Day -5: Number Systems & Logical Symbols & Defining a Function & Finding Domain and Range & One-to-One and Onto Functions & Operations and Composition of Functions & Symmetry and Inverse Functions

Number Systems & Logical Symbols

Before diving into Calculus, we must speak the language of mathematics. This includes identifying different sets of numbers and understanding logical quantifiers.

Core Theorem

Common Notation:
- ℕ: Natural Numbers (positive integers)
- ℤ: Integers
- ℚ: Rational Numbers
- ℝ: Real Numbers
- ℂ: Complex Numbers

Logical Quantifiers:
- ∀: 'For all'
- ∃: 'There exists'
- ∃!: 'There exists a unique'

Common Pitfalls

⚠
Natural Numbers AmbiguityIn some contexts, ℕ includes zero. However, in this curriculum (as per the notes), ℕ refers strictly to positive integers. Always check the boundary conditions!

Defining a Function

A function is a specific type of relationship between two sets where every input has exactly one output.

Core Theorem

A function

f: A → B

is a correspondence such that:

∀

a in A,

∃!

b in B such that

f(a) = b

.
- Domain (

A

): The set of all possible inputs.
- Codomain (

B

): The set where outputs land.
- Range (

f(A)

): The actual set of values produced by the function
(

f(A) ⊆ f(B)

).

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Easy

A rectangular box with no lid has a volume of

10 m^3

. The length of the base is twice its width. The material for the base costs

10/m^2

, and the side material costs

6/m^2

.
Express the total cost

C

as a function of the base width

w

.

How to think about it

Key idea: Volume $V = l * w * h$ . Use the volume to replace $h$ in the cost equation.

Full solution

1. Establish Variables

Let width be

w

, then length

l = 2w

. Let height be

h

. Volume

V = w(2w)h

=

2w^2 * h = 10

,

h = 10/2w^2

.

2. Calculate Areas

Base area

A_{base} = 2w^2

. Side areas consist of two

w * h

faces and two

2w * h

faces.

A_{sides} = 2(wh) + 2(2wh) = 6wh

.

3. Formulate Cost Function

C(w) = 10(2w^2) + 6(6wh)

。Substitute

h = \frac{5}{w^2}

:

C(w) = 20w^2 + 36w(\frac{5}{w^2}) = 20w^2 + \frac{180}{w}

。

Worked Example 02Hard

A right triangle has an area of 25. Express the length of the hypotenuse

h

as a function of the perimeter

p

.

How to think about it

Key idea: Use $a^2 + b^2 = h^2$ and $Area = \frac{1}{2}ab = 25$ . Remember $p = a + b + h$ .

Full solution

1. Establish System of Equations

From the area:

\frac{1}{2}ab = 25 \implies ab = 50

. From the Pythagorean theorem:

a^2 + b^2 = h^2

. From the perimeter:

a + b + h = p \implies a + b = p - h

.

2. Algebraic Substitution

Square the perimeter relation:

(a + b)^2 = (p - h)^2

. Expanding gives

a^2 + b^2 + 2ab = (p - h)^2

. Substitute

a^2 + b^2 = h^2

and

ab = 50

:

h^2 + 2(50) = (p - h)^2

.

3. Solve for h

Expand the right side:

h^2 + 100 = p^2 - 2ph + h^2

. Subtract

h^2

from both sides:

100 = p^2 - 2ph

. Rearrange to isolate

h

:

2ph = p^2 - 100

. Final function:

h(p) = \frac{p^2 - 100}{2p}

.

Step-by-Step SOP

1
Testing for a Function
Check if any $x$ in the domain maps to more than one $y$ . If it does, it's a relation, not a function.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

A taxi fare is

75

for the first

2\text{ km}

, and

5

for every additional

500\text{ m}

. Express the fare

C

as a function of distance

d

(in km).

Need a hint?

This is a step function (piecewise). Note that

500\text{ m} = 0.5\text{ km}

. Use the ceiling function

\lceil x \rceil

to account for any partial segments.

Show solution

1. Define the Step Logic

For the first

2\text{ km}

, the cost is fixed: if

0 < d \leq 2

, then

C = 75

. For

d > 2

, we only charge for the distance exceeding

2\text{ km}

, which is

d - 2

.

2. Formulate the Piecewise Function

The number of

0.5\text{ km}

segments is calculated using the ceiling function

\lceil \frac{d-2}{0.5} \rceil

. The complete function is:

C(d) = \begin{cases} 75, & 0 < d \leq 2 \\ 75 + 5 \lceil \frac{d-2}{0.5} \rceil, & d > 2 \end{cases}

Practice 02Medium

If a function

f(x)

is defined by an algebraic expression but the domain is not specified, how do we find it?

Need a hint?

Think about where the expression is mathematically 'meaningful'.

Show solution

1. The Natural Domain

The domain is the set of all real

x

values for which the expression is defined. Common restrictions: denominators cannot be zero, and values under an even root must be non-negative.

Common Pitfalls

⚠
Range vs. CodomainThe codomain is the 'target' set (usually ℝ), while the range is the set of values the function *actually* hits. They are not always the same!

Finding Domain and Range

The domain of a real-valued function f(x) is the set of all x for which the expression is mathematically defined and results in a real number.

Core Theorem

Key Restrictions:
1. Even Roots: Inside must be

≥0

.
2. Denominators: Must be

≠ 0

.
3. Logarithms: Inside must be

> 0

.
4. Trig: Specific domains for

tan(x),sec(x)

, etc.

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

Find the domain and range of

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

.

How to think about it

Key idea: The expression inside the square root must be greater than or equal to zero, and the square root function itself always yields a non-negative result.

Full solution

1. Domain Calculation

Solve

2 + x - x^2 \ge 0

. Multiplying by

-1

gives

x^2 - x - 2 \le 0

. Factoring the quadratic:

(x-2)(x+1) \le 0

. The domain is

[-1, 2]

.

2. Range Calculation

The expression

g(x) = -x^2 + x + 2

is a downward parabola. Its maximum occurs at

x = -\frac{b}{2a} = \frac{1}{2}

. The maximum value is

g(0.5) = 2.25 = \frac{9}{4}

. Since the minimum value of

g(x)

within the domain is

0

, the range of

f(x) = \sqrt{g(x)}

is

[0, \sqrt{\frac{9}{4}}] = [0, 1.5]

.

Step-by-Step SOP

1
System of Inequalities
If a function is a sum of multiple parts, find the domain of each part separately and then find the intersection (common values) of all sets.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Hard

Find the domain of

f(x) = \sqrt{x^2 - 1} + \frac{1}{\sqrt{4 - x^2}}

.

Need a hint?

The domain must satisfy all conditions simultaneously: expressions inside square roots must be

\ge 0

, and denominators cannot be

0

.

Show solution

1. Analyze the First Term

For

\sqrt{x^2 - 1}

, we require

x^2 - 1 \ge 0 \implies x^2 \ge 1

. Taking the square root gives

|x| \ge 1

, which means

x \in (-\infty, -1] \cup [1, \infty)

.

2. Analyze the Second Term

For

\frac{1}{\sqrt{4 - x^2}}

, the expression inside the root must be positive and non-zero:

4 - x^2 > 0 \implies x^2 < 4

. This gives

|x| < 2

, which means

x \in (-2, 2)

.

3. Find the Intersection

The domain is the intersection of both conditions:

{x \mid (x \le -1 \text{ or } x \ge 1) \text{ and } (-2 < x < 2)}

. This results in

(-2, -1] \cup [1, 2)

.

Practice 02Hard

Find the domain of

f(x) = \sqrt{\sin\sqrt{x}}

.

Need a hint?

Work from the innermost function to the outermost. Consider the periodicity of the sine function.

Show solution

1. Inner Root Condition

For the innermost square root

\sqrt{x}

to be defined, the radicand must be non-negative:

x \ge 0

.

2. Outer Root and Trigonometric Logic

For the outer square root to be defined, we need

\sin\sqrt{x} \ge 0

. The sine function

\sin \theta

is non-negative when its argument

\theta

falls in the intervals

[2k\pi, (2k+1)\pi]

for any integer

k

.

3. Combine and Solve for x

Since we already know

\sqrt{x} \ge 0

, we only consider

k = 0, 1, 2, \dots

. The condition becomes

2k\pi \le \sqrt{x} \le (2k+1)\pi

. Squaring all sides gives the domain:

(2k\pi)^2 \le x \le ((2k+1)\pi)^2

for

k \in \{0, 1, 2, \dots\}

.

Common Pitfalls

⚠
The Denominator vs. Root TrapWhen a square root is in the denominator, remember that the inner value must be strictly GREATER than zero ( $>0$ ), not $\ge 0$ .

One-to-One and Onto Functions

We classify functions based on how they map elements between the domain and the codomain.

Core Theorem

1. One-to-One (Injective): If

x_1 \neq x_2

, then

f(x_1) \neq f(x_2)

.
2. Onto (Surjective): If the Range equals the Codomain (

f(A) = B

).

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

Prove

y = x^3

is one-to-one using algebraic factorization.

How to think about it

Key idea: Use the difference of cubes formula: $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$ .

Full solution

1. Factorization

Assume

x_1^3 = x_2^3

. Moving terms to one side:

x_1^3 - x_2^3 = 0

. By factoring, we get

(x_1 - x_2)(x_1^2 + x_1x_2 + x_2^2) = 0

.

2. Analyzing the Quadratic Factor

For the second factor, we use 'completing the square':

x_1^2 + x_1x_2 + x_2^2 = (x_1 + \frac{1}{2}x_2)^2 + \frac{3}{4}x_2^2

. This expression is only zero if

x_1 = 0

and

x_2 = 0

.

3. Conclusion

Otherwise, we must have

x_1 - x_2 = 0

, which means

x_1 = x_2

. Thus, the function is one-to-one.

Step-by-Step SOP

1
Horizontal Line Test
Use this to visually check if a function is one-to-one (the line should hit the graph at most once).

Operations and Composition of Functions

Functions can be combined using arithmetic operations or by nesting one within another (composition). A critical aspect is determining the resulting domain.

Core Theorem

1. Arithmetic Operations:
-

(f \pm g)(x) = f(x) \pm g(x)

, Domain:

Dom(f) \cap Dom(g)

-

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

, Domain:

Dom(f) \cap Dom(g)

-

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

, Domain:

Dom(f) \cap Dom(g) \cap \{x \mid g(x) \neq 0\}

2. Composition:
-

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

, Domain: \{x \in Dom(g) \mid g(x) \in Dom(f)\}

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

Why is the domain of

h(x) = x \cdot \frac{1}{x}

not all real numbers?
Find the domain of

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

.

How to think about it

Key idea: The domain is determined by the *original* expression before simplification.

Full solution

1. Simplification Trap

For

h(x) = x \cdot \frac{1}{x}

, even though it simplifies to

1

for

x \neq 0

, the original term

\frac{1}{x}

is undefined at

x=0

. Thus,

Dom(h) = \mathbb{R} - \{0\}

.

2. Nested Fractions

For

f(x)

, we must ensure:
1)

x+1 \neq 0 \implies x \neq -1

.
2) The entire denominator

1+\frac{1}{x+1} \neq 0

.
Solving

1+\frac{1}{x+1} = 0 \implies \frac{1}{x+1} = -1 \implies x+1 = -1 \implies x = -2

.
Domain:

\mathbb{R} - \{-1, -2\}

.

Worked Example 02Easy

Decompose

F(x) = \cos^2(x+9)

into

F = f \circ g \circ h

.

How to think about it

Key idea: Identify the 'inner', 'middle', and 'outer' layers of the operation.

Full solution

1. Layer Identification

Inner layer:

h(x) = x+9

. Middle layer:

g(x) = \cos(x)

. Outer layer:

f(x) = x^2

. Then

f(g(h(x))) = (\cos(x+9))^2 = \cos^2(x+9)

.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

Given

f_0(x) = \frac{x}{x+1}

and

f_{n+1} = f_0 \circ f_n

, find a general formula for

f_n(x)

.

Need a hint?

Compute the first few terms (

f_1, f_2

) and look for a pattern.

Show solution

1. Pattern Recognition

f_1(x) = f_0(f_0(x)) = \frac{\frac{x}{x+1}}{\frac{x}{x+1} + 1} = \frac{x}{x + (x+1)} = \frac{x}{2x+1}

.

2. Inductive Step

f_2(x) = f_0(f_1(x)) = \frac{\frac{x}{2x+1}}{\frac{x}{2x+1} + 1} = \frac{x}{x + (2x+1)} = \frac{x}{3x+1}

.
By observation,

f_n(x) = \frac{x}{(n+1)x + 1}

.

Practice 02Medium

Given

g(x) = 2x+1

and

(f \circ g)(x) = 4x^2 + 4x + 7

, find

f(x)

.

Need a hint?

Use substitution: let

u = g(x)

.

Show solution

1. Variable Substitution

Let

u = 2x + 1

, which means

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

.

2. Solve for f(u)

f(u) = 4(\frac{u-1}{2})^2 + 4(\frac{u-1}{2}) + 7 = (u-1)^2 + 2(u-1) + 7 = u^2 - 2u + 1 + 2u - 2 + 7 = u^2 + 6

. Thus,

f(x) = x^2 + 6

.

Common Pitfalls

⚠
Domain of Composite FunctionsDo not just look at the final simplified form of $f(g(x))$ . You must exclude any values that make $g(x)$ undefined, even if they seem 'fine' after plugging into $f$ .

Symmetry and Inverse Functions

Understanding the symmetry of functions (Even/Odd) and the conditions for the existence of inverses is fundamental for sketching graphs and solving equations in Calculus.

Core Theorem

1. Parity (Even/Odd):
- Even Function:

f(-x) = f(x)

, symmetric about the

y

-axis.
- Odd Function:

f(-x) = -f(x)

, symmetric about the origin.

2. Inverse Functions (

f^{-1}

):
- Exists if

f

is one-to-one.
-

f^{-1}(y) = x \iff f(x) = y

.
- The domain of

f^{-1}

is the range of

f

, and vice versa.
- Graphs of

f

and

f^{-1}

are reflections across the line

y = x

.

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Easy

If

g(x)

is an odd function, what can we conclude about

g(0)

? What if

g(x)

is an even function?

How to think about it

Key idea: Plug $x = 0$ into the definitions $g(-x) = -g(x)$ and $g(-x) = g(x)$ .

Full solution

1. Odd Function Case

g(-0) = -g(0) \implies g(0) = -g(0) \implies 2g(0) = 0

. Therefore, if

0 \in Dom(g)

, then

g(0) = 0

.

2. Even Function Case

g(-0) = g(0) implies g(0) = g(0)

. This is always true, so

g(0)

can be any value in the range.

Worked Example 02Medium

Determine the parity of

f(x) = x|x|

.

How to think about it

Key idea: Test $f(-x)$ and see if it equals $f(x)$ or $-f(x)$ .

Full solution

1. Substitution

f(-x) = (-x)|-x|

.

2. Simplification

Since

|-x| = |x|

, we have

f(-x) = -x|x| = -f(x)

.

3. Conclusion

Since

f(-x) = -f(x)

, the function is an odd function.

Step-by-Step SOP

1
Determining Parity
Always start by replacing $x$ with $(-x)$ in the original expression and simplify.
2
Finding Inverse Algebraically
1. Set $y = f(x)$ . 2. Exchange $x$ and $y$ . 3. Solve for $y$ . 4. Replace $y$ with $f^{-1}(x)$ .

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

Find the inverse of

f(x) = x^3 + 2

.

Need a hint?

Swap

x

and

y

, then solve for the new

y

.

Show solution

1. Swap Variables

Let

x = y^3 + 2

.

2. Isolate y

y^3 = x - 2 \implies y = \sqrt[3]{x-2}

.

3. Result

f^{-1}(x) = \sqrt[3]{x-2}

.

Practice 02Hard

Under what condition does

sin(x)

have an inverse?

Need a hint?

A function must be one-to-one to have an inverse.

Show solution

1. Analyze Domain

On

[0, \pi]

,

\sin(x)

is not one-to-one (e.g.,

\sin(\pi/6) = \sin(5\pi/6)

).

2. Restrict Domain

On

[-\pi/2, \pi/2]

,

\sin(x)

is strictly increasing and therefore one-to-one.

3. Define Inverse

We can define

\sin^{-1}: [-1, 1] \to [-\pi/2, \pi/2]

.

Common Pitfalls

⚠
Inverse Notation Confusion $f^{-1}(x)$ is NOT the same as $\frac{1}{f(x)}$ . The former is the inverse function, while the latter is the reciprocal.
⚠
Parity is not BinaryA function can be neither even nor odd (e.g., $f(x) = x^3 - 3x^2$ ). Don't assume every function must fall into one of these categories.

Worked Examples

1. Establish Variables

2. Calculate Areas

3. Formulate Cost Function

1. Establish System of Equations

2. Algebraic Substitution

3. Solve for h

Testing for a Function

Practice — Your Turn

1. Define the Step Logic

2. Formulate the Piecewise Function

1. The Natural Domain

Worked Examples

1. Domain Calculation

2. Range Calculation

System of Inequalities

Practice — Your Turn

1. Analyze the First Term

2. Analyze the Second Term

3. Find the Intersection

1. Inner Root Condition

2. Outer Root and Trigonometric Logic

3. Combine and Solve for x

Worked Examples

1. Factorization

2. Analyzing the Quadratic Factor

3. Conclusion

Horizontal Line Test

Worked Examples

1. Simplification Trap

2. Nested Fractions

1. Layer Identification

Practice — Your Turn

1. Pattern Recognition

2. Inductive Step

1. Variable Substitution

2. Solve for f(u)

Worked Examples

1. Odd Function Case

2. Even Function Case

1. Substitution

2. Simplification

3. Conclusion

Determining Parity

Finding Inverse Algebraically

Practice — Your Turn

1. Swap Variables

2. Isolate y

3. Result

1. Analyze Domain

2. Restrict Domain

3. Define Inverse

Gary Chang