Inverse Functions

Overview

An inverse reverses a mapping. The important H2 distinction is:

• every function has a reversed relation obtained by swapping the coordinates of its graph;
• only a one-one function has an inverse function.

Use this branch with Functions and Domains, Ranges, and One-One Functions.

Inverse Relation Versus Inverse Function

Suppose $f$ maps input $x$ to output $y$.

The inverse relation reverses the ordered pairs in the graph of $f$: if $(x,y)$ lies on the graph of $f$, then $(y,x)$ lies on the inverse relation.

But this reversed relation is a function only if each output of $f$ came from exactly one input.

That is why one-one behaviour matters.

Example:

$$f(x)=x^2,x\in \mathbb{R}.$$

The output $1$ comes from both $x=1$ and $x=-1$. Reversing the graph would give both $(1,1)$ and $(1,-1)$, so the inverse relation sends one input to two outputs and is not a function.

If instead

$$g(x)=x^2,x\ge 0,$$

then each output in $[0,\infty )$ comes from exactly one input. Hence $g^{-1}$ exists.

Existence Condition

The inverse function $f^{-1}$ exists if and only if $f$ is one-one on its stated domain.

Do not write only:

$f^{-1}$ exists by the horizontal line test.

The more precise reasoning is:

• every horizontal line cuts the graph at most once;
• therefore $f$ is one-one;
• therefore the inverse relation is a function;
• hence $f^{-1}$ exists.

The horizontal line test is evidence for one-one behaviour. The actual condition is one-one-ness.

Domain-Range Swap

If $f^{-1}$ exists, then

$$D_{f^{-1}}=R_f,R_{f^{-1}}=D_f.$$

This is not a formula to memorise separately. It follows from reversing the mapping:

• the original outputs become the new inputs;
• the original inputs become the new outputs.

Example:

$$f:[0,\infty )\to \mathbb{R},f(x)=x^2.$$

Then

$$R_f=[0,\infty ).$$

So

$$f^{-1}:[0,\infty )\to [0,\infty ),f^{-1}(x)=\sqrt{x}.$$

Graph Relationship

If $f^{-1}$ exists, the graph of $y=f^{-1}(x)$ is the reflection of the graph of $y=f(x)$ in the line $y=x$.

The coordinate reason is:

$$(a,f(a))⟷(f(a),a).$$

So inverse functions swap input and output coordinates.

How to read this figure. The graph uses $f(x)=x^2$ only on $x\ge 0$, so the original function is one-one. Each point $(a,f(a))$ becomes $(f(a),a)$ on the inverse; for example, $(2,4)$ becomes $(4,2)$. The reflected curve is $f^{-1}(x)=\sqrt{x}$, whose domain is the original range $[0,\infty )$. Equal scales on both axes are required for the line $y=x$ to be a genuine visual mirror.

When sketching inverse graphs:

• use equal scale on both axes;
• reflect endpoints accurately;
• swap domain and range;
• label the line $y=x$ if it is part of the argument.

Finding the Rule of an Inverse

Before finding $f^{-1}$, first establish that $f$ is one-one on its stated domain.

Then:

1. write $y=f(x)$;
2. rearrange to make $x$ the subject;
3. replace the old output variable by the new input variable;
4. state the domain of the inverse using $D_{f^{-1}}=R_f$.

The familiar instruction “swap $x$ and $y$” is a shorthand for reversing ordered pairs. It does not by itself prove that the reversed relation is a function; one-one-ness must be established first.

Example:

$$f:\mathbb{R}\to \mathbb{R},f(x)=2x-2.$$

This function is one-one because it is linear with non-zero gradient.

Write

$$y=2x-2.$$

Make $x$ the subject:

$$x=\frac{y+2}{2}.$$

Therefore

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

Since $R_f=\mathbb{R}$,

$$D_{f^{-1}}=\mathbb{R}.$$

Branch Choice After Restricting Domain

The most common inverse-function error is choosing the wrong branch after solving a square.

Example:

$$f(x)=(x+1{)}^2+2,x\le -1.$$

First, the stated domain makes $f$ one-one.

Now find the inverse:

$$y=(x+1{)}^2+2.$$

Then

$$y-2=(x+1{)}^2.$$

So

$$x+1=\pm \sqrt{y-2}.$$

But the original domain requires

$$x\le -1,$$

so

$$x+1\le 0.$$

Therefore choose the negative branch:

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

Hence

$$f^{-1}(x)=-1-\sqrt{x-2},x\ge 2.$$

The branch is chosen from the original domain, not from preference.

Checking an inverse rule

After obtaining a candidate inverse, check both compositions on their correct domains:

$$f^{-1}(f(x))=x(x\in D_f),$$

and

$$f(f^{-1}(x))=x(x\in R_f).$$

This detects algebraic errors and incorrect branch choices. A formula that works only for some allowed inputs is not the inverse function on the stated domain.

Composition with an Inverse

If $f^{-1}$ exists, then reversing after applying $f$ gives the original input:

$$f^{-1}f(x)=x,x\in D_f.$$

Also,

$$f{f}^{-1}(x)=x,x\in R_f.$$

These two identity functions may have different domains. Therefore, as functions,

$$f^{-1}f$$

and

$$f{f}^{-1}$$

need not be the same function unless their domains are the same.

This is an anchor-note emphasis: equality of functions requires checking domain, not just formula.

Example: let

$$f:[-2,2]\to [-1,7],f(x)=2x+3.$$

Then

$$f^{-1}f(x)=x,x\in [-2,2],$$

whereas

$$f{f}^{-1}(x)=x,x\in [-1,7].$$

Both rules simplify to $x$, but the two identity functions have different domains. They are therefore not equal as functions.

Intercepts and Endpoints Under Inversion

Reflection in $y=x$ swaps coordinate roles:

• an $x$-intercept $(a,0)$ of $f$ becomes the $y$-intercept $(0,a)$ of $f^{-1}$;
• a $y$-intercept $(0,b)$ of $f$ becomes the $x$-intercept $(b,0)$ of $f^{-1}$;
• an included endpoint remains included after its coordinates are swapped;
• an excluded endpoint remains excluded.

This lets you sketch many inverse graphs without first finding an algebraic inverse rule.

Reciprocal Warning

The notation $f^{-1}$ means inverse function, not reciprocal.

In general,

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

For example, if $f(x)=2x-2$, then

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

not

$$\frac{1}{2x-2}.$$

Common Pitfalls

• Finding an inverse rule before proving that the inverse function exists.
• Saying “by horizontal line test” without explaining one-one behaviour.
• Forgetting $D_{f^{-1}}=R_f$.
• Choosing the wrong square-root branch after restricting the domain.
• Sketching inverse graphs with unequal axis scales.
• Treating $f^{-1}(x)$ as a reciprocal.
• Assuming $f^{-1}f$ and $f{f}^{-1}$ are the same without checking domains.

Related

• Functions
• Domains, Ranges, and One-One Functions
• Composite Functions
• Graphing Techniques