Resources For Teachers For Tutors For Students & Parents Pricing
Year 12 Maths - Methods (Unit 3 & Unit 4) Algebra and equations

Inverse functions and existence conditions

20 practice questions 0 video lessons Theory + worked examples
Create a free accountTrack your progress and save your work as you go.
Create free account

Theory

The inverse of a function \(f\) reverses its rule, sending each output back to the input it came from, so that \(f\big(f^{-1}(x)\big)=x\). An inverse function exists only when \(f\) is one-to-one (it passes the horizontal-line test); its graph is the reflection of \(f\) in the line \(y=x\), with domain and range swapped. The rule is found by swapping \(x\) and \(y\) and solving for \(y\).

The inverse of a function \(f\), written \(f^{-1}\), undoes what \(f\) does: if \(f\) maps \(a\) to \(b\), then \(f^{-1}\) maps \(b\) back to \(a\). Applying one after the other returns the starting value, so \(f\big(f^{-1}(x)\big)=x\) and \(f^{-1}\big(f(x)\big)=x\). Because inputs and outputs are exchanged, the domain and range swap: \(\operatorname{dom}(f^{-1})=\operatorname{ran}(f)\) and \(\operatorname{ran}(f^{-1})=\operatorname{dom}(f)\).

An inverse function exists only if \(f\) is one-to-one — no two different inputs share the same output. Graphically this is the horizontal-line test: if every horizontal line meets the graph at most once, \(f\) is one-to-one and \(f^{-1}\) is itself a function. A many-to-one function, such as \(y=x^{2}\), fails the test, so it has no inverse function on its full domain.

When a function is many-to-one, we can restrict the domain to a piece on which it is one-to-one — for \(y=x^{2}\), taking \(x\ge 0\) gives a one-to-one branch with inverse \(y=\sqrt{x}\). The graph of \(f^{-1}\) is the reflection of \(f\) in the line \(y=x\): each point \((a,b)\) on \(f\) becomes \((b,a)\) on \(f^{-1}\).

Key idea. \(f^{-1}\) exists as a function precisely when \(f\) is one-to-one. Then \(f\big(f^{-1}(x)\big)=x\), the graphs reflect in \(y=x\), and \(\operatorname{dom}(f^{-1})=\operatorname{ran}(f)\).
Inverse as reflection in y=xThe curve y=e to the x and its inverse y=ln x are mirror images in the dashed line y=x; the point (0,1) on the graph of f maps to (1,0) on the graph of the inverse. x y y=eˣ y=ln x y=x (0,1) (1,0)
\(y=e^{x}\) and its inverse \(y=\ln x\) reflect in \(y=x\); \((0,1)\) maps to \((1,0)\)
Restricted x squared and its inverseThe parabola y=x squared restricted to x greater than or equal to 0 is one-to-one; its inverse y=root x is the reflection of that branch in the line y=x. x y y=x² y=√x y=x
\(y=x^{2}\) restricted to \(x\ge 0\) is one-to-one; its inverse is \(y=\sqrt{x}\)

Composing a function with its inverse returns the input — this is the defining property:

\[f\big(f^{-1}(x)\big)=x \qquad f^{-1}\big(f(x)\big)=x\]
f(f1(x))=x

Reflecting in \(y=x\) swaps coordinates, and the domain and range exchange:

\[(a,b)\in f \iff (b,a)\in f^{-1}\]
(a,b)f(b,a)f1
\[\operatorname{dom}(f^{-1})=\operatorname{ran}(f) \qquad \operatorname{ran}(f^{-1})=\operatorname{dom}(f)\]
dom(f1)=ran(f)

To find the rule, exchange \(x\) and \(y\) in \(y=f(x)\) and solve for \(y\):

\[y=f(x)\ \longrightarrow\ x=f(y)\ \longrightarrow\ y=f^{-1}(x)\]
x=f(y)y=f1(x)
Existence condition. \(f^{-1}\) is a function only if \(f\) is one-to-one (passes the horizontal-line test). If \(f\) is many-to-one, restrict the domain first, then \(\operatorname{dom}(f^{-1})=\operatorname{ran}(f)\).

How to find an inverse function

  1. Check it is one-to-one. Apply the horizontal-line test. If \(f\) is many-to-one, restrict the domain to a one-to-one piece before continuing.
  2. Write \(y=f(x)\). Replace \(f(x)\) with \(y\) so both variables are visible.
  3. Swap \(x\) and \(y\). This performs the reflection in \(y=x\); the equation becomes \(x=f(y)\).
  4. Solve for \(y\). Make \(y\) the subject; the result is \(f^{-1}(x)\).
  5. State the domain. Set \(\operatorname{dom}(f^{-1})=\operatorname{ran}(f)\) — the inverse only accepts values \(f\) actually outputs.
Check your answer. Verify with \(f\big(f^{-1}(x)\big)=x\): substituting the inverse into \(f\) should collapse to \(x\). If it does not, re-check the algebra.
Example 1 — inverse of a linear function
Find the inverse of \(f(x)=2x-1\), and verify \(f\big(f^{-1}(x)\big)=x\).
Solution
Write \(y=f(x)\), then swap \(x\) and \(y\):
\(y\)\(=\)\(2x-1\)
\(x\)\(=\)\(2y-1\)
Solve for \(y\):
\(x+1\)\(=\)\(2y\)
\(y\)\(=\)\(\dfrac{x+1}{2}\)
Check: substitute into \(f\):
\(f\big(f^{-1}(x)\big)\)\(=\)\(2\!\left(\dfrac{x+1}{2}\right)-1\)
\(=\)\(x\ \checkmark\)
\(\therefore\) \(f^{-1}(x)=\dfrac{x+1}{2}\)
A linear function and its inverseThe line y=2x-1 and its inverse y=(x+1)/2 are reflections of each other in the line y=x. x y y=2x−1 y=½(x+1) y=x
f1(x)=x+12
Example 2 — restricting the domain
Explain why \(f(x)=x^{2}\) has no inverse function on \(\mathbb{R}\). Restrict the domain to make it one-to-one, then find the inverse.
Solution
Existence — apply the horizontal-line test:

On \(\mathbb{R}\), \(f\) is many-to-one (e.g. \(f(2)=f(-2)=4\)), so a horizontal line can cut the graph twice. No inverse function exists there.

Restrict to a one-to-one branch \(x\ge 0\); its range is \(y\ge 0\):
\(y\)\(=\)\(x^{2},\ x\ge 0\)
\(x\)\(=\)\(y^{2}\)
Solve for \(y\), taking the non-negative root:
\(y\)\(=\)\(\sqrt{x}\)
\(\therefore\) \(f^{-1}(x)=\sqrt{x}\), with \(\operatorname{dom}(f^{-1})=[0,\infty)\)
f1(x)=x
Example 3 — square root with a shift
Find the inverse of \(f(x)=\sqrt{x-1}+2\), \(x\ge 1\), and state its domain.
Solution
Find the range of \(f\) (it becomes the domain of \(f^{-1}\)):

\(\sqrt{x-1}\ge 0\), so \(f(x)\ge 2\); the range of \(f\) is \([2,\infty)\).

Write \(y=f(x)\) and swap \(x\) and \(y\):
\(y\)\(=\)\(\sqrt{x-1}+2\)
\(x\)\(=\)\(\sqrt{y-1}+2\)
Solve for \(y\) — isolate the root, then square:
\(x-2\)\(=\)\(\sqrt{y-1}\)
\((x-2)^{2}\)\(=\)\(y-1\)
\(y\)\(=\)\((x-2)^{2}+1\)
\(\therefore\) \(f^{-1}(x)=(x-2)^{2}+1\), \(x\ge 2\)
f1(x)=(x2)2+1
Example 4 — inverse of a rational function
Find the inverse of \(f(x)=\dfrac{2x-3}{x+1}\).
Solution
Write \(y=f(x)\) and swap \(x\) and \(y\):
\(y\)\(=\)\(\dfrac{2x-3}{x+1}\)
\(x\)\(=\)\(\dfrac{2y-3}{y+1}\)
Clear the fraction and collect the \(y\) terms:
\(x(y+1)\)\(=\)\(2y-3\)
\(xy+x\)\(=\)\(2y-3\)
\(xy-2y\)\(=\)\(-3-x\)
\(y(x-2)\)\(=\)\(-(x+3)\)
Divide to make \(y\) the subject:
\(y\)\(=\)\(\dfrac{x+3}{2-x}\)
\(\therefore\) \(f^{-1}(x)=\dfrac{x+3}{2-x}\)
f1(x)=x+32x

Common pitfalls

\(f^{-1}\) is not \(\dfrac{1}{f}\). The \(-1\) is inverse-function notation, not an index. \(f^{-1}(x)\) undoes \(f\); it is generally not the reciprocal \(\dfrac{1}{f(x)}\).
Do not skip the one-to-one check. A many-to-one function has no inverse function until the domain is restricted. Swapping \(x\) and \(y\) on \(y=x^{2}\) alone gives \(x=y^{2}\), which is not a function without the condition \(y\ge 0\).
Do not forget to swap domain and range. The inverse only accepts the values \(f\) outputs, so \(\operatorname{dom}(f^{-1})=\operatorname{ran}(f)\). Stating the rule without its domain is an incomplete answer.

Frequently asked questions

What is an inverse function?

It reverses \(f\): if \(f\) maps \(a\) to \(b\), then \(f^{-1}\) maps \(b\) back to \(a\). Applying one after the other returns the start, so \(f\big(f^{-1}(x)\big)=x\) and \(f^{-1}\big(f(x)\big)=x\).

When does a function have an inverse function?

Exactly when it is one-to-one — no two inputs share an output. Use the horizontal-line test: if every horizontal line meets the graph at most once, an inverse function exists.

How do you find the rule of an inverse function?

Write \(y=f(x)\), swap \(x\) and \(y\), then solve for \(y\). For \(f(x)=2x-1\) this gives \(x=2y-1\), so \(f^{-1}(x)=\dfrac{x+1}{2}\).

How is the graph of the inverse related to the graph of f?

\(f^{-1}\) is the reflection of \(f\) in the line \(y=x\). Each point \((a,b)\) on \(f\) becomes \((b,a)\) on \(f^{-1}\), so the graphs are mirror images across that line.

Why do you sometimes restrict the domain?

A many-to-one function such as \(f(x)=x^{2}\) fails the horizontal-line test and has no inverse function. Restricting the domain, e.g. to \(x\ge 0\), keeps a one-to-one piece that does have an inverse.

Is f inverse the same as one over f?

No. \(f^{-1}\) is the inverse function, not the reciprocal. The \(-1\) is not an index, so \(f^{-1}(x)\) is generally different from \(\dfrac{1}{f(x)}\).