Function | NEB Class 11 Mathematics Notes

Chapter Overview

A Function is a specialized type of relation where every element in the starting set (domain) is connected to exactly one element in the target set (co-domain). Unlike general relations, functions follow a strict rule: no input can produce multiple outputs. Think of a function as a 'machine' that takes an input value and produces a single, definite output value. This chapter bridges your understanding from relations to functions, exploring how to identify, classify, and work with functions using various representations.

What this chapter covers

Relations and Functions: Differences and ConnectionsDomain, Co-domain, and RangeFunction Representations (Mapping, Ordered Pairs, Tables, Graphs, Equations)Vertical Line Test for Function IdentificationClassification of Functions (One-to-one, Onto, Bijective)Algebra of Functions (Sum, Difference, Product, Quotient)Composition of FunctionsInverse Functions and Finding InversesTypes of Functions (Algebraic, Transcendental, Logarithmic, Exponential)Logarithmic Functions and Properties

Learning Objectives

1

Clearly distinguish between a relation and a function using formal definitions and practical examples.

2

Accurately identify and calculate the domain, co-domain, and range of functions presented in various forms.

3

Represent functions using multiple methods: mapping diagrams, ordered pairs, tables, graphs, and algebraic equations.

4

Apply the Vertical Line Test to determine whether a given graph represents a function.

5

Classify functions as injective (one-to-one), surjective (onto), or bijective, and prove these properties algebraically.

6

Perform algebraic operations on functions including addition, subtraction, multiplication, and division.

7

Understand function composition $(g \circ f)(x)$ and compute composite functions accurately.

8

Find inverse functions and verify the correctness of inverses using composition properties.

9

Work with logarithmic and exponential functions, applying key properties and formulas.

10

Solve complex function-related problems from NEB examination papers.

The Essence of Functions: Order and Uniqueness

Key Insight

The core principle that distinguishes a function from a general relation is uniqueness of output. Every input must map to exactly one output. This seemingly simple rule creates a mathematical structure so powerful that it becomes the foundation of calculus, coordinate geometry, and all higher mathematics.

Think of it this way: A function is like a ticket machine at a cinema. You insert a ticket number (input), and the machine outputs exactly one movie (output). No ticket number can produce two different movies. This uniqueness is what makes functions reliable, predictable, and useful for modeling real-world phenomena.

Critical distinction: A relation allows one input to connect to multiple outputs. But a function forbids this. The moment you see an input with multiple outputs, you know it is not a function.

In NEB exams, the most common way to test your understanding of this concept is through the Vertical Line Test. If a vertical line intersects a graph at more than one point, it is not a function. This visual test embodies the uniqueness principle perfectly.

Key Definitions

Relation

A relation from set $A$ to set $B$ is a subset of the Cartesian product $A \times B$. In other words, a relation is any collection of ordered pairs $(a, b)$ where $a \in A$ and $b \in B$. A relation has no restrictions; one element of $A$ can relate to zero, one, or many elements of $B$.

Function

A function $f: A \to B$ is a special relation where every element in set $A$ (the domain) is associated with exactly one element in set $B$ (the co-domain). Formally, if $f(x) = y$ and $f(x) = z$, then $y = z$. This uniqueness of output is the defining characteristic of a function.

Domain

The domain of a function $f: A \to B$ is the set $A$. It consists of all possible input values for which the function is defined. When finding the domain, we must identify which values are allowed (e.g., denominators cannot be zero, values under square roots must be non-negative).

Co-domain

The co-domain of a function $f: A \to B$ is the set $B$. It is the 'target set' into which the function is intended to map all elements of the domain. The co-domain may include elements that are never actually 'reached' by the function.

Range

The range of a function is the set of all actual outputs (images) produced by the function. It is a subset of the co-domain. Formally, Range$(f) = \{f(x) : x \in \text{Domain}(f)\}$. The range contains only values that are actually mapped to; the co-domain may contain additional elements.

Image and Pre-image

If $f(x) = y$, then $y$ is called the image of $x$ under the function $f$, and $x$ is called a pre-image of $y$. Every element in the domain has an image, but not every element in the co-domain has a pre-image.

One-to-One Function (Injective)

A function $f: A \to B$ is one-to-one (injective) if different inputs produce different outputs. Formally, if $f(x_1) = f(x_2)$, then $x_1 = x_2$. Equivalently, if $x_1 \neq x_2$, then $f(x_1) \neq f(x_2)$. Graphically, the Horizontal Line Test can identify injective functions.

Onto Function (Surjective)

A function $f: A \to B$ is onto (surjective) if every element in the co-domain $B$ has at least one pre-image in the domain $A$. In other words, Range$(f) =$ Co-domain$(B)$. Every element in $B$ is the image of some element in $A$.

Bijective Function

A function is bijective if it is both one-to-one (injective) and onto (surjective). Only bijective functions have inverses. A bijection establishes a perfect 'pairing' between elements of the domain and co-domain, with no element left unpaired.

Inverse Function

If $f: A \to B$ is a bijection, then the inverse function $f^{-1}: B \to A$ is defined such that $f^{-1}(b) = a$ if and only if $f(a) = b$. The inverse 'undoes' the function. We have the property: $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$ for all appropriate $x$.

Function Composition

If $f: A \to B$ and $g: B \to C$ are two functions, their composition $(g \circ f): A \to C$ is defined by $(g \circ f)(x) = g(f(x))$. We read this as '$g$ composed with $f$'. The function $f$ is applied first, then $g$ is applied to the result.

Relation vs. Function: The Critical Differences

Understanding how relations differ from functions is fundamental. Every function is a relation, but not every relation is a function. Use this comparison to identify which is which.

Relation

$$General Connection$$

Definition: Any subset of the Cartesian product $A \times B$.

Uniqueness: One element from $A$ can connect to zero, one, or many elements from $B$.

Example: 'Is the parent of' relation. Person $x$ can be the parent of multiple persons (children).

Representation: Can show one-to-many, many-to-one, and many-to-many connections.

Testing Visually: No special test required; simply look for all connections.

VS

Function

$$Strict Rule$$

Definition: A relation where every input has exactly one output.

Uniqueness: Every element from $A$ must connect to exactly one element from $B$. Never zero outputs, never multiple outputs.

Example: '$f(x) = x + 1$ function'. Input 2 always produces output 3. Never two outputs.

Representation: Only one-to-one connections allowed (from input perspective).

Testing Visually: Use the Vertical Line Test. If a vertical line cuts the graph more than once, it is not a function.

Memory Aid - V.L.T.: The Vertical Line Test is your visual check. Draw a vertical line anywhere on the graph. If it intersects the curve at exactly one point everywhere, it is a function. If at any position it intersects at two or more points, it is not a function.

Formula Sheet & Key Properties

Logarithm Product Rule

\log_a(xy) = \log_a x + \log_a y

The logarithm of a product equals the sum of the logarithms. This is one of the most important properties for simplifying logarithmic expressions.

1Let $\log_a x = m$, which means $x = a^m$.

2Let $\log_a y = n$, which means $y = a^n$.

3Now multiply the two exponential forms: $xy = a^m \cdot a^n$.

4Using the law of exponents: $xy = a^{m+n}$.

5Convert back to logarithmic form: $\log_a(xy) = m + n$.

6Substitute our original definitions: $\log_a(xy) = \log_a x + \log_a y$. (Proved)

Logarithm Quotient Rule

\log_a\left(\frac{x}{y}\right) = \log_a x - \log_a y

The logarithm of a quotient equals the difference of the logarithms.

1Let $\log_a x = m$, so $x = a^m$.

2Let $\log_a y = n$, so $y = a^n$.

3Form the quotient: $\frac{x}{y} = \frac{a^m}{a^n}$.

4Using the law of exponents: $\frac{x}{y} = a^{m-n}$.

5Convert to logarithmic form: $\log_a\left(\frac{x}{y}\right) = m - n$.

6Substitute: $\log_a\left(\frac{x}{y}\right) = \log_a x - \log_a y$. (Proved)

Logarithm Power Rule

\log_a(x^p) = p \log_a x

The logarithm of a power equals the exponent times the logarithm of the base.

1Let $\log_a x = m$, which means $x = a^m$.

2Raise both sides to the power $p$: $x^p = (a^m)^p = a^{mp}$.

3Convert to logarithmic form: $\log_a(x^p) = mp$.

4Substitute $m = \log_a x$: $\log_a(x^p) = p \cdot \log_a x$. (Proved)

Change of Base Formula

\log_a x = \frac{\log_b x}{\log_b a}

Convert logarithms from one base to another. Commonly used to calculate logarithms with calculators (base 10 or natural log).

1Let $y = \log_a x$, which means $a^y = x$.

2Take logarithm (base $b$) of both sides: $\log_b(a^y) = \log_b x$.

3Apply the power rule: $y \cdot \log_b a = \log_b x$.

4Solve for $y$: $y = \frac{\log_b x}{\log_b a}$.

5Substitute back: $\log_a x = \frac{\log_b x}{\log_b a}$. (Proved)

Special Logarithm Values

\log_a 1 = 0, \quad \log_a a = 1, \quad \log_a a^p = p

These fundamental values simplify calculations and appear frequently in exams.

1For $\log_a 1 = 0$: We need $a^? = 1$. Since $a^0 = 1$ for any $a > 0, a \neq 1$, we have $\log_a 1 = 0$.

2For $\log_a a = 1$: We need $a^? = a$. Since $a^1 = a$, we have $\log_a a = 1$.

3For $\log_a a^p = p$: We need $a^? = a^p$. Since the exponent is $p$, we have $\log_a a^p = p$.

Algebra of Functions

(f \pm g)(x) = f(x) \pm g(x), \quad (f \cdot g)(x) = f(x) \cdot g(x), \quad \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}, g(x) \neq 0

These are operations you can perform when combining two functions.

1Sum/Difference: The sum (or difference) of two functions is computed by adding (or subtracting) their outputs for the same input. This is straightforward and intuitive.

2Product: The product of two functions is computed by multiplying their outputs for the same input.

3Quotient: The quotient is computed by dividing the output of $f$ by the output of $g$. We must exclude values where $g(x) = 0$ to avoid division by zero.

Function Composition

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

Function composition applies functions in sequence. The function $f$ is applied first, then $g$ is applied to the result.

1Composition $(g \circ f)(x)$ means: first compute $f(x)$, then apply $g$ to that result.

2For example, if $f(x) = x + 1$ and $g(x) = x^2$, then $(g \circ f)(x) = g(f(x)) = g(x+1) = (x+1)^2$.

3Note that $(g \circ f)(x) \neq (f \circ g)(x)$ in general. Composition is not commutative.

Step-by-Step: Finding Domain and Range

When faced with a function in the exam, follow this logical sequence to find its domain and range correctly.

Does the function contain a square root, cube root, or even root?

Yes

The expression under an even root must be $\geq 0$. Set up an inequality and solve.

No

Continue to next step ↓

Does the function have a denominator (fraction)?

Yes

The denominator must never equal zero. Set denominator = 0, solve, and exclude those values from domain.

No

Continue to next step ↓

Does the function contain logarithms?

Yes

The argument of a logarithm must be strictly positive (> 0). Set up inequalities for all logarithmic arguments.

No

Continue to next step ↓

Are there any other restrictions mentioned in the problem?

Yes

Apply those restrictions to the domain.

No

If no restrictions apply, the domain is all real numbers ($\mathbb{R}$). Now find the range.

To find the range, can you solve for $x$ in terms of $y$ (where $y = f(x)$)?

Yes

If you can express $x = g(y)$, then the restrictions that prevent valid $x$ values tell you the excluded $y$ values. These exclusions form the complement of the range.

No

Sketch the graph or analyze the behavior. Identify the minimum and maximum values (if they exist).

Top 5 Memory Tricks for Functions and Logarithms

Trick: These memory devices will help you recall key concepts and formulas quickly during exams.

Steps to Remember:

Trick 1 - V.L.T. (Vertical Line Test): A function means Vertical lines should cut the graph at exactly one point. Remember 'V' for 'Vertical' and 'V' for 'Valid function'. If the vertical line cuts twice, the graph is not a valid function.
Trick 2 - B-I-O for Bijective: Bijective = Injective (one-one) + Onto (onto). A bijective function must be both. Think 'B' stands for 'Both'.
Trick 3 - 'Log Multiply = Add': When you see $\log(A \times B)$, remember it becomes $\log A + \log B$. Multiplication inside the log becomes addition outside. 'Multiply to Add' rhyme helps recall this.
Trick 4 - 'Inverse = Swap and Solve': To find $f^{-1}(x)$ from $f(x)$: (1) Write $y = f(x)$, (2) Swap $x$ and $y$, (3) Solve for $y$. This three-step process never changes.
Trick 5 - 'Range is Real; Co-domain is Target': The co-domain is the 'intended target' (what the problem says), but the range is 'what actually happens' (what the function produces). Range $\subseteq$ Co-domain always.

Mnemonic: Remember: Functions demand UNIQUENESS of output. Domain = Starting Set. Range = Actual Outputs. Codomain = Intended Target. VLT = Function Check Tool.

Common Mistakes in NEB Function Exams

These errors are made by many students and result in significant mark loss. Study them carefully to avoid repeating them.

Mistake 1 - Confusing Range with Co-domain: A student writes: 'The function $f(x) = x^2$ has range $\mathbb{R}$.' This is incorrect! The range of $f(x) = x^2$ is $[0, \infty)$ (only non-negative values). The co-domain might be $\mathbb{R}$, but the range excludes negative numbers. Always compute the actual outputs, not just the stated target set.
Mistake 2 - Forgetting Domain Restrictions: Given $f(x) = \sqrt{x-3}$, a student writes: 'Domain = $\mathbb{R}$'. Wrong! The expression under the square root must be non-negative: $x - 3 \geq 0$, so $x \geq 3$. Domain = $[3, \infty)$. Always check for square roots, fractions, and logarithms.
Mistake 3 - Incorrect Inverse Procedure: A student tries to find the inverse of $f(x) = 2x + 3$ but writes: '$f^{-1}(x) = \frac{x}{2} - 3$'. The error is not swapping correctly. The correct method: $y = 2x + 3 \Rightarrow x = 2y + 3 \Rightarrow x - 3 = 2y \Rightarrow y = \frac{x-3}{2}$. So $f^{-1}(x) = \frac{x-3}{2}$.
Mistake 4 - Wrong Order in Composition: Given $f(x) = x+1$ and $g(x) = x^2$, a student computes $g \circ f$ as $f(g(x)) = x^2 + 1$. This is wrong! $(g \circ f)(x)$ means $g(f(x)) = g(x+1) = (x+1)^2 = x^2 + 2x + 1$. The notation $(g \circ f)$ means apply $f$ first, then $g$.
Mistake 5 - Logarithm Domain Errors: Given $f(x) = \log(x^2 - 1)$, a student writes: 'Domain = $x > 0$'. Incorrect! The argument must be positive: $x^2 - 1 > 0 \Rightarrow (x-1)(x+1) > 0 \Rightarrow x < -1$ or $x > 1$. Domain = $(-\infty, -1) \cup (1, \infty)$. Note: $x^2 > 1$ is the correct condition, not $x > 0$.
Mistake 6 - Not Checking if Inverse Exists: Students often try to find the inverse of a function that is not bijective. The inverse of $f(x) = x^2$ does not exist on $\mathbb{R} \to \mathbb{R}$ because it is not one-to-one. You must first verify that the function is bijective (or restrict the domain to make it bijective) before finding an inverse.

Solved Examples (Easy → Medium → Hard)

1

Find $f(3)$ if $f(x) = 2x^2 - 5x + 1$.

Easy

Solution

To find $f(3)$, we substitute $x = 3$ into the function.

Given: $f(x) = 2x^2 - 5x + 1$

$f(3) = 2(3)^2 - 5(3) + 1$

$f(3) = 2(9) - 15 + 1$

$f(3) = 18 - 15 + 1$

$f(3) = 4$

Therefore, the output of the function when the input is 3 is 4.

∴ $f(3) = 4$

2

Identify the domain of $f(x) = \frac{1}{x-2}$.

Easy

Solution

For a fraction to be defined, the denominator can never equal zero.

Given: $f(x) = \frac{1}{x-2}$

The denominator is $(x-2)$. Setting it equal to zero:
$x - 2 = 0$
$x = 2$

This means the function is undefined at $x = 2$. All other real numbers are valid inputs.

Therefore, the domain excludes 2 and includes all other real numbers.

In set notation: Domain = $\mathbb{R} - \{2\}$ or $(-\infty, 2) \cup (2, \infty)$.

∴ Domain = $\mathbb{R} - \{2\}$ or $(-\infty, 2) \cup (2, \infty)$

3

Given $f = \{(1, 2), (2, 3), (3, 4)\}$, find the range of $f$.

Easy

Solution

The range of a function is the set of all output values (the second coordinates of the ordered pairs).

Given the function as ordered pairs: $f = \{(1, 2), (2, 3), (3, 4)\}$

The outputs are: 2 (from the pair $(1, 2)$), 3 (from the pair $(2, 3)$), and 4 (from the pair $(3, 4)$).

Therefore, the range is the set containing all these output values.

Range = $\{2, 3, 4\}$

∴ Range = $\{2, 3, 4\}$

4

Evaluate $\log_3 81$.

Easy

Solution

We need to find the power to which 3 must be raised to get 81.

The problem asks: $3^? = 81$

Let's express 81 as a power of 3:
$3^1 = 3$
$3^2 = 9$
$3^3 = 27$
$3^4 = 81$

So $3^4 = 81$.

Therefore, $\log_3 81 = 4$ because the exponent is 4.

∴ $\log_3 81 = 4$

5

Find the inverse of the function $f(x) = 3x + 5$.

Medium

Solution

To find the inverse function, we follow the standard procedure: write $y = f(x)$, swap $x$ and $y$, and solve for $y$.

Step 1: Write the equation with $y$:
$y = 3x + 5$

Step 2: Swap the variables $x$ and $y$ (this is the key step that 'reverses' the function):
$x = 3y + 5$

Step 3: Solve for $y$ (now $y$ is the dependent variable):
$x = 3y + 5$
$x - 5 = 3y$
$y = \frac{x - 5}{3}$

Step 4: The inverse function is:
$f^{-1}(x) = \frac{x - 5}{3}$

Verification: Let's check: $f(f^{-1}(x)) = f\left(\frac{x-5}{3}\right) = 3 \cdot \frac{x-5}{3} + 5 = (x-5) + 5 = x$. ✓

∴ $f^{-1}(x) = \frac{x - 5}{3}$

6

Find the domain of $f(x) = \sqrt{x - 4}$.

Medium

Solution

For a square root (or any even root) to produce a real number, the expression under the root must be non-negative.

Given: $f(x) = \sqrt{x - 4}$

The expression under the square root is $(x - 4)$. This must be greater than or equal to zero:
$x - 4 \geq 0$

Solving this inequality:
$x \geq 4$

This means the function is defined for all real numbers greater than or equal to 4. The value $x = 4$ is included because $\sqrt{4-4} = \sqrt{0} = 0$, which is a valid real number.

Therefore, the domain in interval notation is $[4, \infty)$.

∴ Domain = $[4, \infty)$

7

If $f(x) = 2x + 1$ and $g(x) = x^2$, find $(g \circ f)(x)$.

Medium

Solution

The composition $(g \circ f)(x) = g(f(x))$ means: first apply $f$ to get an intermediate result, then apply $g$ to that result.

Given: $f(x) = 2x + 1$ and $g(x) = x^2$

Step 1: Calculate $f(x)$:
$f(x) = 2x + 1$

Step 2: Now apply $g$ to this result. We substitute the entire expression $f(x) = 2x + 1$ into $g$:
$(g \circ f)(x) = g(f(x)) = g(2x + 1)$

Step 3: The function $g$ squares its input. So we square $(2x + 1)$:
$(g \circ f)(x) = (2x + 1)^2$

Step 4: Expand using the square of a binomial formula $(a+b)^2 = a^2 + 2ab + b^2$:
$(2x + 1)^2 = (2x)^2 + 2(2x)(1) + 1^2$
$= 4x^2 + 4x + 1$

Therefore, $(g \circ f)(x) = 4x^2 + 4x + 1$.

∴ $(g \circ f)(x) = 4x^2 + 4x + 1$

8

Determine the range of $f(x) = \frac{1}{x+1}$ for $x \in \mathbb{R}, x \neq -1$.

Medium

Solution

To find the range, we set $y = f(x)$ and solve for $x$ in terms of $y$. The values of $y$ for which we can solve are exactly the range values.

Given: $y = \frac{1}{x+1}$

Step 1: Rearrange to isolate the denominator:
$y(x + 1) = 1$

Step 2: Expand:
$yx + y = 1$

Step 3: Collect terms with $x$:
$yx = 1 - y$

Step 4: Solve for $x$:
$x = \frac{1 - y}{y}$

Step 5: For $x$ to be real, we need to avoid division by zero. The denominator $y$ cannot equal zero:
$y \neq 0$

This means that no matter what real number we choose for $y$ (except 0), we can find a corresponding $x$. Therefore, the range is all real numbers except 0.

Range = $\mathbb{R} - \{0\}$ or $(-\infty, 0) \cup (0, \infty)$.

∴ Range = $\mathbb{R} - \{0\}$ or $(-\infty, 0) \cup (0, \infty)$

9

Prove that $f(x) = 2x + 3$ is a one-to-one (injective) function on $\mathbb{R}$.

Hard

Solution

To prove a function is one-to-one, we must show that if $f(x_1) = f(x_2)$, then necessarily $x_1 = x_2$. This means different inputs cannot produce the same output.

Given: $f(x) = 2x + 3$

Proof:

Assume $f(x_1) = f(x_2)$ for some $x_1, x_2 \in \mathbb{R}$.

Then:
$2x_1 + 3 = 2x_2 + 3$

Subtract 3 from both sides (subtraction preserves equality):
$2x_1 = 2x_2$

Divide both sides by 2 (division preserves equality for non-zero divisors):
$x_1 = x_2$

We have shown that if two inputs produce the same output, they must be the same input. This is the definition of a one-to-one function.

Therefore, $f(x) = 2x + 3$ is injective (one-to-one) on $\mathbb{R}$. ✓

∴ Proved using the definition: $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$.

10

Find the domain of $f(x) = \sqrt{9 - x^2}$.

Hard

Solution

For the square root to be defined (to produce a real number), the expression under the root must be non-negative.

Given: $f(x) = \sqrt{9 - x^2}$

We need: $9 - x^2 \geq 0$

Step 1: Rearrange the inequality:
$9 - x^2 \geq 0$
$9 \geq x^2$
$x^2 \leq 9$

Step 2: Take square roots of both sides. When taking the square root of an inequality, we get:
$|x| \leq 3$

Step 3: Interpret the absolute value inequality. The condition $|x| \leq 3$ means the distance from $x$ to zero is at most 3:
$-3 \leq x \leq 3$

Step 4: Both boundary points are included because the original inequality had 'equal to' ($\geq$):
- At $x = -3$: $f(-3) = \sqrt{9 - 9} = \sqrt{0} = 0$ ✓ (defined)
- At $x = 3$: $f(3) = \sqrt{9 - 9} = \sqrt{0} = 0$ ✓ (defined)

Therefore, the domain is the closed interval from -3 to 3.

Domain = $[-3, 3]$

∴ Domain = $[-3, 3]$

11

Simplify $\log_2 32 + \log_2 8 - \log_2 4$ using logarithm properties.

Hard

Solution

We use the logarithm properties to combine and simplify this expression.

Given: $\log_2 32 + \log_2 8 - \log_2 4$

Method 1: Using Properties Directly

Step 1: Apply the product rule $\log_a x + \log_a y = \log_a(xy)$:
$\log_2 32 + \log_2 8 = \log_2(32 \times 8) = \log_2 256$

Step 2: Apply the quotient rule $\log_a x - \log_a y = \log_a(\frac{x}{y})$:
$\log_2 256 - \log_2 4 = \log_2\left(\frac{256}{4}\right) = \log_2 64$

Step 3: Evaluate $\log_2 64$. We ask: $2^? = 64$
$2^1 = 2$
$2^2 = 4$
$2^3 = 8$
$2^4 = 16$
$2^5 = 32$
$2^6 = 64$

So $\log_2 64 = 6$.

Method 2: Converting to Powers (Verification)

$32 = 2^5$, so $\log_2 32 = 5$
$8 = 2^3$, so $\log_2 8 = 3$
$4 = 2^2$, so $\log_2 4 = 2$

Therefore: $5 + 3 - 2 = 6$ ✓

The answer is 6.

∴ $\log_2 32 + \log_2 8 - \log_2 4 = 6$

12

Prove that $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^3$ is bijective.

Hard

Solution

To prove a function is bijective, we must show two things: (1) it is one-to-one (injective), and (2) it is onto (surjective).

Given: $f(x) = x^3$, with domain and co-domain both being $\mathbb{R}$.

Part 1: Prove $f$ is One-to-One (Injective)

We must show: if $f(x_1) = f(x_2)$, then $x_1 = x_2$.

Assume $f(x_1) = f(x_2)$:
$x_1^3 = x_2^3$

Take the cube root of both sides (cube root is a one-to-one function):
$\sqrt[3]{x_1^3} = \sqrt[3]{x_2^3}$
$x_1 = x_2$

Therefore, $f$ is injective. ✓

Part 2: Prove $f$ is Onto (Surjective)

We must show: for every $y \in \mathbb{R}$ (the co-domain), there exists an $x \in \mathbb{R}$ (the domain) such that $f(x) = y$.

Let $y$ be any real number. We need to find $x$ such that $x^3 = y$.

The solution is: $x = \sqrt[3]{y}$

Since every real number has a real cube root (unlike even roots), this $x$ is always a real number in the domain $\mathbb{R}$.

For this $x$: $f(x) = f(\sqrt[3]{y}) = (\sqrt[3]{y})^3 = y$

Therefore, every $y$ in the co-domain has a pre-image in the domain. Thus, $f$ is surjective. ✓

Conclusion:

Since $f$ is both injective and surjective, it is bijective. ✓

∴ Proved by showing $f$ is both one-to-one and onto.

MCQ Practice

Mixed Level — MCQ

1

Which of the following relations is a function?

Correct!

Incorrect. Correct: $\{(1, 2), (2, 3), (3, 4)\}$

Solution

A relation is a function only if every input appears at most once (and every input in the domain appears exactly once). Check each relation for repeated first coordinates. Only the second option has no repeated inputs.

∴ $\{(1, 2), (2, 3), (3, 4)\}$

2

The domain of $f(x) = \frac{1}{x-3}$ is:

Correct!

Incorrect. Correct: $\mathbb{R} - \{3\}$

Solution

For a fraction, the denominator cannot equal zero. Setting $x - 3 = 0$ gives $x = 3$. This is the only value to exclude. The domain includes all real numbers except 3.

∴ $\mathbb{R} - \{3\}$

3

If $f(x) = 2x + 1$ and $g(x) = x^2 - 1$, then $(f \circ g)(2) = ?$

Correct!

Incorrect. Correct: $7$

Solution

$(f \circ g)(2)$ means apply $g$ first, then apply $f$ to the result. Step 1: $g(2) = 2^2 - 1 = 4 - 1 = 3$. Step 2: $f(3) = 2(3) + 1 = 6 + 1 = 7$.

∴ $7$

4

The range of $f(x) = x^2 + 1$ for $x \in \mathbb{R}$ is:

Correct!

Incorrect. Correct: $[1, \infty)$

Solution

The function $f(x) = x^2 + 1$ has $x^2 \geq 0$, so $f(x) \geq 1$. At $x = 0$, we get $f(0) = 1$ (the minimum). As $x$ increases or decreases from zero, $x^2$ increases, making $f(x)$ larger. There is no upper bound.

∴ $[1, \infty)$

5

$\log_2(16) = ?$

Correct!

Incorrect. Correct: $4$

Solution

The logarithm $\log_2(16)$ asks: to what power must we raise 2 to get 16? Computing: $2^1=2$, $2^2=4$, $2^3=8$, $2^4=16$. The answer is 4.

∴ $4$

6

Which statement about logarithms is true?

Correct!

Incorrect. Correct: $\log_a(xy) = \log_a x + \log_a y$

Solution

The Product Rule states that the log of a product equals the sum of the logs. This is one of the three main logarithm properties (Product, Quotient, Power).

∴ $\log_a(xy) = \log_a x + \log_a y$

7

The inverse function of $f(x) = \frac{x+2}{3}$ is:

Correct!

Incorrect. Correct: $f^{-1}(x) = 3x - 2$

Solution

To find the inverse: write $y = \frac{x+2}{3}$, swap variables to get $x = \frac{y+2}{3}$, then solve for $y$: $3x = y + 2 \Rightarrow y = 3x - 2$.

∴ $f^{-1}(x) = 3x - 2$

8

A function is onto (surjective) if:

Correct!

Incorrect. Correct: Every element in the co-domain has at least one pre-image

Solution

The definition of 'onto' is that the range equals the co-domain. This means every element in the target set is reached by at least one input value. This is distinct from 'one-to-one', which is about avoiding repeated outputs.

∴ Every element in the co-domain has at least one pre-image

9

If $f(x) = \sqrt{x}$, what is the domain?

Correct!

Incorrect. Correct: $[0, \infty)$

Solution

For the square root to be real, the argument must be non-negative: $x \geq 0$. This includes zero since $\sqrt{0} = 0$ is valid. Domain = $[0, \infty)$.

∴ $[0, \infty)$

10

The value of $\log_a 1$ for any valid base $a$ is:

Correct!

Incorrect. Correct: $0$

Solution

The logarithm $\log_a 1$ asks: $a^? = 1$. Since any number to the power of 0 equals 1, the answer is always 0, regardless of the base.

∴ $0$

11

Which of the following functions is bijective on $\mathbb{R} \to \mathbb{R}$?

Correct!

Incorrect. Correct: $f(x) = 2x + 5$

Solution

A bijection requires both one-to-one and onto properties. Linear functions with non-zero slope are bijective on $\mathbb{R}$. The function $f(x) = 2x + 5$ satisfies both: if $f(a) = f(b)$, then $2a + 5 = 2b + 5 \Rightarrow a = b$ (one-to-one), and for any $y \in \mathbb{R}$, we can solve $2x + 5 = y$ to get $x = \frac{y-5}{2}$ (onto).

∴ $f(x) = 2x + 5$

Short Questions (2-4 Marks Each)

1

Define a function and explain how it differs from a relation.

[2 marks]

Solution

Definition of Function: A function $f: A \to B$ is a relation in which every element of the domain $A$ is associated with exactly one element in the co-domain $B$.

Definition of Relation: A relation from $A$ to $B$ is any subset of the Cartesian product $A \times B$. It simply pairs elements from $A$ with elements from $B$ with no restrictions.

Key Difference: A relation allows one element of $A$ to relate to zero, one, or many elements of $B$. A function requires that every element of $A$ relates to exactly one element of $B$.

Example to Illustrate: Consider the relation 'shares a birthday with' among students. This is a relation but not a function because one student can share a birthday with multiple other students. However, the relation '$f(x) = x + 1$' is a function because each input $x$ produces exactly one output $x + 1$.

∴ A function is a special relation where every input has exactly one output. Relations have no such restriction.

2

Find the range of $f(x) = 2x + 3$ where $x \in \{1, 2, 3, 4\}$.

[2 marks]

Solution

The range consists of all the outputs (images) produced when we apply the function to each element of the domain.

Given: $f(x) = 2x + 3$ and domain = $\{1, 2, 3, 4\}$

Calculate $f(x)$ for each input:
- $f(1) = 2(1) + 3 = 2 + 3 = 5$
- $f(2) = 2(2) + 3 = 4 + 3 = 7$
- $f(3) = 2(3) + 3 = 6 + 3 = 9$
- $f(4) = 2(4) + 3 = 8 + 3 = 11$

The outputs are: $5, 7, 9, 11$

Therefore, Range = $\{5, 7, 9, 11\}$

∴ Range = $\{5, 7, 9, 11\}$

3

Verify that $f(x) = 3x + 1$ and $g(x) = \frac{x-1}{3}$ are inverse functions.

[3 marks]

Solution

Two functions are inverses if and only if their compositions both yield the identity function. We must verify both $f(g(x)) = x$ and $g(f(x)) = x$.

Check 1: Compute $f(g(x))$

$f(g(x)) = f\left(\frac{x-1}{3}\right)$

$= 3 \cdot \frac{x-1}{3} + 1$

$= (x - 1) + 1$

$= x$ ✓

Check 2: Compute $g(f(x))$

$g(f(x)) = g(3x + 1)$

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

$= \frac{3x}{3}$

$= x$ ✓

Since $f(g(x)) = x$ and $g(f(x)) = x$, we have verified that $f$ and $g$ are inverse functions.

∴ Verified: Both $f(g(x)) = x$ and $g(f(x)) = x$.

4

Find the domain of $f(x) = \frac{1}{\sqrt{x-2}}$.

[3 marks]

Solution

This function has both a square root (in the denominator) and a denominator. Both create restrictions.

Restriction 1: Expression under the square root must be non-negative

For $\sqrt{x-2}$ to be real: $x - 2 \geq 0$
This gives: $x \geq 2$

Restriction 2: Denominator cannot be zero

We need $\sqrt{x-2} \neq 0$

This means: $x - 2 \neq 0$
So: $x \neq 2$

Combine the restrictions

We need $x \geq 2$ AND $x \neq 2$

This simplifies to: $x > 2$

Therefore, Domain = $(2, \infty)$ (open interval, excluding 2)

∴ Domain = $(2, \infty)$

5

Simplify $\log_5 25 + \log_5 5 - \log_5 1$.

[2 marks]

Solution

We'll evaluate each logarithm separately using the definition and special values.

Given: $\log_5 25 + \log_5 5 - \log_5 1$

First term: $\log_5 25$
We need: $5^? = 25$
Since $5^2 = 25$, we have $\log_5 25 = 2$

Second term: $\log_5 5$
We need: $5^? = 5$
Since $5^1 = 5$, we have $\log_5 5 = 1$

Third term: $\log_5 1$
We need: $5^? = 1$
Since $5^0 = 1$, we have $\log_5 1 = 0$

Now combine:
$2 + 1 - 0 = 3$

∴ $\log_5 25 + \log_5 5 - \log_5 1 = 3$

6

Determine whether $f(x) = x^3 - 2$ is one-to-one.

[2 marks]

Solution

A function is one-to-one if different inputs produce different outputs. We can verify this by showing: if $f(x_1) = f(x_2)$, then $x_1 = x_2$.

Given: $f(x) = x^3 - 2$

Assume $f(x_1) = f(x_2)$:

$x_1^3 - 2 = x_2^3 - 2$

Add 2 to both sides:
$x_1^3 = x_2^3$

Take the cube root of both sides (cube root is a one-to-one function):
$\sqrt[3]{x_1^3} = \sqrt[3]{x_2^3}$

$x_1 = x_2$

We've shown that if two inputs produce the same output, they must be identical. Therefore, $f(x) = x^3 - 2$ is one-to-one (injective).

∴ Yes, $f(x) = x^3 - 2$ is one-to-one.

7

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

[3 marks]

Solution

The composition $(f \circ g)(3)$ means: apply $g$ to 3, then apply $f$ to the result.

Given: $f(x) = \sqrt{x+1}$ and $g(x) = x^2 - 3$

Step 1: Calculate $g(3)$

$g(3) = 3^2 - 3 = 9 - 3 = 6$

Step 2: Apply $f$ to the result

$(f \circ g)(3) = f(g(3)) = f(6)$

$f(6) = \sqrt{6+1} = \sqrt{7}$

Therefore, $(f \circ g)(3) = \sqrt{7}$

∴ $(f \circ g)(3) = \sqrt{7}$

8

State the domain and range of $f(x) = \log_2(x)$.

[2 marks]

Solution

Domain: The argument of a logarithm must be strictly positive (> 0). So we need $x > 0$.
Domain = $(0, \infty)$

Range: As $x$ varies over all positive real numbers, the logarithm can take any real value:
- When $x$ is very small (approaching 0), $\log_2(x) \to -\infty$
- When $x = 1$, $\log_2(1) = 0$
- When $x = 2$, $\log_2(2) = 1$
- When $x$ is very large, $\log_2(x) \to +\infty$

Range = $\mathbb{R}$ (all real numbers)

∴ Domain = $(0, \infty)$; Range = $\mathbb{R}$

9

Prove using the definition that $f(x) = \frac{1}{x}$ is onto (surjective) on $\mathbb{R}^* \to \mathbb{R}^*$ (where $\mathbb{R}^*$ denotes non-zero reals).

[3 marks]

Solution

To prove a function is onto, we must show that for every element $y$ in the co-domain, there exists an element $x$ in the domain such that $f(x) = y$.

Given: $f(x) = \frac{1}{x}$ with domain = $\mathbb{R}^* = \mathbb{R} - \{0\}$ and co-domain = $\mathbb{R}^*$

Proof:

Let $y$ be any element in the co-domain $\mathbb{R}^*$. This means $y$ is any non-zero real number.

We need to find an $x$ in the domain $\mathbb{R}^*$ such that $f(x) = y$.

Set up the equation: $\frac{1}{x} = y$

Solve for $x$: $x = \frac{1}{y}$

Check: Since $y \neq 0$ (as $y \in \mathbb{R}^*$), the value $\frac{1}{y}$ is well-defined and non-zero. So $x = \frac{1}{y} \in \mathbb{R}^*$ (in the domain).

Verify: $f(x) = f\left(\frac{1}{y}\right) = \frac{1}{\frac{1}{y}} = y$ ✓

Since for every $y$ in the co-domain we can find a corresponding $x$ in the domain, the function is onto (surjective).

∴ Proved: For each $y \in \mathbb{R}^*$, we have $x = \frac{1}{y} \in \mathbb{R}^*$ such that $f(x) = y$.

Long Questions (5-8 Marks Each)

1

Prove that $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 2x + 3$ is bijective and find its inverse function.

[8 marks]

Solution

To prove bijectivity, we must verify two properties: (1) the function is one-to-one (injective), and (2) the function is onto (surjective). Then we'll find the inverse.

Part 1: Prove Injectivity (One-to-One)

We must show: if $f(x_1) = f(x_2)$, then $x_1 = x_2$.

Assume $f(x_1) = f(x_2)$:
$2x_1 + 3 = 2x_2 + 3$

Subtract 3 from both sides:
$2x_1 = 2x_2$

Divide by 2:
$x_1 = x_2$

This proves that equal outputs imply equal inputs. Therefore, $f$ is injective. ✓

Part 2: Prove Surjectivity (Onto)

We must show: for every $y \in \mathbb{R}$, there exists an $x \in \mathbb{R}$ such that $f(x) = y$.

Let $y$ be any real number. We need to find $x$ such that:
$2x + 3 = y$

Solving for $x$:
$2x = y - 3$
$x = \frac{y - 3}{2}$

This $x$ is a real number for every real $y$ (no division by zero, no square roots of negatives). So $x \in \mathbb{R}$ is in the domain.

Verification: $f(x) = f\left(\frac{y-3}{2}\right) = 2 \cdot \frac{y-3}{2} + 3 = (y-3) + 3 = y$ ✓

This proves that every $y$ in the co-domain is the image of some $x$ in the domain. Therefore, $f$ is surjective. ✓

Part 3: Conclude Bijectivity

Since $f$ is both injective and surjective, $f$ is bijective. ✓

Part 4: Find the Inverse Function

Since $f$ is bijective, the inverse $f^{-1}$ exists.

Method: Write $y = f(x)$, swap $x$ and $y$, solve for $y$.

Starting: $y = 2x + 3$

Swap: $x = 2y + 3$

Solve for $y$:
$x - 3 = 2y$
$y = \frac{x - 3}{2}$

Therefore: $f^{-1}(x) = \frac{x - 3}{2}$

Verification:

Check $f(f^{-1}(x)) = x$:
$f(f^{-1}(x)) = f\left(\frac{x-3}{2}\right) = 2 \cdot \frac{x-3}{2} + 3 = (x-3) + 3 = x$ ✓

Check $f^{-1}(f(x)) = x$:
$f^{-1}(f(x)) = f^{-1}(2x+3) = \frac{(2x+3)-3}{2} = \frac{2x}{2} = x$ ✓

∴ Proved bijective. Inverse: $f^{-1}(x) = \frac{x-3}{2}$

2

Find the domain and range of $f(x) = \sqrt{x^2 - 4}$, and sketch the function.

[6 marks]

Solution

Part 1: Find the Domain

For a square root to be real, the expression under the root must be non-negative:
$x^2 - 4 \geq 0$

Factor:
$(x - 2)(x + 2) \geq 0$

This inequality holds when both factors are positive or both are negative:
- Both positive: $x \geq 2$ and $x \geq -2$ → $x \geq 2$
- Both negative: $x \leq 2$ and $x \leq -2$ → $x \leq -2$

Combining: $x \leq -2$ or $x \geq 2$

Domain = $(-\infty, -2] \cup [2, \infty)$

Part 2: Find the Range

To find the range, analyze the behavior of the function:
- When $x = 2$: $f(2) = \sqrt{4 - 4} = 0$
- When $x = -2$: $f(-2) = \sqrt{4 - 4} = 0$
- When $x = 3$: $f(3) = \sqrt{9 - 4} = \sqrt{5} \approx 2.24$
- When $x = -3$: $f(-3) = \sqrt{9 - 4} = \sqrt{5} \approx 2.24$
- As $x \to \infty$: $x^2 - 4 \to \infty$, so $f(x) \to \infty$
- As $x \to -\infty$: $x^2 - 4 \to \infty$, so $f(x) \to \infty$

The minimum value is 0 (achieved at $x = \pm 2$). There is no maximum; the function increases without bound.

Range = $[0, \infty)$

Part 3: Sketch

The function:
- Has two separate branches (due to the domain restriction)
- Left branch: exists for $x \leq -2$, starts at point $(-2, 0)$ and extends upward as $x$ decreases
- Right branch: exists for $x \geq 2$, starts at point $(2, 0)$ and extends upward as $x$ increases
- Both branches are parts of the same curve $y = \sqrt{x^2 - 4}$
- The function is symmetric about the $y$-axis (even function)
- No portion of the curve exists between $x = -2$ and $x = 2$

∴ Domain = $(-\infty, -2] \cup [2, \infty)$; Range = $[0, \infty)$; Graph shows two symmetric branches from $(-2, 0)$ and $(2, 0)$.

3

Given $f(x) = x + 1$ and $g(x) = 2x - 3$, find $(f \circ g)(x)$, $(g \circ f)(x)$, and show that $f \circ g \neq g \circ f$.

[5 marks]

Solution

Part 1: Find $(f \circ g)(x) = f(g(x))$

The composition $(f \circ g)(x)$ means: apply $g$ first, then apply $f$ to the result.

Step 1: Start with $g(x) = 2x - 3$

Step 2: Substitute this entire expression into $f$:
$(f \circ g)(x) = f(g(x)) = f(2x - 3)$

Step 3: Apply the rule for $f$. Since $f(x) = x + 1$, we replace the input with $(2x-3)$:
$f(2x - 3) = (2x - 3) + 1 = 2x - 2$

Therefore: $(f \circ g)(x) = 2x - 2$

Part 2: Find $(g \circ f)(x) = g(f(x))$

The composition $(g \circ f)(x)$ means: apply $f$ first, then apply $g$ to the result. Notice the order is reversed!

Step 1: Start with $f(x) = x + 1$

Step 2: Substitute this into $g$:
$(g \circ f)(x) = g(f(x)) = g(x + 1)$

Step 3: Apply the rule for $g$. Since $g(x) = 2x - 3$, we replace the input with $(x+1)$:
$g(x + 1) = 2(x + 1) - 3 = 2x + 2 - 3 = 2x - 1$

Therefore: $(g \circ f)(x) = 2x - 1$

Part 3: Compare and Show $f \circ g \neq g \circ f$

We have:
- $(f \circ g)(x) = 2x - 2$
- $(g \circ f)(x) = 2x - 1$

These are not equal. For example:
- $(f \circ g)(0) = 2(0) - 2 = -2$
- $(g \circ f)(0) = 2(0) - 1 = -1$

Since $-2 \neq -1$, we have proven that $(f \circ g)(x) \neq (g \circ f)(x)$.

Conclusion: Function composition is NOT commutative. The order in which you compose functions matters significantly.

∴ $(f \circ g)(x) = 2x - 2$; $(g \circ f)(x) = 2x - 1$; They are not equal.

4

Prove the logarithm product rule: $\log_a(xy) = \log_a x + \log_a y$ for all positive $x, y$ and valid base $a$.

[5 marks]

Solution

We'll prove this fundamental logarithm property using exponential notation and the laws of exponents.

Given: We want to prove $\log_a(xy) = \log_a x + \log_a y$

Proof:

Let $\log_a x = m$. By the definition of logarithm, this means:
$a^m = x$ ... (Equation 1)

Let $\log_a y = n$. By the definition of logarithm, this means:
$a^n = y$ ... (Equation 2)

Now multiply equations (1) and (2):
$a^m \cdot a^n = x \cdot y = xy$

Using the law of exponents $a^p \cdot a^q = a^{p+q}$:
$a^{m+n} = xy$

Convert this back to logarithmic form. By the definition of logarithm, if $a^{m+n} = xy$, then:
$\log_a(xy) = m + n$

Now substitute back the original definitions of $m$ and $n$:
$\log_a(xy) = \log_a x + \log_a y$

This completes the proof. ✓

Example to Illustrate:

Verify with $a = 2$, $x = 4$, $y = 8$:
- LHS: $\log_2(4 \times 8) = \log_2(32) = 5$ (since $2^5 = 32$)
- RHS: $\log_2 4 + \log_2 8 = 2 + 3 = 5$ ✓

The property is verified!

∴ Proved using exponential form and the law $a^m \cdot a^n = a^{m+n}$.

5

Find the inverse of $f(x) = \frac{x-1}{x+2}$ and verify that $f(f^{-1}(x)) = x$.

[6 marks]

Solution

Part 1: Find the Inverse

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

To find the inverse, we write $y = f(x)$, swap $x$ and $y$, and solve for $y$.

Step 1: Write the equation with $y$:
$y = \frac{x-1}{x+2}$

Step 2: Swap $x$ and $y$:
$x = \frac{y-1}{y+2}$

Step 3: Solve for $y$. Multiply both sides by $(y+2)$:
$x(y + 2) = y - 1$

Step 4: Expand:
$xy + 2x = y - 1$

Step 5: Collect all terms with $y$ on one side:
$xy - y = -1 - 2x$

Step 6: Factor out $y$:
$y(x - 1) = -1 - 2x$

Step 7: Solve for $y$:
$y = \frac{-1 - 2x}{x - 1}$

Step 8: Simplify by factoring out -1 from the numerator:
$y = \frac{-(1 + 2x)}{x - 1} = \frac{-2x - 1}{x - 1}$

Or equivalently: $y = \frac{-(2x + 1)}{x - 1}$

Therefore: $f^{-1}(x) = \frac{-2x - 1}{x - 1}$ or $f^{-1}(x) = \frac{-(2x+1)}{x-1}$

Part 2: Verify $f(f^{-1}(x)) = x$

We must compute $f\left(\frac{-2x-1}{x-1}\right)$ and show it simplifies to $x$.

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

Simplify the numerator:
$\frac{-2x-1}{x-1} - 1 = \frac{-2x-1}{x-1} - \frac{x-1}{x-1} = \frac{-2x-1-(x-1)}{x-1} = \frac{-2x-1-x+1}{x-1} = \frac{-3x}{x-1}$

Simplify the denominator:
$\frac{-2x-1}{x-1} + 2 = \frac{-2x-1}{x-1} + \frac{2(x-1)}{x-1} = \frac{-2x-1+2(x-1)}{x-1} = \frac{-2x-1+2x-2}{x-1} = \frac{-3}{x-1}$

Now divide:
$\frac{\frac{-3x}{x-1}}{\frac{-3}{x-1}} = \frac{-3x}{x-1} \times \frac{x-1}{-3} = \frac{-3x \times (x-1)}{(x-1) \times (-3)} = \frac{-3x}{-3} = x$ ✓

We have verified that $f(f^{-1}(x)) = x$.

∴ $f^{-1}(x) = \frac{-2x-1}{x-1}$; Verified: $f(f^{-1}(x)) = x$

6

Solve $\log_x 4 + \log_x 8 = \log_x 64$. Find the value of $x$.

[5 marks]

Solution

We need to solve the equation $\log_x 4 + \log_x 8 = \log_x 64$.

Step 1: Apply the product rule of logarithms to the left side

Recall: $\log_a p + \log_a q = \log_a(pq)$

$\log_x 4 + \log_x 8 = \log_x(4 \times 8) = \log_x 32$

So the equation becomes:
$\log_x 32 = \log_x 64$

Step 2: Use the fact that if $\log_a p = \log_a q$, then $p = q$

If $\log_x 32 = \log_x 64$, then:
$32 = 64$

But this is false! Let me reconsider the problem statement... Actually, the equation as originally stated leads to a contradiction if we apply logarithm properties directly. Let me verify by solving more carefully.

Actually, wait—let me check if the logarithm base appears in both sides. We have $\log_x$ which is the variable!

Let's rewrite the equation more carefully:
$\log_x 4 + \log_x 8 = \log_x 64$

Using the product rule:
$\log_x(4 \times 8) = \log_x 64$
$\log_x 32 = \log_x 64$

This would imply $32 = 64$, which is impossible.

Hold on—perhaps there's a typo in the original problem. Let me solve a corrected version:

Corrected Problem: $\log_x 4 + \log_x 8 = 3$

Using the product rule:
$\log_x(4 \times 8) = 3$
$\log_x 32 = 3$

Convert to exponential form:
$x^3 = 32$

Take the cube root:
$x = \sqrt[3]{32} = \sqrt[3]{2^5} = 2^{5/3}$

We can also write this as:
$x = 2 \cdot \sqrt[3]{4}$ or $x \approx 3.17$

Let's verify: If $x = 2^{5/3}$, then $\log_x 32 = \log_{2^{5/3}} 2^5 = \frac{5}{5/3} = 3$ ✓

∴ If the original equation is $\log_x 4 + \log_x 8 = 3$, then $x = 2^{5/3}$ or $x = \sqrt[3]{32}$

One-Page Revision Cheat Sheet

1

Definition of Function: Every input has exactly one output. No input can produce multiple outputs. This 'uniqueness of output' is the core definition.

2

Vertical Line Test: If a vertical line cuts a graph at more than one point anywhere, it is not a function. Use this to quickly reject non-functions.

3

Domain: All possible inputs. Exclude: (1) values that make denominators zero, (2) values that make even roots negative, (3) values that make logarithm arguments non-positive.

4

Range: All actual outputs. Set $y = f(x)$, solve for $x$, and determine which $y$ values are possible. Range $\subseteq$ Co-domain always.

5

One-to-One (Injective): Different inputs produce different outputs. Test: if $f(a)=f(b)$, does this force $a=b$?

6

Onto (Surjective): Every element in the co-domain is an actual output. Range = Co-domain. For any $y$ in co-domain, can you find an $x$ with $f(x)=y$?

7

Bijective = One-to-One AND Onto: Only bijective functions have inverses. Check both properties before claiming an inverse exists.

8

Finding Inverse: Write $y=f(x)$, swap $x$ and $y$, solve for $y$. The resulting $y$ is $f^{-1}(x)$. Verify: $f(f^{-1}(x))=x$ and $f^{-1}(f(x))=x$.

9

Function Composition $(g \circ f)(x)$: Apply $f$ first, then $g$ to the result. $(g \circ f)(x) = g(f(x))$. Generally, $g \circ f \neq f \circ g$ (not commutative).

10

Logarithm Product Rule: $\log_a(xy) = \log_a x + \log_a y$ (multiply inputs → add logs)

11

Logarithm Quotient Rule: $\log_a(x/y) = \log_a x - \log_a y$ (divide inputs → subtract logs)

12

Logarithm Power Rule: $\log_a(x^p) = p \log_a x$ (exponent comes down as multiplier)

13

Special Logarithm Values: $\log_a 1 = 0$, $\log_a a = 1$, $\log_a a^p = p$. Always true, memorize.

14

Change of Base: $\log_a x = \frac{\log_b x}{\log_b a}$ (use to convert between bases, especially to base 10 or natural log)

15

Algebra of Functions: $(f+g)(x) = f(x)+g(x)$, $(fg)(x)=f(x)g(x)$, $(f/g)(x)=f(x)/g(x)$ (where $g(x) \neq 0$)