Run a function backwards and, provided it never sends two different inputs to the same output, you get a genuine undo button.
9 min read3 worked examples
One-one vs many-one
Recall that a function maps every element of its domain to exactly one element of its range. Whether a function can be inverted comes down to one property: how many inputs share each output.
One-one vs many-one
A function is one-one (or injective) if every output comes from exactly one input — no two different $x$-values ever give the same $f(x)$. Graphically, any horizontal line crosses the graph at most once. A function is many-one if at least one output can be reached from more than one input, e.g. $f(x)=x^2$ has $f(2)=f(-2)=4$, so a horizontal line like $y=4$ crosses the graph twice.
This distinction matters because, as the next section shows, only a one-one function has a genuine inverse.
Inverse functions
The inverse of $f$, written $f^{-1}$, is the function that exactly "undoes" $f$: it reverses the mapping, sending each output of $f$ back to the input that produced it.
$f^{-1}f(x) = x \quad\text{and}\quad ff^{-1}(x) = x$
Only one-one functions have a true inverse
For $f^{-1}$ to be a genuine function, each input to $f^{-1}$ (i.e. each output of $f$) must map back to exactly one value — which is only guaranteed if $f$ itself is one-one. A many-one function like $f(x)=x^2$ ($x\in\mathbb{R}$) has no inverse as it stands, since $f^{-1}(4)$ would have to be both $2$ and $-2$. The fix is to restrict the domain of $f$ so it becomes one-one — e.g. restricting $f(x)=x^2$ to $x\ge0$ gives $f^{-1}(x)=\sqrt{x}$, a genuine function.
The domain of $f^{-1}$ equals the range of $f$, and the range of $f^{-1}$ equals the domain of $f$ — the roles of domain and range simply swap. To find $f^{-1}(x)$ algebraically: write $y=f(x)$, rearrange to make $x$ the subject, then swap $x$ and $y$.
Resulty = f⁻¹(x) is the reflection of y = f(x) in the line y = x
1
If $(a,b)$ lies on $y=f(x)$, then $f(a)=b$.
Definition of a point on the graph.
2
Since $f^{-1}$ reverses $f$, $f^{-1}(b)=a$.
Direct consequence of the inverse relationship.
3
So $(b,a)$ lies on $y=f^{-1}(x)$ — the coordinates of every point are swapped.
Reinterpreting the statement $f^{-1}(b)=a$ as a coordinate pair.
4
Swapping the coordinates $(a,b)\to(b,a)$ is exactly the geometric effect of reflecting in the line $y=x$.
Standard reflection rule for swapping x and y coordinates.
5
Therefore the entire graph of $y=f^{-1}(x)$ is the reflection of $y=f(x)$ in the line $y=x$.
True for every point, hence true for the whole curve.
▪ proven
Worked example 1
Finding an inverse function and its domain
$f(x) = \dfrac{3x+2}{x-1}$ for $x\in\mathbb{R}, x\ne1$. Find $f^{-1}(x)$ and state its domain.
MethodWrite y = f(x), rearrange to make x the subject, then swap x and y
1
Let $y = \dfrac{3x+2}{x-1}$.
Start from the defining equation.
2
$y(x-1) = 3x+2 \Rightarrow yx - y = 3x+2$.
Multiply both sides by (x − 1) to clear the fraction.
3
$yx - 3x = y+2 \Rightarrow x(y-3) = y+2$.
Collect all x-terms on one side and factorise.
4
$x = \dfrac{y+2}{y-3}$.
Make x the subject.
5
Swap $x$ and $y$: $f^{-1}(x) = \dfrac{x+2}{x-3}$.
Swapping variables reflects the roles of input and output.
6
The domain of $f^{-1}$ equals the range of $f$. Since $f$ can never output $y=3$ (that value would need $x(0)=5$, impossible), the range of $f$ — and so the domain of $f^{-1}$ — is $x\in\mathbb{R}, x\ne3$.
Domain/range swap between a function and its inverse.
▪ solved
Examiner's eye
Notice the excluded value flipped from $x\ne1$ (domain of $f$) to $x\ne3$ (domain of $f^{-1}$) — exactly as expected, since the domain of $f^{-1}$ is the range of $f$, not the domain of $f$. Always re-derive the domain of an inverse rather than copying the original one across.
Worked example 2
Restricting a domain to create an inverse · Challenge
$f(x) = (x-2)^2+3$ for $x\in\mathbb{R}$. Explain why $f$ has no inverse as defined, then find the largest domain of the form $x\ge k$ on which $f$ is one-one, and state $f^{-1}(x)$ on this restricted domain.
MethodLocate the vertex to find where the function stops being one-one, then invert on the restricted branch
1
$f(x)=(x-2)^2+3$ is a quadratic with vertex at $(2,3)$, opening upwards.
Completed-square form directly gives the vertex.
2
As a full quadratic on $\mathbb{R}$, $f$ is many-one: e.g. $f(1)=4$ and $f(3)=4$ — two different inputs share an output.
A horizontal line through the vertex region crosses the graph twice, so f fails the one-one test and has no inverse function.
3
Restricting the domain to $x\ge2$ (the vertex onward) keeps only the increasing branch, which is one-one.
The largest such restriction starts exactly at the vertex, x = 2.
4
Let $y=(x-2)^2+3 \Rightarrow y-3=(x-2)^2 \Rightarrow x-2=\pm\sqrt{y-3}$.
Rearrange to make x the subject.
5
Since $x\ge2$, take the positive root: $x = 2+\sqrt{y-3}$.
The restricted domain tells us which sign of the square root to keep.
6
Swap $x$ and $y$: $f^{-1}(x) = 2+\sqrt{x-3}$, with domain $x\ge3$ (the range of the restricted $f$).
Swap variables and state the domain as the range of the original restricted function.
▪ solved
Look deeper
This is the general strategy for inverting any many-one function on an A-level paper: locate the turning point (or point of symmetry), restrict the domain to one side of it so the function becomes one-one, then invert as normal. The restriction point is always where the derivative is zero for a smooth curve like a quadratic.
Worked example 3
Finding f⁻¹(a) directly, without finding f⁻¹(x) first
$f(x)=\dfrac{3x-1}{x+2}$ for $x\in\mathbb{R}, x\ne-2$. Find $f^{-1}(4)$.
MethodSolve f(x) = 4 directly — no need to find the full inverse first
1
$f^{-1}(4)$ is the value of $x$ for which $f(x)=4$, so solve $\dfrac{3x-1}{x+2}=4$ directly.
By definition, f⁻¹(a) is whichever x satisfies f(x) = a.
2
$3x-1 = 4(x+2) = 4x+8$.
Multiply both sides by (x + 2) to clear the fraction.
3
$3x-1 = 4x+8 \Rightarrow -9 = x$.
Collect terms and solve for x.
4
So $f^{-1}(4) = -9$.
Check: f(-9) = (-27-1)/(-9+2) = -28/-7 = 4. ✓
▪ solved
A useful shortcut
When only a single value of $f^{-1}$ is needed, solving $f(x)=a$ directly is faster than deriving the entire formula for $f^{-1}(x)$ and then substituting. (Here the full inverse is $f^{-1}(x)=\dfrac{2x+1}{3-x}$ — substituting $x=4$ gives $\dfrac{9}{-1}=-9$, matching the direct method.)
Common mistake
Assuming every function has an inverse. Always check one-one-ness first (or confirm the domain has already been suitably restricted) before attempting to find $f^{-1}$ — a many-one function has no inverse until its domain is cut down.
Check yourself
$f(x)=\dfrac{x+1}{x-2}$ for $x\in\mathbb{R}, x\ne2$. Find $f^{-1}(x)$ and state its domain. (Hint: write y = f(x), clear the fraction, collect the x-terms, then swap x and y.)
Method
1
Let $y=\dfrac{x+1}{x-2} \Rightarrow y(x-2)=x+1 \Rightarrow yx-2y=x+1$.
Clear the fraction by multiplying both sides by (x − 2).