Invertible Functions & Finding the Inverse

MATInverses

A function is invertible exactly when it is a bijection, and its inverse $f^{-1}$ undoes $f$ : if $y=f(x)$ then $x=f^{-1}(y)$ . ISC problems give a function on a restricted domain/codomain (so it becomes a bijection) and ask you to prove invertibility and find $f^{-1}$ explicitly.

The reliable method is to set $y=f(x)$ , solve for $x$ in terms of $y$ , and present $f^{-1}$ — taking care with the domain when a square root forces a sign choice.

Inverse relationship

y = f(x) \iff x = f^{-1}(y)

Defined only when

f

is a bijection; swap the roles of input and output.

Inverse cancels the function

(f^{-1} \circ f)(x) = x, \quad (f \circ f^{-1})(y) = y

These two identities are the standard verification that a candidate is the inverse.

Invertibility criterion

f^{-1} : B \to A \ \text{exists} \iff f \ \text{is a bijection}

f:A \to B

must be one-one and onto for

f^{-1}

to exist.

Rational example

f(x) = \dfrac{2x+1}{3x+4}, \quad f^{-1}(y) = \dfrac{4y-1}{2-3y}

f:\mathbb{R}\setminus\{-\tfrac{4}{3}\} \to \mathbb{R}\setminus\{\tfrac{2}{3}\}

; cross-multiply and solve for

x

Completed-square quadratic

f(x) = x^2 - 6x + 10 = (x-3)^2 + 1, \quad f^{-1}(y) = 3 + \sqrt{y-1}

f:[3,\infty) \to [1,\infty)

; the restricted domain forces the

+

root so

x \ge 3

Method: write $y=f(x)$ , solve algebraically for $x$ , then rewrite as $f^{-1}(y)$ (optionally rename $y$ as $x$ ).
Domain of $f^{-1}$ equals the range of $f$ , and range of $f^{-1}$ equals the domain of $f$ — state these to complete the answer.
For quadratics, complete the square first; e.g. $f(x)=x^2+4$ on $[0,\infty)$ gives $f^{-1}(y)=\sqrt{y-4}$ with domain $[4,\infty)$ .
When a square root appears, the restricted domain dictates which sign to take: domain $[3,\infty)$ forces $x=3+\sqrt{y-1}$ , not the minus root.
For the bill $C(x)=20x+150$ , the inverse is $C^{-1}(y)=\dfrac{y-150}{20}$ , so a bill of $450$ gives $C^{-1}(450)=15$ GB.
Confirm invertibility by either proving $f$ is a bijection or verifying $(f^{-1} \circ f)(x)=x$ and $(f \circ f^{-1})(y)=y$ .
The graph of $f^{-1}$ is the reflection of the graph of $f$ in the line $y=x$ .

Where the marks go

Writing $f^{-1}(x)=\dfrac{1}{f(x)}$ — the inverse function is NOT the reciprocal.
Choosing the wrong sign of the square root for a quadratic inverse, ignoring the restricted domain that fixes it.
Forgetting to state the domain of $f^{-1}$ (= range of $f$ ), which is required for a complete ISC answer.
Claiming a function is invertible without first establishing it is a bijection on the given domain and codomain.

How the board asks it

Numericalthe method: set $y=f(x)$ , solve for $x$ , rewrite as $f^{-1}$
If $f:\mathbb{R}\to\mathbb{R}$ is defined by $f(x)=\dfrac{3x+2}{4}$ , find its inverse function $f^{-1}(x)$ .
Derive / proveinvertibility criterion (bijection)
Show that the function $f:\mathbb{R}\to\mathbb{R}$ given by $f(x)=4x+3$ is invertible, and hence find $f^{-1}$ .
Numericalcompleted-square quadratic on a restricted domain with sign choice
Let $f:[0,\infty)\to[4,\infty)$ be defined by $f(x)=x^2+4$ . Prove that $f$ is invertible and find $f^{-1}$ , stating its domain.
Applicationlinear inverse model
A broadband bill is $C(x)=20x+150$ rupees for $x$ GB of data. Find $C^{-1}(y)$ and use it to determine the data used when the bill is $450$ rupees.
Diagram / graphgraph of $f^{-1}$ as reflection in $y=x$
Given $f(x)=2x-1$ on $\mathbb{R}$ , draw the graphs of $f$ and $f^{-1}$ on the same axes and state the line of symmetry relating them.

Practise this topic

PreviousComposition of Functions

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.