Composite and Inverse Functions

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# Composite and Inverse Functions ## 🎯 Learning Objectives By the end of this topic, you will be able to: 1. **Compute** the composition $f \circ g$ of two functions 2.

Composite and Inverse Functions

🎯 Learning Objectives

By the end of this topic, you will be able to:

Compute the composition $f \circ g$ of two functions
Determine the domain of composite functions
Test whether a function has an inverse using the horizontal line test
Find the inverse of a function algebraically
Verify inverse relationships using $f(f^{-1}(x)) = f^{-1}(f(x)) = x$

📋 Prerequisites

Functions — domain, range, injective, surjective, bijective
Exponential Functions — exponentials as injective functions

📖 Core Content

7.1 Composite Functions

7.1.1 Intuition: Function Chains

A composite function is applying one function, then another. Like a factory assembly line: raw material goes through machine

g

, then its output goes through machine

f

. If

g

doubles a number and

f

squares it, then

f(g(x))

means: first double, then square.

g(f(x))

means: first square, then double — a different result!

💡 Why this matters: Composition is how we build complex transformations from simple ones. In ML, a neural network is a composition of linear transformations and activation functions.

7.1.2 Definition

The composition of

f

with

g

is:

(f \circ g)(x) = f(g(x))

Domain:

\{x \mid x \in \text{domain}(g) \text{ AND } g(x) \in \text{domain}(f)\}

Key: Apply

g

first, then

f

to the result. (Diagram)

7.1.3 Worked Examples

Example 1.1 (Easy): Let

f(x) = x^2

and

g(x) = x + 3

. Find

(f \circ g)(2)

and

(g \circ f)(2)

. Step 1:

(f \circ g)(2) = f(g(2)) = f(5) = 25

Step 2:

(g \circ f)(2) = g(f(2)) = g(4) = 7

\boxed{(f \circ g)(2) = 25,\ (g \circ f)(2) = 7}

Example 1.2 (Medium): For

f(x) = \sqrt{x}

and

g(x) = 2x - 1

, find

(f \circ g)(x)

and its domain. Step 1:

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

Step 2 — Domain: Need

2x - 1 \geq 0 \implies x \geq 1/2

Also

2x - 1

must be in domain of

f = [0, \infty)

\boxed{(f \circ g)(x) = \sqrt{2x - 1},\ \text{domain } [1/2, \infty)}

Example 1.3 (Hard): Given

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

and

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

, find

(f \circ g)(x)

and its domain. Step 1:

(f \circ g)(x) = f\left(\frac{x+1}{x}\right) = \frac{1}{\frac{x+1}{x} - 1} = \frac{1}{\frac{x+1 - x}{x}} = \frac{1}{\frac{1}{x}} = x

Step 2 — Domain:

g

is undefined at

x = 0

g(x) = 1

would make denominator

0

f

; solve

\frac{x+1}{x} = 1 \implies x+1 = x \implies 1 = 0

(impossible, so no extra restriction).

\boxed{(f \circ g)(x) = x,\ \text{domain } x \neq 0}

7.2 Inverse Functions

7.2.1 Intuition: Undoing a Function

An inverse function

f^{-1}

"undoes" what

f

did. If

f

is the "double" function (

f(x) = 2x

), then

f^{-1}

is the "half" function (

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

). The inverse reverses the mapping: if

f(a) = b

, then

f^{-1}(b) = a

💡 Why this matters: Inverses let us solve equations. If $y = f(x)$ and we want $x$ in terms of $y$ , we use $x = f^{-1}(y)$ . Logarithms are inverses of exponentials — used everywhere in data science.

7.2.2 Condition for Existence

A function has an inverse iff it is bijective (both injective and surjective). For functions with codomain = range, surjectivity is automatic if we set codomain = range. So the key condition is injective (one-to-one). Horizontal Line Test: If every horizontal line hits the graph at most once, the function is injective and has an inverse.

Function	HLT Pass?	Has Inverse?
$f(x) = 2x + 3$	✓	Yes
$f(x) = x^2$ (on $\mathbb{R}$ )	✗	No
$f(x) = x^2$ (on $[0,\infty)$ )	✓	Yes ( $\sqrt{x}$ )
$f(x) = e^x$	✓	Yes ( $\ln x$ )

7.2.3 Finding the Inverse Algebraically

Recipe:

Write $y = f(x)$
Swap $x$ and $y$
Solve for $y$
The result is $y = f^{-1}(x)$
Verify: $f(f^{-1}(x)) = f^{-1}(f(x)) = x$

7.2.4 Worked Examples

Example 2.1 (Easy): Find the inverse of

f(x) = 3x - 5

. Step 1:

y = 3x - 5

Step 2: Swap:

x = 3y - 5

Step 3: Solve:

x + 5 = 3y \implies y = \frac{x+5}{3}

Step 4:

f^{-1}(x) = \frac{x+5}{3}

Verify:

f(f^{-1}(x)) = 3\left(\frac{x+5}{3}\right) - 5 = x + 5 - 5 = x

✓

\boxed{f^{-1}(x) = \frac{x+5}{3}}

Example 2.2 (Medium): Find the inverse of

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

. Step 1:

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

Step 2: Swap:

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

Step 3: Solve:

x(y-1) = 2y+1 \implies xy - x = 2y + 1 \implies xy - 2y = x + 1 \implies y(x-2) = x+1 \implies y = \frac{x+1}{x-2}

Step 4:

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

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

Example 2.3 (Hard): Find the inverse of

f(x) = \sqrt{2x - 3}

. Step 1:

y = \sqrt{2x - 3}

(domain

x \geq 3/2

, range

y \geq 0

) Step 2: Swap:

x = \sqrt{2y - 3}

Step 3: Square:

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

Step 4: Domain of

f^{-1}

= range of

f = [0, \infty)

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

for

x \geq 0

. Verify:

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

for

x \geq 0

✓

\boxed{f^{-1}(x) = \frac{x^2 + 3}{2},\ x \geq 0}

7.3 Domain and Range of Inverse

Key relationship:

Domain of $f^{-1}$ = Range of $f$
Range of $f^{-1}$ = Domain of $f$ (Diagram)

📐 Key Formulas — Summary Table

Concept	Formula	Notes
Composition	$(f \circ g)(x) = f(g(x))$	Apply $g$ first
Composition domain	$\{x \in D_g \mid g(x) \in D_f\}$	Careful with restrictions
Inverse definition	$f(f^{-1}(x)) = f^{-1}(f(x)) = x$
Inverse existence	$f$ must be bijective	Injective necessary
Finding inverse	Swap $x,y$ , solve for $y$	Algebraic method
Inverse domain	$D_{f^{-1}} = R_f$
Inverse range	$R_{f^{-1}} = D_f$
Graph of inverse	Reflection across $y = x$	Symmetry property

⚠️ Common Pitfalls

Pitfall 1: Confusing $f^{-1}$ with $(f(x))^{-1} = 1/f(x)$

Mistake: Thinking

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

. Correct:

f^{-1}

is the inverse function (undoes

f

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

is the reciprocal. These are completely different! Example:

f(x) = 2x

→

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

, not

1/(2x)

Pitfall 2: Composition Order

Mistake: Computing

(f \circ g)(x)

g(f(x))

. Correct:

(f \circ g)(x) = f(g(x))

— apply

g

first, then

f

. The notation

f \circ g

means "

f

after

g

Pitfall 3: Forgetting Domain Restrictions for Inverses

Mistake: Finding

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

for

f(x) = x^2

without restricting domain. Correct:

f(x) = x^2

\mathbb{R}

has no inverse (fails HLT). On

[0, \infty)

, its inverse is

\sqrt{x}

. On

(-\infty, 0]

, its inverse is

-\sqrt{x}

Pitfall 4: Swapping $x$ and $y$ Before Considering Domain

Mistake: Finding an algebraic expression for

f^{-1}

but ignoring domain restrictions. Correct: After solving, check the domain of

f^{-1}

= range of original

f

📝 Practice Questions

Q1: If $f(x) = x + 2$ and $g(x) = x^2$ , find $(f \circ g)(3)$ and $(g \circ f)(3)$ .

Strategy Hint: Apply inner function first.

$(f \circ g)(3) = f(9) = 11$ $(g \circ f)(3) = g(5) = 25$

$\boxed{11,\ 25}$ Q2: Find $(f \circ g)(x)$ for $f(x) = \frac{1}{x}$ and $g(x) = x-1$ .

$(f \circ g)(x) = \frac{1}{x-1}$

$\boxed{\frac{1}{x-1}}$ Q3: Find the inverse of $f(x) = 4x + 7$ .

$y = 4x + 7 \to x = 4y + 7 \to y = \frac{x-7}{4}$

$\boxed{f^{-1}(x) = \frac{x-7}{4}}$ Q4: Does $f(x) = x^2 + 1$ on $\mathbb{R}$ have an inverse? Why or why not?

Strategy Hint: Check HLT.

$f(-2) = 5 = f(2)$ → not injective → no inverse.

$\boxed{\text{No — fails HLT}}$ Q5: Find the inverse of $f(x) = \frac{x-2}{x+3}$ .

$y = \frac{x-2}{x+3}$ Swap: $x = \frac{y-2}{y+3} \to x(y+3) = y-2 \to xy + 3x = y - 2 \to xy - y = -3x - 2 \to y(x-1) = -(3x+2) \to y = \frac{3x+2}{1-x}$

$\boxed{f^{-1}(x) = \frac{3x+2}{1-x}}$ Q6: If $f(x) = 2x$ and $g(x) = 3x - 1$ , find $(f \circ g)(x)$ .

$(f \circ g)(x) = f(3x-1) = 2(3x-1) = 6x - 2$

$\boxed{6x - 2}$ Q7: Verify that $f(x) = \frac{x}{x-1}$ and $g(x) = \frac{x}{x-1}$ are inverses.

Strategy Hint: Check $f(g(x)) = x$ .

$f(g(x)) = \frac{\frac{x}{x-1}}{\frac{x}{x-1} - 1} = \frac{\frac{x}{x-1}}{\frac{x - (x-1)}{x-1}} = \frac{\frac{x}{x-1}}{\frac{1}{x-1}} = x$ ✓

$\boxed{\text{Yes, they are inverses}}$ Q8: What is the domain of $(f \circ g)(x)$ if $f(x) = \sqrt{x}$ and $g(x) = x-5$ ?

$g(x) = x-5$ has domain all reals. But $f$ requires non-negative input. So need $x-5 \geq 0 \to x \geq 5$ .

$\boxed{[5, \infty)}$ Q9: Find the inverse of $f(x) = e^{x+2}$ .

$y = e^{x+2}$ Swap: $x = e^{y+2}$ Take $\ln$ : $\ln x = y + 2 \to y = \ln x - 2$

$\boxed{f^{-1}(x) = \ln x - 2}$ Q10: If $f$ and $g$ are inverses, what is $f(g(5))$ ?

By definition of inverse: $f(g(x)) = x$ , so $f(g(5)) = 5$ .

$\boxed{5}$ Q11: Find the inverse of $f(x) = x^3 - 1$ .

$y = x^3 - 1$ Swap: $x = y^3 - 1 \to y^3 = x + 1 \to y = \sqrt[3]{x+1}$

$\boxed{f^{-1}(x) = \sqrt[3]{x+1}}$ Q12: For $f(x) = \frac{1}{x}$ , find $f^{-1}(x)$ . What do you notice?

$y = 1/x$ , swap: $x = 1/y \to y = 1/x$ $f^{-1}(x) = 1/x = f(x)$ — the function is self-inverse (involution).

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

🔗 Cross-References

Previous topic: Exponential Functions
Next topic: Logarithms — inverses of exponentials
Related: Functions — injective, surjective, bijective concepts
Across courses: BSMA1003 Maths 2 (inverse functions in calculus, inverse trigonometric functions) Join Discord Previous5.1 Exponential FunctionsNext6.1 Logarithmic Functions & Properties

What this note is really saying

Composite and Inverse Functions

🎯 Learning Objectives

📋 Prerequisites

📖 Core Content

7.1 Composite Functions

7.1.1 Intuition: Function Chains

7.1.2 Definition

7.1.3 Worked Examples

7.2 Inverse Functions

7.2.1 Intuition: Undoing a Function

7.2.2 Condition for Existence

7.2.3 Finding the Inverse Algebraically

7.2.4 Worked Examples

7.3 Domain and Range of Inverse

📐 Key Formulas — Summary Table

⚠️ Common Pitfalls

Pitfall 1: Confusing f−1f^{-1}f−1 with (f(x))−1=1/f(x)(f(x))^{-1} = 1/f(x)(f(x))−1=1/f(x)

Pitfall 2: Composition Order

Pitfall 3: Forgetting Domain Restrictions for Inverses

Pitfall 4: Swapping xxx and yyy Before Considering Domain

📝 Practice Questions

🔗 Cross-References

Intelligence Hub

Pitfall 1: Confusing $f^{-1}$ with $(f(x))^{-1} = 1/f(x)$

Pitfall 4: Swapping $x$ and $y$ Before Considering Domain