1.3 Inverse Functions

Condition for Existence

An inverse function $f^{- 1}$ "undoes" the action of $f$ . If $f (x) = y$ , then $f^{- 1} (y) = x$ .

However, for this to work unambiguously, the function must be a One-to-One mapping. If multiple inputs give the same output (Many-to-One), we wouldn't know which input to go back to!

The Horizontal Line Test

A graph represents a one-to-one function if no horizontal line intersects the graph more than once.
Example: $y = x^{2}$ (Parabola) fails this test, so it has no inverse (unless we restrict the domain to $x \geq 0$ ).

Domain and Range Relationship

When we seek the inverse, the inputs and outputs swap roles. This leads to a fundamental property:

Function	Domain (Inputs)	Range (Outputs)
$f (x)$	Set $A$	Set $B$
$f^{- 1} (x)$	Set $B$	Set $A$

Notice how the Domain and Range swap positions.

Finding $f^{- 1} (x)$ Algebraically

To find the formula for an inverse function, follow these four steps:

1Write the function as

y = f (x)

2Interchange (swap)

x

and

y

3Rearrange the equation to make

y

the subject

4Replace

y

with

f^{- 1} (x)

💡 Tip for Quadratics

If $f (x)$ involves a quadratic like $x^{2} - 4 x$ , you must complete the square first (e.g., $(x - 2)^{2} - 4$ ) so that $x$ appears only once. This makes it possible to rearrange for $x$ .

Graphical Relationship

The graphs of $y = f (x)$ and $y = f^{- 1} (x)$ are reflections of each other in the line $y = x$ .

The line $y = x$ acts as a mirror.
Any point $(a, b)$ on $f$ becomes $(b, a)$ on $f^{- 1}$ .
Domain and range swap visually on the axes.
If graphs intersect, they (usually) intersect on the line $y = x$ .

Worked Examples

Find the inverse of the function

f (x) = \frac{2 x + 3}{x - 1}

, where

x \neq 1

. State the domain and range of

f^{- 1} (x)

1. Set y = f(x):

y = \frac{2 x + 3}{x - 1}

2. Swap x and y:

x = \frac{2 y + 3}{y - 1}

3. Rearrange to make y the subject:

Multiply by $ (y-1) $:

x (y - 1) = 2 y + 3

Expand:

x y - x = 2 y + 3

Group y terms:

x y - 2 y = x + 3

Factorize y:

y (x - 2) = x + 3

Divide:

y = \frac{x + 3}{x - 2}

4. Write in inverse notation:

f^{- 1} (x) = \frac{x + 3}{x - 2}

Domain and Range:

Domain of f^{- 1} = Range of f

For

f (x)

, as

x \to \infty

y \to 2

. So Range of

f

y \neq 2

Thus,

Domain of f^{- 1} : x \in R, x \neq 2

Range of f^{- 1} = Domain of f = y \in R, y \neq 1

The function

g (x) = (x - 3)^{2} + 2

is defined for the domain

x \geq 3

. Find an expression for

g^{- 1} (x)

and sketch the graphs of

g

and

g^{- 1}

on the same axes.

Finding the inverse:

Let

y = (x - 3)^{2} + 2

Swap variables:

x = (y - 3)^{2} + 2

Rearrange:

x - 2 = (y - 3)^{2}

Square root:

y - 3 = \pm \sqrt{x - 2}

Since domain of

g

x \geq 3

, the range of

g^{- 1}

must be

y \geq 3

. We take the positive root.

y = 3 + \sqrt{x - 2}

So,

g^{- 1} (x) = 3 + \sqrt{x - 2}

Graph Sketching:

The graph shows

g (x)

(blue) reflected in

y = x

to give

g^{- 1} (x)

(pink). Note how the vertex

(3, 2)

reflects to

(2, 3)

Functions

f

and

g

are defined by

f (x) = 3 x - 1

and

g (x) = \frac{2}{x}

x \neq 0

. Find the value of

x

for which

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

Step 1: Find $f^{- 1} (x)$

Let

y = 3 x - 1 ⟹ y + 1 = 3 x ⟹ x = \frac{y + 1}{3}

f^{- 1} (x) = \frac{x + 1}{3}

Step 2: Equate to $g (x)$

\frac{x + 1}{3} = \frac{2}{x}

Cross multiply:

x (x + 1) = 6

x^{2} + x - 6 = 0

Step 3: Solve Quadratic

(x + 3) (x - 2) = 0

x = - 3 or x = 2

The function

h (x) = 4 - e^{2 x}

is defined for

x \in R

. Find

h^{- 1} (x)

and state its domain.

Let

y = 4 - e^{2 x}

Swap:

x = 4 - e^{2 y}

Isolate exponential:

e^{2 y} = 4 - x

Take ln:

2 y = \ln (4 - x)

y = \frac{1}{2} \ln (4 - x)

h^{- 1} (x) = \frac{1}{2} \ln (4 - x)

Domain:

For ln to be defined, argument must be positive:

4 - x > 0

implies

x < 4

Domain of

h^{- 1}

x < 4

Given

f (x) = \ln (x + 2)

for

x > - 2

, explain why

f

has an inverse and find

f^{- 1} (3)

Existence:

The function

f (x) = \ln (x + 2)

is a strictly increasing function (derivative is

1 / (x + 2)

which is always positive for

x > - 2

). Therefore, it is a one-to-one function, so the inverse exists.

Finding value:

We could find the inverse function first, or solve directly:

Let

f^{- 1} (3) = a

. Then

f (a) = 3

\ln (a + 2) = 3

Exponentiate:

a + 2 = e^{3}

a = e^{3} - 2

f^{- 1} (3) = e^{3} - 2 \approx 18.086

The function

k (x) = 1 + \sqrt{x}

for

x \geq 0

. Write down the range of

k

and find an expression for

k^{- 1} (x)

Range of k:

x \geq 0

\sqrt{x} \geq 0

, so

1 + \sqrt{x} \geq 1

Range:

y \geq 1

Inverse:

Let

y = 1 + \sqrt{x}

. Swap:

x = 1 + \sqrt{y}

\sqrt{y} = x - 1

Square both sides:

y = (x - 1)^{2}

Important: Domain of

k^{- 1}

is Range of

k

, so

x \geq 1

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

The function

f (x) = 2 x^{2} - 8 x + 5

is defined for

x > 2

. (i) Express in form

a (x - b)^{2} + c

. (ii) Find

f^{- 1} (x)

(i) Completing the Square:

Factor out 2:

2 (x^{2} - 4 x) + 5

2 [(x - 2)^{2} - 4] + 5

2 (x - 2)^{2} - 8 + 5 = 2 (x - 2)^{2} - 3

Vertex is at

(2, - 3)

(ii) Inverse:

Let

y = 2 (x - 2)^{2} - 3

. Swap:

x = 2 (y - 2)^{2} - 3

x + 3 = 2 (y - 2)^{2}

(y - 2)^{2} = \frac{x + 3}{2}

y - 2 = \pm \sqrt{\frac{x + 3}{2}}

Since domain

x > 2

, range of inverse

y > 2

. We take positive root.

f^{- 1} (x) = 2 + \sqrt{\frac{x + 3}{2}}

The function

g (x) = \ln (2 x + 1)

is defined for

x > - 0.5

. (i) Find

g^{- 1} (x)

. (ii) Solve

g^{- 1} (x) = 4

(i) Finding Inverse:

Let

y = \ln (2 x + 1)

. Swap:

x = \ln (2 y + 1)

Exponentiate both sides:

e^{x} = 2 y + 1

2 y = e^{x} - 1

g^{- 1} (x) = \frac{e^{x} - 1}{2}

(ii) Solving Equation:

\frac{e^{x} - 1}{2} = 4

e^{x} - 1 = 8 ⟹ e^{x} = 9

x = \ln (9) \approx 2.197

Let

h (x) = e^{x} + 1

for

x \in R

. Sketch

h (x)

and

h^{- 1} (x)

on the same axes, showing intercepts.

Analysis:

h (x)

is an exponential growth shifted up by 1. Horizontal asymptote

y = 1

. Y-intercept

(0, 2)

h^{- 1} (x)

acts as a Logarithmic function. Vertical asymptote

x = 1

. X-intercept

(2, 0)

Inverse Formula:

x = e^{y} + 1 ⟹ e^{y} = x - 1 ⟹ h^{- 1} (x) = \ln (x - 1)

10.

Functions

f (x) = x - 4

and

g (x) = x^{2}

(domain

x \in R

). Find

(f g)^{- 1} (5)

. Explain why

(g f)^{- 1}

does not exist.

Part 1: Find $(f g)^{- 1} (5)$

First find

f g (x) = f (g (x)) = f (x^{2}) = x^{2} - 4

Let

y = x^{2} - 4

. Swap:

x = y^{2} - 4

Wait!

f g (x)

is a parabola (Many-to-One). Inverse does not exist generally.

However, if we restrict domain, or check the specific value:

It asks for the pre-image of 5.

x^{2} - 4 = 5 ⟹ x^{2} = 9 ⟹ x = \pm 3

Since there are two answers, the inverse function is not well-defined without a domain restriction.

Part 2: Why $(g f)^{- 1}$ does not exist

g f (x) = g (x - 4) = (x - 4)^{2}

This is a parabola with vertex at

(4, 0)

. It is Many-to-One.

Therefore, it does not have an inverse.

11.

The function

f (x) = \frac{x}{x + 2}

x \neq - 2

. (i) Find

f^{- 1} (x)

. (ii) Find values of

x

such that

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

(i) Finding Inverse:

Let

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

. Swap:

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

x (y + 2) = y

x y + 2 x = y

2 x = y - x y

2 x = y (1 - x)

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

(ii) Solving $f (x) = f^{- 1} (x)$ :

\frac{x}{x + 2} = \frac{2 x}{1 - x}

x (1 - x) = 2 x (x + 2)

x - x^{2} = 2 x^{2} + 4 x

3 x^{2} + 3 x = 0

3 x (x + 1) = 0

x = 0

x = - 1

12.

Given

f (x) = 2 - \sqrt{x - 1}

for

x \geq 1

. properties. Sketch

f

and

f^{- 1}

Analysis:

Domain

x \geq 1

. Range

y \leq 2

(since square root is positive).

Inverse:

Let

y = 2 - \sqrt{x - 1}

. Swap:

x = 2 - \sqrt{y - 1}

\sqrt{y - 1} = 2 - x

y - 1 = (2 - x)^{2}

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

Domain of Inverse is Range of f:

x \leq 2

1.3 Inverse Functions

Condition for Existence

The Horizontal Line Test

Domain and Range Relationship

Finding f−1(x) Algebraically

💡 Tip for Quadratics

Graphical Relationship

Worked Examples

Finding $f^{- 1} (x)$ Algebraically