MHT-CET Maths · Sets, Relations and Functions

Inverse Functions — Finding f⁻¹ and Solving f(x) = f⁻¹(x)

Swap the roles: set y = f(x), solve for x in terms of y, and rename — a linear-fractional (ax + b)/(cx + d) inverts to (dx − b)/(−cx + a); for an increasing f, f(x) = f⁻¹(x) reduces to f(x) = x.

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Why this matters

9 PYQs at 11% HARD. Of the seven set from 2021 to 2025, four were 'find f⁻¹' for a linear-fractional, a quadratic-type or a composite function, two used the self-inverse condition f∘f = x, and one solved f(x) = f⁻¹(x). The HARD one adds two inverses and solves a quadratic. The method never changes; the checks that matter are the branch of a square root and the domain the inverse must land in.

Concept 1 of 4: Inverse of (ax + b)/(cx + d): Solve for x, and the Swap-and-Negate Shortcut

Write

y = \dfrac{ax + b}{cx + d}

, cross-multiply, gather the

x

terms, divide. The result is always

\dfrac{dy - b}{-cy + a}

: swap

a

and

d

, negate

b

and

c

Definition

$f(x) = \dfrac{2x - 3}{3x - 4}$ : $3xy - 4y = 2x - 3 \Rightarrow x(3y - 2) = 4y - 3 \Rightarrow f^{-1}(x) = \dfrac{4x - 3}{3x - 2}$ .
Shortcut check: $a = 2, b = -3, c = 3, d = -4$ : $\dfrac{dx - b}{-cx + a} = \dfrac{-4x + 3}{-3x + 2} = \dfrac{4x - 3}{3x - 2}$ . Same.
$f(x) = \dfrac{x - 3}{x - 2}$ : $f^{-1}(x) = \dfrac{2x - 3}{x - 1}$ . $g(x) = 3x - 2$ : $g^{-1}(x) = \dfrac{x + 2}{3}$ . Their sum $= \dfrac{19}{6}$ gives $2x^2 - 5x - 3 = 0$ , roots $-\tfrac12$ and $3$ , sum $\tfrac52$ .
The inverse's domain excludes $x = \dfrac{a}{c}$ , the value $f$ never takes — the same number as the onto exception on the function-types page.

Linear-fractional inverse

f(x) = \frac{ax + b}{cx + d} \ \Rightarrow\ f^{-1}(x) = \frac{dx - b}{-cx + a}

Worked example

Find

f^{-1}(x)

for

f(x) = \dfrac{5x + 1}{2x - 7}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 14th May Shift 2 · Q107Moderate

Example 1 · Sets, Relations and Functions · Inverse Functions — Finding f⁻¹ and Solving f(x) = f⁻¹(x)

f(x) = \frac{2x-3}{3x-4}, x \neq \frac{4}{3}

, then

f^{-1}(x)

Taking the reciprocal

f^{-1}

is not

\dfrac1f

\dfrac{3x - 4}{2x - 3}

-type options are the reciprocal; the inverse swaps and negates the coefficients instead.

Drill 1 more on inverse of (ax + b)/(cx + d): solve for x, and the swap-and-negate shortcut

Concept 2 of 4: Self-Inverse: f(f(x)) = x Fixes the Parameter

A function equal to its own inverse satisfies

f(f(x)) = x

. For

\dfrac{ax + b}{cx + d}

that is exactly

a + d = 0

— compose once and compare coefficients.

Definition

$f(x) = \dfrac{a - x}{a + x}$ : $f(f(x)) = \dfrac{a - \frac{a - x}{a + x}}{a + \frac{a - x}{a + x}} = \dfrac{a^2 + ax - a + x}{a^2 + ax + a - x}$ . Equal to $x$ for all $x$ forces $a = 1$ ; then $f(x) = \dfrac{1 - x}{1 + x}$ and $f\!\left(-\tfrac15\right) = \dfrac{6/5}{4/5} = \dfrac32$ .
$f(x) = \dfrac{\alpha x}{x + 1}$ is self-inverse iff $\alpha = -1$ (composite page).
Test: $\dfrac{ax + b}{cx + d}$ is an involution iff $a = -d$ (or $f$ is the identity). $\dfrac{x + 4}{x - 1}$ : yes; $\dfrac{2x - 3}{3x - 4}$ : no.
Two functions that are inverses of each other satisfy $f(g(x)) = g(f(x)) = x$ , which is how $\dfrac{3x + 4}{5x - 7}$ and $\dfrac{7x + 4}{5x - 3}$ commute.

Involution test

f(f(x)) = x \ \iff\ f = f^{-1} \qquad \frac{ax + b}{cx + d} \text{ self-inverse} \iff a + d = 0

Worked example

For what value of

k

f(x) = \dfrac{3x + 2}{5x + k}

its own inverse?

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 1 · Q128Moderate

Example 2 · Sets, Relations and Functions · Inverse Functions — Finding f⁻¹ and Solving f(x) = f⁻¹(x)

For a suitable chosen real constant

a

, let a function

f: \mathbb{R} \setminus \{-a\} \to \mathbb{R}

be defined by

f(x) = \dfrac{a-x}{a+x}

. Further suppose that for any real number

x \ne -a

and

f(x) \ne -a

(f \circ f)(x) = x

. Then

f\left(-\dfrac{1}{5}\right)

is equal to

Solving f(x) = x instead of f(f(x)) = x

Fixed points of

f

are not the same as

f

being an involution. The parameter comes from the identity

f(f(x)) = x

holding for EVERY

x

Concept 3 of 4: Inverse With a Square Root: Choose the Branch From the Domain

When solving

y = f(x)

for

x

produces a

\pm

, the stated domain of

f

picks the sign.

f(x) = x + \dfrac1x

[1, \infty)

keeps the '+' root; a composite

(f \circ g)^{-1}

is found by inverting the composite formula, not by composing inverses in the wrong order.

Definition

$y = x + \dfrac1x \Rightarrow x^2 - yx + 1 = 0 \Rightarrow x = \dfrac{y \pm \sqrt{y^2 - 4}}{2}$ ; for $x \ge 1$ take '+': $f^{-1}(x) = \dfrac{x + \sqrt{x^2 - 4}}{2}$ .
$(f \circ g)(x) = 3x^2 + 7$ (with $f(x) = 3x + 10$ , $g(x) = x^2 - 1$ ): $x = \left(\dfrac{y - 7}{3}\right)^{1/2}$ , so $(f \circ g)^{-1}(x) = \left(\dfrac{x - 7}{3}\right)^{1/2}$ on $x \ge 0$ .
$(f \circ g)(x) = 2x^3 + 7$ : $(f \circ g)^{-1}(-9)$ solves $2x^3 + 7 = -9$ , $x^3 = -8$ , $x = -2$ .
$(f \circ g)^{-1} = g^{-1} \circ f^{-1}$ : the order reverses. Inverting the composed formula directly avoids the reversal error.

Order of inverses

(f \circ g)^{-1} = g^{-1} \circ f^{-1}

Worked example

f : [0, \infty) \to [3, \infty)

f(x) = x^2 + 3

, find

f^{-1}(x)

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 11th May Shift 2 · Q106Moderate

Example 3 · Sets, Relations and Functions · Inverse Functions — Finding f⁻¹ and Solving f(x) = f⁻¹(x)

f:[1,\infty)\to[2,\infty)

is given by

f(x) = x+\frac{1}{x}

then

f^{-1}(x)

equals

Keeping the '−' branch

\dfrac{x - \sqrt{x^2 - 4}}{2}

is option (C), and it is at most

1

for

x \ge 2

— it lands OUTSIDE the domain

[1, \infty)

except at the endpoint. The domain chooses the sign.

Drill 2 more on inverse with a square root: choose the branch from the domain

Concept 4 of 4: Solving f(x) = f⁻¹(x): For an Increasing f, Solve f(x) = x

The graphs of

f

and

f^{-1}

are reflections in

y = x

; for an increasing

f

they can only meet ON that line, so

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

reduces to

f(x) = x

Definition

$f(x) = (x + 1)^2 - 1$ , $x \ge -1$ , is increasing. $f(x) = x \Rightarrow x^2 + x = 0 \Rightarrow x \in \{0, -1\}$ .
Why the reduction is legitimate: if $f(x) = f^{-1}(x) = y$ with $y > x$ , increasing $f$ gives $f(y) > f(x) = y$ but $f(y) = f(f^{-1}(x)) = x < y$ , a contradiction; likewise $y < x$ .
The option with complex numbers $\dfrac{-3 \pm i\sqrt3}{2}$ comes from solving $f(x) = f^{-1}(x)$ by brute force, $(x + 1)^2 - 1 = \sqrt{x + 1} - 1$ , and squaring; those roots are not real and $x \ge -1$ is required.
For a DECREASING $f$ the reduction fails ( $f(x) = \dfrac1x$ meets its inverse everywhere); check monotonicity first.

Fixed points

f \text{ increasing}:\quad f(x) = f^{-1}(x) \iff f(x) = x

Worked example

f(x) = x^2 + 2x

x \ge -1

, find all

x

with

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

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 15th May Shift 1 · Q126Moderate

Example 4 · Sets, Relations and Functions · Inverse Functions — Finding f⁻¹ and Solving f(x) = f⁻¹(x)

Let

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

, then the set

\{x: f(x)=f^{-1}(x)\}

Including the complex roots

Option (C) lists two non-real numbers alongside

0

and

-1

. The set is real-valued and domain-restricted; the fixed-point equation

x^2 + x = 0

has exactly the two real solutions.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

Inverse of (ax + b)/(cx + d): Solve for x, and the Swap-and-Negate Shortcut
Linear-fractional inverse
$f(x) = \frac{ax + b}{cx + d} \ \Rightarrow\ f^{-1}(x) = \frac{dx - b}{-cx + a}$
Self-Inverse: f(f(x)) = x Fixes the Parameter
Involution test
$f(f(x)) = x \ \iff\ f = f^{-1} \qquad \frac{ax + b}{cx + d} \text{ self-inverse} \iff a + d = 0$
Inverse With a Square Root: Choose the Branch From the Domain
Order of inverses
$(f \circ g)^{-1} = g^{-1} \circ f^{-1}$
Solving f(x) = f⁻¹(x): For an Increasing f, Solve f(x) = x
Fixed points
$f \text{ increasing}:\quad f(x) = f^{-1}(x) \iff f(x) = x$

Watch out for (4)

Taking the reciprocal
→ Inverse of (ax + b)/(cx + d): Solve for x, and the Swap-and-Negate Shortcut
Solving f(x) = x instead of f(f(x)) = x
→ Self-Inverse: f(f(x)) = x Fixes the Parameter
Keeping the '−' branch
→ Inverse With a Square Root: Choose the Branch From the Domain
Including the complex roots
→ Solving f(x) = f⁻¹(x): For an Increasing f, Solve f(x) = x

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Related notes

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