Lecture Note 1: Functions of a Real Variable

1. Introduction to Functions

A function is a rule that assigns each input value from one set to exactly one output value in another set.

If X and Y are sets of real numbers, a real-valued function f from X to Y assigns every x in X to one and only one value y in Y. The value y is written as f(x), which is read as "f of x."

The set X is called the domain of the function. It contains all values that may be used as inputs. The set of all actual output values produced by the function is called the range. The input variable x is called the independent variable, while the output variable y is called the dependent variable.

A relation is not necessarily a function. It becomes a function only when each input has exactly one output.

Example 1

If f(x) = x² - 7, evaluate f(3), f(-2), f(3a), and f(b - 1).

Solution:

f(3) = 3² - 7 = 9 - 7 = 2

f(-2) = (-2)² - 7 = 4 - 7 = -3

f(3a) = (3a)² - 7 = 9a² - 7

f(b - 1) = (b - 1)² - 7

= b² - 2b + 1 - 7

= b² - 2b - 6

Key Takeaways

• A function gives exactly one output for each valid input.
• The domain is the set of permitted input values.
• The range is the set of output values actually produced.
• f(x) is not multiplication; it means the value of the function f at x.
• To evaluate a function, replace x everywhere with the given input.

2. Function Notation, Explicit Form, and Implicit Form

Function notation is a convenient way of showing dependence between two variables. In y = f(x), the value of y depends on the value chosen for x.

An equation may define a function explicitly or implicitly.

An explicit function is written directly in the form y = f(x). For example:

y = x² - 4x + 7

An implicit equation relates x and y without first isolating y. For example:

x² + 2y = 1

To write it explicitly, solve for y:

2y = 1 - x²

y = ^{1 - x2}⁄₂

Therefore, the function may be written as:

f(x) = ^{1 - x2}⁄₂

Example 2

If f(x) = 2x² - 5x + 1 and h ≠ 0, simplify:

^{f(a + h) - f(a)}⁄_h

Solution:

First find f(a + h):

f(a + h) = 2(a + h)² - 5(a + h) + 1

= 2(a² + 2ah + h²) - 5a - 5h + 1

= 2a² + 4ah + 2h² - 5a - 5h + 1

Also:

f(a) = 2a² - 5a + 1

Now substitute:

^{f(a + h) - f(a)}⁄_h = ^{2a2 + 4ah + 2h2 - 5a - 5h + 1 - (2a2 - 5a + 1)}⁄_h

= ^{4ah + 2h2 - 5h}⁄_h

= ^{h(4a + 2h - 5)}⁄_h

= 4a + 2h - 5

Exam Points

• f(a + h) means substitute a + h wherever x appears.
• Always expand carefully before subtracting f(a).
• In a difference quotient, factor h from the numerator before cancelling.
• h must not be zero because division by zero is undefined.

3. Domain and Range of a Function

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

The range of a function is the set of all possible output values produced by the function.

A domain may be explicitly stated or implied by the formula.

An explicitly stated domain is given directly with the function. For example:

f(x) = ¹⁄_{x² - 4}, 4 ≤ x ≤ 5

Here, the domain is 4 ≤ x ≤ 5.

An implied domain is the largest set of real numbers for which the formula makes sense. For example:

g(x) = ¹⁄_{x² - 4}

Here, x cannot be 2 or -2 because those values make the denominator zero. Therefore, the domain is all real numbers except x = -2 and x = 2.

Rules for Finding Domains

1. For polynomial functions, the domain is all real numbers.
2. For rational functions, exclude values that make the denominator zero.
3. For square-root functions, the expression inside the square root must be greater than or equal to zero.
4. For composite functions, check both the inner and outer functions.

Example 3

Find the domain and range of:

f(x) = √(x + 2)

Solution:

The expression inside the square root must be non-negative:

x + 2 ≥ 0

x ≥ -2

So the domain is:

[-2, ∞)

Since square roots are always non-negative, the range is:

[0, ∞)

Example 4

Find the domain and range of:

g(x) = ¹⁄_{x² - x}

Solution:

Factor the denominator:

x² - x = x(x - 1)

The denominator must not be zero:

x(x - 1) ≠ 0

So:

x ≠ 0 and x ≠ 1

The domain is:

(-∞, 0) ∪ (0, 1) ∪ (1, ∞)

To find the range, let:

y = ¹⁄_{x² - x}

Then:

x² - x = ¹⁄_y

So:

x² - x - ¹⁄_y = 0

For real x-values to exist, the discriminant must be non-negative:

1 + ⁴⁄_y ≥ 0

This gives:

y > 0 or y ≤ -4

Therefore, the range is:

(-∞, -4] ∪ (0, ∞)

Key Takeaways

• Domain concerns input values.
• Range concerns output values.
• A denominator must never be zero.
• A square root in real-valued functions must not contain a negative radicand.
• The range may require deeper analysis than the domain.

4. Piecewise-Defined Functions

A piecewise-defined function is a function described by different formulas on different parts of its domain.

For example:

f(x) = 1 - x, if x < 1

f(x) = √(x - 1), if x ≥ 1

This means that the rule used depends on the value of x.

Example 5

Given:

f(x) = 1 - x, if x < 1

f(x) = √(x - 1), if x ≥ 1

Evaluate f(-2), f(0), and f(5). Then find the domain and range.

Solution:

Since -2 < 1:

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

Since 0 < 1:

f(0) = 1 - 0 = 1

Since 5 ≥ 1:

f(5) = √(5 - 1) = √4 = 2

The function is defined for x < 1 and x ≥ 1. These intervals cover all real numbers. Therefore, the domain is:

(-∞, ∞)

For x < 1, the expression 1 - x is positive. For x ≥ 1, the expression √(x - 1) is non-negative. The smallest output is 0, which occurs at x = 1. Therefore, the range is:

[0, ∞)

Exam Points

• Always choose the formula based on the condition attached to x.
• Do not use the wrong branch of a piecewise function.
• To find the domain, combine all intervals where the function is defined.
• To find the range, examine the output values from each branch.

5. Composite Functions

A composite function is formed when the output of one function becomes the input of another function.

If f and g are functions, then the composite of f with g is written as:

(f ∘ g)(x) = f(g(x))

This means that g is applied first, and then f is applied to the result.

The domain of f ∘ g is the set of all x-values in the domain of g for which g(x) is in the domain of f.

In general:

(f ∘ g)(x) ≠ (g ∘ f)(x)

Example 6

Given:

f(x) = x²

g(x) = x - 3

Find f ∘ g and g ∘ f.

Solution:

(f ∘ g)(x) = f(g(x))

= f(x - 3)

= (x - 3)²

Also:

(g ∘ f)(x) = g(f(x))

= g(x²)

= x² - 3

Therefore:

(f ∘ g)(x) = (x - 3)²

(g ∘ f)(x) = x² - 3

These are not the same function.

Example 7

Let:

f(x) = √x

g(x) = √(2 - x)

Find each composite function and its domain.

1. f ∘ g

(f ∘ g)(x) = f(g(x))

= f(√(2 - x))

= √(√(2 - x))

For this to be defined:

2 - x ≥ 0

x ≤ 2

Domain:

(-∞, 2]

2. g ∘ f

(g ∘ f)(x) = g(f(x))

= g(√x)

= √(2 - √x)

For √x to be defined:

x ≥ 0

For √(2 - √x) to be defined:

2 - √x ≥ 0

√x ≤ 2

x ≤ 4

Therefore, the domain is:

[0, 4]

3. f ∘ f

(f ∘ f)(x) = f(f(x))

= f(√x)

= √(√x)

The domain is:

[0, ∞)

4. g ∘ g

(g ∘ g)(x) = g(g(x))

= g(√(2 - x))

= √(2 - √(2 - x))

First:

2 - x ≥ 0, so x ≤ 2

Second:

2 - √(2 - x) ≥ 0

√(2 - x) ≤ 2

2 - x ≤ 4

x ≥ -2

Therefore, the domain is:

[-2, 2]

Key Takeaways

• In f(g(x)), work from inside to outside.
• The inner function is evaluated first.
• The output of the inner function must be allowed as an input of the outer function.
• f ∘ g and g ∘ f are generally different.
• Composite-function domains require careful restriction.

6. Even and Odd Functions

Even and odd functions describe symmetry in a function.

A function f is even if:

f(-x) = f(x)

for every x in the domain.

The graph of an even function is symmetric about the y-axis.

A function f is odd if:

f(-x) = -f(x)

for every x in the domain.

The graph of an odd function is symmetric about the origin.

If neither condition is satisfied, the function is neither even nor odd.

Example 8

Determine whether:

f(x) = x⁵ + x

is even, odd, or neither.

Solution:

f(-x) = (-x)⁵ + (-x)

= -x⁵ - x

= -(x⁵ + x)

= -f(x)

Therefore, f is odd.

Example 9

Determine whether:

g(x) = 1 - x⁴

is even, odd, or neither.

Solution:

g(-x) = 1 - (-x)⁴

= 1 - x⁴

= g(x)

Therefore, g is even.

Example 10

Determine whether:

h(x) = 2x - x²

is even, odd, or neither.

Solution:

h(-x) = 2(-x) - (-x)²

= -2x - x²

This is not equal to h(x), and it is not equal to -h(x). Therefore, h is neither even nor odd.

Exam Points

• Replace x with -x and simplify.
• If the result is exactly f(x), the function is even.
• If the result is exactly -f(x), the function is odd.
• If neither result occurs, the function is neither even nor odd.
• Even functions have y-axis symmetry.
• Odd functions have origin symmetry.

7. Periodic Functions

A periodic function is a function whose values repeat at regular intervals.

A function f is periodic if there exists a positive constant p such that:

f(x + p) = f(x)

for every x in the domain.

The smallest positive value of p that satisfies this condition is called the period of the function.

Examples

The sine function is periodic:

y = sin x

Its period is:

2π

This means:

sin(x + 2π) = sin x

The tangent function is also periodic:

y = tan x

Its period is:

π

This means:

tan(x + π) = tan x

Key Takeaways

• Periodic functions repeat their values.
• The period is the length of one complete cycle.
• sin x and cos x have period 2π.
• tan x has period π.
• Periodic functions are central in trigonometry, waves, and oscillatory motion.

8. Exam Formula and Concept Summary

Function

A function assigns each input exactly one output.

Function Notation

f(x) means the value of f at x.

Domain

The domain is the set of permitted input values.

Range

The range is the set of output values actually produced.

Square-Root Restriction

For √A to be real:

A ≥ 0

Rational-Function Restriction

For ^A⁄_B to be defined:

B ≠ 0

Composite Function

(f ∘ g)(x) = f(g(x))

Domain of f ∘ g:

x must be in the domain of g, and g(x) must be in the domain of f.

Even Function

f(-x) = f(x)

Odd Function

f(-x) = -f(x)

Periodic Function

f(x + p) = f(x), where p > 0.

9. Practice Questions

1. Evaluate f(x) = 2x - 3 at f(0), f(-3), and f(6).
2. Evaluate f(x) = √(x + 3) at f(-2), f(6), and f(1).
3. Evaluate f(x) = cos(2x) at f(0), f(π/2), and f(2π).
4. Evaluate f(x) = x²(x - 4) at f(4), f(3/2), and f(3).
5. If f(x) = x³, simplify ^{f(x + h) - f(x)}⁄_h, where h ≠ 0.
6. Find the domain and range of f(x) = √(4 - x²).
7. Find the domain and range of g(x) = √(x - 5).
8. Find the domain and range of h(x) = ^{x + 3}⁄₂.
9. Find the domain of g(t) = ^{4 - t2}⁄_{2 - t}.
10. For the piecewise function f(x) = x + 2, if x < 0, and f(x) = 1 - x, if x ≥ 0, evaluate f(-3), f(0), and f(4). Then determine the domain and range.
11. Determine whether each function is even, odd, or neither: f(x) = x³ - x, g(x) = x² + 5, and h(x) = x² + x.
12. Given f(x) = x + 1 and g(x) = x², find (f ∘ g)(x), (g ∘ f)(x), and the domain of each composite function.

10. Revision Checklist

Before an examination, make sure you can:

• Define a function correctly.
• Distinguish between domain and range.
• Evaluate f(a), f(-a), f(a + h), and f(x + h).
• Find domains of square-root and rational functions.
• Work with piecewise-defined functions.
• Compute composite functions.
• Find domains of composite functions.
• Test whether a function is even, odd, or neither.
• Identify simple periodic functions and their periods.