

Let

$f(x)=2x+1,\qquad g(x)=x^2-4.$

Solve $fg(x)=19$.

First form the composite:

$fg(x)=f(x^2-4)=2(x^2-4)+1=2x^2-7.$

Now solve:

$2x^2-7=19,$

$2x^2=26,$

$x^2=13,$

$x=\pm\sqrt{13}.$

Both answers are valid unless the domain has been restricted.

Example 10: use an inverse to solve an equation

Let $f(x)=x^3+4$. Solve $f(x)=31$.

The inverse is

$f^{-1}(x)=\sqrt[3]{x-4}.$

Apply $f^{-1}$ to both sides:

$x=f^{-1}(31)=\sqrt[3]{31-4}=3.$

This is the function version of performing inverse operations on an equation.

Example 11: find an unknown function

Suppose

$f(x)=2x-5$

and

$fg(x)=6x+7.$

Find $g(x)$.

Since $fg(x)=f(g(x))$, applying $f$ to $g(x)$ gives

$2g(x)-5=6x+7.$

Therefore,

$2g(x)=6x+12,$

so

$g(x)=3x+6.$

Alternatively, apply $f^{-1}$ to the composite:

$g(x)=f^{-1}(6x+7)=\frac{(6x+7)+5}{2}=3x+6.$

Common misconceptions

• Reading $fg(x)$ from left to right. The inside function acts first, so $fg(x)=f(g(x))$.
• Treating $f^{-1}(x)$ as $1/f(x)$. An inverse reverses a function; a reciprocal divides $1$ by its output.
• Replacing only part of an expression. When substituting $g(x)$ into $f$, use brackets around the entire expression before simplifying.
• Assuming every function has an inverse. A function must be one-to-one on its stated domain.
• Writing $\pm$ automatically when inverting a quadratic. The domain restriction determines which square-root branch is valid.
• Forgetting domains after simplifying. Restrictions from denominators, roots and the original function still apply.
• Checking only $ff^{-1}(x)$. When domains are delicate, check both $ff^{-1}(x)$ and $f^{-1}f(x)$ on their appropriate domains.

Quick checks

1. Calculate composites

Given

$f(x)=x-3,\qquad g(x)=2x^2,$

find $fg(x)$ and $gf(x)$.

Answer

$fg(x)=f(2x^2)=2x^2-3.$

$gf(x)=g(x-3)=2(x-3)^2=2x^2-12x+18.$

2. Find an inverse

Find the inverse of

$f(x)=\frac{x+4}{3}.$

Answer

From $y=(x+4)/3$,

$3y=x+4\quad\Longrightarrow\quad x=3y-4.$

Therefore,

$f^{-1}(x)=3x-4.$

3. Use a restricted domain

The function

$f(x)=x^2-6x+11$

has domain $x\ge3$. Find $f^{-1}(x)$ and state its domain.

Answer

Complete the square:

$f(x)=(x-3)^2+2.$

Since $x\ge3$, take the positive square root when reversing the function:

$y=(x-3)^2+2,$

$x=3+\sqrt{y-2}.$

Hence

$f^{-1}(x)=3+\sqrt{x-2},\qquad x\ge2.$

4. Check a domain

Let

$f(x)=\frac1x,\qquad g(x)=x-4.$

Find $fg(x)$ and state its domain.

Answer

$fg(x)=f(x-4)=\frac1{x-4}.$

The input to $f$ cannot be zero, so $x-4\ne0$. The domain is therefore $x\ne4$.

What to learn next

Composite and inverse functions support later work with function transformations, exponential functions, and inverse trigonometric functions. Before moving on, make sure you can explain why composition order matters and why a domain restriction may be necessary before an inverse exists.