9662e.dev

Introduction to Functions

A function is a relationship between inputs and outputs where each input is related to exactly one output. Functions are one of the most fundamental concepts in mathematics, serving as building blocks for more complex mathematical structures.

A function

f

can be represented in various ways:

Function Notation

Standard Notation

f(x) = \text{[expression involving x]}

Where

x

is the input and

f(x)

is the output.

Mapping Notation

f: X \to Y

Where

X

is the domain (set of inputs) and

Y

is the codomain (possible outputs).

Set Notation

f = \{(x, y) \in X \times Y : y = f(x)\}

A function as a set of ordered pairs where each first component appears exactly once.

The key property of a function is that each input value has exactly one corresponding output value. This is sometimes called the vertical line test in graphical representations: any vertical line should intersect the graph of a function at most once.

Domain and Range

Two important concepts associated with functions are domain and range:

• The domain of a function is the set of all possible input values for which the function is defined.

• The range (or image) of a function is the set of all possible output values that the function produces.

If not explicitly stated, the domain of a function is assumed to be the largest possible set of real numbers for which the function produces real outputs without undefined operations.

Common Domain Restrictions

Division by Zero

\text{For } f(x) = \frac{1}{x} \text{, the domain is } \{x \in \mathbb{R} : x \neq 0\}

Even-root Functions

\text{For } f(x) = \sqrt{x} \text{, the domain is } \{x \in \mathbb{R} : x \geq 0\}

Logarithmic Functions

\text{For } f(x) = \log(x) \text{, the domain is } \{x \in \mathbb{R} : x > 0\}

Problem: Find the domain and range of the function

f(x) = \sqrt{4-x^2}

.

1

For the domain, we need

4-x^2 \geq 0

since we can't take the square root of a negative number.

2

Solving

4-x^2 \geq 0

:

-x^2 \geq -4

x^2 \leq 4

-2 \leq x \leq 2

3

For the range, we note that

0 \leq 4-x^2 \leq 4

when

x

is in the domain.

0 \leq \sqrt{4-x^2} \leq 2

4

Therefore:

Domain:

[-2,2]

Range:

[0,2]

Types of Functions

Functions can be classified based on various properties and behaviors:

Linear Functions

f(x) = mx + b

Where

m

is the slope and

b

is the y-intercept. The graph is a straight line.

Quadratic Functions

f(x) = ax^2 + bx + c, \quad a \neq 0

The graph forms a parabola. Its vertex is at

x = -\frac{b}{2a}

.

Polynomial Functions

f(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0, \quad a_n \neq 0

A function defined by a polynomial expression with degree

n

.

Rational Functions

f(x) = \frac{P(x)}{Q(x)}

A ratio of two polynomial functions, where

Q(x) \neq 0

.

Exponential Functions

f(x) = a \cdot b^x, \quad a \neq 0, b > 0, b \neq 1

Base

b

raised to power

x

. Grows very quickly when

b > 1

.

Logarithmic Functions

f(x) = \log_b(x), \quad b > 0, b \neq 1, x > 0

Inverse of exponential function. Base

b

logarithm of

x

.

Trigonometric Functions

f(x) = \sin(x), \cos(x), \tan(x), \text{etc.}

Periodic functions related to the geometry of circles and triangles.

Problem: Determine the type of function:

f(x) = 3x^2 - 4x + 5

.

1

This function has the form

f(x) = ax^2 + bx + c

where

a = 3

,

b = -4

, and

c = 5

.

2

Since the highest power of

x

is 2, and

a = 3 \neq 0

, this is a quadratic function.

3

The graph of this function is a parabola that opens upward (since

a > 0

).

Function Properties

Functions can have various properties that describe their behavior:

Key Function Properties

Even and Odd Functions

• Even:

f(-x) = f(x)

for all

x

in the domain

• Odd:

f(-x) = -f(x)

for all

x

in the domain

Even functions are symmetric about the y-axis; odd functions are symmetric about the origin.

Increasing and Decreasing Functions

• Increasing:

f(x_1) < f(x_2)

whenever

x_1 < x_2

• Decreasing:

f(x_1) > f(x_2)

whenever

x_1 < x_2

One-to-One Functions

A function

f

is one-to-one (injective) if distinct inputs always yield distinct outputs:

f(x_1) = f(x_2) \implies x_1 = x_2

Graphically, a function is one-to-one if it passes the horizontal line test.

Onto Functions

A function

f: X \to Y

is onto (surjective) if every element in the codomain

Y

is the image of at least one element in the domain

X

.

In other words, the range equals the codomain.

Bijective Functions

A function is bijective if it is both one-to-one and onto.

Only bijective functions have inverses that are also functions.

Problem: Determine whether the function

f(x) = x^3

is even, odd, or neither.

1

To check if

f

is even, substitute

-x

into the function:

f(-x) = (-x)^3 = -x^3

2

Compare this with

f(x)

:

f(-x) = -x^3

f(x) = x^3

3

Since

f(-x) = -f(x)

for all

x

,

f(x) = x^3

is an odd function.

Problem: Determine whether the function

f(x) = x^2 + 3

is one-to-one.

1

Let's check if distinct inputs yield distinct outputs. Let's say

f(x_1) = f(x_2)

.

2

That means:

x_1^2 + 3 = x_2^2 + 3

x_1^2 = x_2^2

x_1 = \pm x_2

3

Since

x_1

could equal

-x_2

when

x_2 \neq 0

, two different inputs could yield the same output.

4

For example,

f(2) = 2^2 + 3 = 7

and

f(-2) = (-2)^2 + 3 = 7

.

5

Therefore,

f(x) = x^2 + 3

is not one-to-one.

Function Operations

Functions can be combined using arithmetic operations as well as composition to create new functions:

Basic Function Operations

Arithmetic Operations

• Sum:

(f + g)(x) = f(x) + g(x)

• Difference:

(f - g)(x) = f(x) - g(x)

• Product:

(f \cdot g)(x) = f(x) \cdot g(x)

• Quotient:

(f / g)(x) = f(x) / g(x)

, where

g(x) \neq 0

The domain of the resulting function is the intersection of the domains of

f

and

g

.

Function Composition

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

The domain of

f \circ g

consists of all

x

in the domain of

g

such that

g(x)

is in the domain of

f

.

Problem: If

f(x) = x^2 - 3

and

g(x) = 2x + 1

, find

(f \circ g)(x)

and

(g \circ f)(x)

.

1

For

(f \circ g)(x)

, substitute

g(x)

into

f

:

(f \circ g)(x) = f(g(x)) = f(2x + 1)

2

Evaluate

f(2x + 1)

:

f(2x + 1) = (2x + 1)^2 - 3 = 4x^2 + 4x + 1 - 3 = 4x^2 + 4x - 2

3

For

(g \circ f)(x)

, substitute

f(x)

into

g(x)

:

(g \circ f)(x)=g(f(x))=g(x^2-3)

2

Evaluate

g(x^2-3)

:

g(x^2-3) = 2(x^2 - 3) + 1 = 2x^2 - 6 + 1 = 2x^2 - 5

Function Transformations

Function transformations allow us to create new functions from existing ones through operations like shifts, stretches, and reflections:

Basic Transformations

Vertical Shifts

y = f(x) + k

Shifts the graph up by

k

units (down if

k < 0

).

Horizontal Shifts

y = f(x - h)

Shifts the graph right by

h

units (left if

h < 0

).

Vertical Stretches and Compressions

y = a \cdot f(x)

Stretches the graph vertically by factor

a

if

|a| > 1

, compresses if

0 < |a| < 1

.

Horizontal Stretches and Compressions

y = f(bx)

Compresses the graph horizontally by factor

b

if

|b| > 1

, stretches if

0 < |b| < 1

.

Reflections

•

y = -f(x)

: Reflection across the x-axis

•

y = f(-x)

: Reflection across the y-axis

Problem: How does the graph of

g(x) = -2f(x-3) + 4

relate to the graph of

f(x)

?

1

Breaking down the transformations from inside to outside:

2

f(x-3)

: Shift the graph of

f(x)

right by 3 units

3

-2f(x-3)

: Vertically stretch by factor 2 and reflect across x-axis

4

-2f(x-3) + 4

: Shift up by 4 units

5

In summary,

g(x)

is the graph of

f(x)

shifted right 3 units, vertically stretched by factor 2, reflected across the x-axis, and shifted up 4 units.

Inverse Functions

If a function

f

is one-to-one, it has an inverse function

f^{-1}

that "undoes" what

f

does.

Inverse Function Properties

Definition

For a one-to-one function

f

, its inverse

f^{-1}

satisfies:

•

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

for all

x

in the domain of

f

•

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

for all

y

in the range of

f

Domain and Range

The domain of

f^{-1}

is the range of

f

, and vice versa.

Graph of Inverse

The graph of

f^{-1}

is the reflection of the graph of

f

across the line

y = x

.

To find the inverse of a function:

1. Replace

f(x)

with

y

2. Interchange

x

and

y

3. Solve for

y

4. Replace

y

with

f^{-1}(x)

Problem: Find the inverse of

f(x) = 3x - 5

.

1

Replace

f(x)

with

y

:

y = 3x - 5

2

Interchange

x

and

y

:

x = 3y - 5

3

Solve for

y

:

x + 5 = 3y

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

4

Therefore,

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

5

To verify:

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

Piecewise Functions

A piecewise function is defined by different expressions for different parts of its domain. You don't need to know this type of function for the IB, but it is useful to have as background knowledge.

f(x) = \begin{cases} \text{expression}_1, & \text{if } x \in \text{domain}_1 \\ \text{expression}_2, & \text{if } x \in \text{domain}_2 \\ \vdots \\ \text{expression}_n, & \text{if } x \in \text{domain}_n \\ \end{cases}

The absolute value function is a common piecewise function:

|x| = \begin{cases} x, & \text{if } x \geq 0 \\ -x, & \text{if } x < 0 \\ \end{cases}

Problem: Evaluate the piecewise function

f(x) = \begin{cases} x^2, & \text{if } x \leq 1 \\ 2x - 1, & \text{if } x > 1 \\ \end{cases}

at

x = 0

,

x = 1

, and

x = 2

.

1

For

x = 0

:

0 \leq 1

, so use

f(x) = x^2

f(0) = 0^2 = 0

2

For

x = 1

:

1 \leq 1

, so use

f(x) = x^2

f(1) = 1^2 = 1

3

For

x = 2

:

2 > 1

, so use

f(x) = 2x - 1

f(2) = 2(2) - 1 = 4 - 1 = 3

Applications of Functions

Functions model countless real-world scenarios and relationships across various fields.

Economics

• Cost functions:

C(x) =

fixed costs + variable costs

• Revenue functions:

R(x) = p \cdot x

, where

p

is price and

x

is quantity

• Profit functions:

P(x) = R(x) - C(x)

Physics

• Position functions:

s(t)

gives position at time

t

• Velocity functions:

v(t) = s'(t)

(derivative of position)

• Force functions:

F(d)

gives force as function of distance

Problem: A company determines that the cost

C(x)

(in dollars) of producing

x

items is

C(x) = 500 + 25x

. The items sell for $40 each. Find the profit function and determine how many items must be sold to break even.

1

The revenue from selling

x

items at $40 each is:

R(x) = 40x

2

The profit function is revenue minus cost:

P(x) = R(x) - C(x) = 40x - (500 + 25x) = 40x - 500 - 25x = 15x - 500

3

To break even, the profit must be zero:

P(x) = 0

15x - 500 = 0

15x = 500

x = 33.33...

4

Since we can't produce a fractional number of items, the company needs to produce and sell at least 34 items to make a profit.

Sequences and Series

Exponents and Logarithms

Functions

Polynomial Functions

Rational Functions

Exponential and Log Functions

Trigonometric Functions

Trigonometric Identities

Solving Triangles

Vectors

Limits and Continuity

Derivatives

Integrals

Descriptive Statistics

Probability Concepts

Probability Distributions

Complex Numbers

Further Calculus

Matrices

Applications and Modeling

IA Preparation (Math)

IA Preparation (Physics)

Functions

Introduction to Functions

Function Notation

Domain and Range

Common Domain Restrictions

Types of Functions

Function Properties

Key Function Properties

Function Operations

Basic Function Operations

Function Transformations

Basic Transformations

Inverse Functions

Inverse Function Properties

Piecewise Functions

Applications of Functions