Linear Functions Revision Notes for Grade 12 NSC Matric Mathematics

Linear Functions

Introduction to inverse functions

An inverse function reverses the effect of the original function. When you have a function that transforms input values into output values, its inverse function transforms those output values back to the original input values. For linear functions of the form $y = axe + q$ , finding the inverse involves a systematic algebraic process.

The inverse of a function $f(x)$ is written as $f^{-1}(x)$ . It's important to understand that the inverse function "undoes" what the original function does. If a function maps a value from its domain to its range, the inverse function maps that same value back from the range to the domain.

infoNote

The notation $f^{-1}(x)$ does not mean $\frac{1}{f(x)}$ . The $-1$ is not an exponent in this context - it specifically indicates the inverse function. This is a common source of confusion for students.

Finding inverse functions

General method for y = axe + q

To find the inverse of any linear function in the form $y = axe + q$ , follow this systematic approach:

Interchange x and y: Replace every $x$ with $y$ and every $y$ with $x$ in the original equation
Make y the subject: Solve the new equation to express $y$ in terms of $x$
Express in function notation: Write the final answer using inverse function notation

The general result is that if $y = axe + q$ , then the inverse function is:

$y = \frac{1}{a}x - \frac{q}{a}$

This formula shows that if a linear function is invertible, its inverse will also be linear.

chatImportant

For a function to have an inverse, it must be one-to-one (injective). All linear functions with $a \neq 0$ are one-to-one, so they all have inverses. If $a = 0$ , the function becomes constant and does not have an inverse.

Step-by-step process

Let's explore this method using detailed worked examples to ensure you understand each stage of the process.

lightbulbExample

Worked Example 1: Finding the inverse of p(x) = -3x + 1

Step 1: Set up the equation

Start with: $y = -3x + 1$

Step 2: Interchange x and y

Replace $x$ with $y$ and $y$ with $x$ : $x = -3y + 1$

Step 3: Make y the subject

Solve for $y$ :

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

Step 4: Express in function notation

Therefore: $p^{-1}(x) = -\frac{x}{3} + \frac{1}{3}$

The graph above shows both the original function $p(x) = -3x + 1$ and its inverse $p^{-1}(x) = -\frac{1}{3}x + \frac{1}{3}$ . Notice how they are reflections of each other about the line $y = x$ .

lightbulbExample

Worked Example 2: Finding the inverse of f(x) = 2x - 3

Step 1: Set up and interchange

Starting with: $y = 2x - 3$

Interchange $x$ and $y$ : $x = 2y - 3$

Step 2: Solve for y

$x = 2y - 3$
$x + 3 = 2y$
$\frac{x + 3}{2} = y$
$y = \frac{x}{2} + \frac{3}{2}$

Step 3: Function notation

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

Properties of inverse functions

Domain and range relationship

Understanding how domain and range behave with inverse functions is crucial for exam success.

For a linear function $y = axe + q$ :

Domain: $\{x : x \in \mathbb{R}\}$ (all real numbers)
Range: $\{y : y \in \mathbb{R}\}$ (all real numbers)

When you find the inverse function:

Domain of $f^{-1}$ : $\{x : x \in \mathbb{R}\}$ (same as range of original function)
Range of $f^{-1}$ : $\{y : y \in \mathbb{R}\}$ (same as domain of original function)

chatImportant

The domain and range swap positions when you move from a function to its inverse. This is a fundamental property that applies to all invertible functions, not just linear ones.

Intercepts relationship

The relationship between intercepts of a function and its inverse follows a specific pattern that's very useful in exams.

For a linear function $y = axe + q$ (where $a \neq 0$ ):

Finding intercepts of the original function:

y-intercept: Set $x = 0$ , so $y = a(0) + q = q$ . Point: $(0, q)$
x-intercept: Set $y = 0$ , so $0 = axe + q$ , which gives $x = -\frac{q}{a}$ . Point: $\left(-\frac{q}{a}, 0\right)$

Finding intercepts of the inverse function $y = \frac{1}{a}x - \frac{q}{a}$ :

y-intercept: Set $x = 0$ , so $y = \frac{1}{a}(0) - \frac{q}{a} = -\frac{q}{a}$ . Point: $\left(0, -\frac{q}{a}\right)$
x-intercept: Set $y = 0$ , so $0 = \frac{1}{a}x - \frac{q}{a}$ , which gives $x = q$ . Point: $(q, 0)$

infoNote

Key insight: The intercepts are mirror images of each other. The x and y coordinates swap positions between the function and its inverse. This provides an excellent way to check your work!

Graphical relationship

The most important visual property of inverse functions is their reflexion relationship.

The graph of $f^{-1}(x)$ is the reflexion of the graph of $f(x)$ about the line $y = x$ . This means:

Every point $(a, b)$ on the graph of $f(x)$ corresponds to the point $(b, a)$ on the graph of $f^{-1}(x)$
The line $y = x$ acts as a "mirror" between the two graphs
Both graphs will intersect the line $y = x$ at the same points

This reflexion property is extremely useful for checking your work and understanding the relationship between functions and their inverses visually.

infoNote

If you're having trouble visualising this reflexion, try plotting a few points from the original function, then plot their "swapped" coordinates. You'll see how the inverse function emerges as the reflexion about $y = x$ .

Exam tips and common mistakes

chatImportant

Common exam traps:

Forgetting to interchange $x$ and $y$ at the beginning
Making algebraic errors when solving for $y$
Not expressing the final answer in proper function notation
Confusing domain and range relationships
Writing $f^{-1}(x)$ when you mean $\frac{1}{f(x)}$

Problem-solving approach:

Always start by clearly writing $y = f(x)$
Systematically interchange variables
Carefully solve for $y$ using proper algebraic techniques
Double-check by substituting values
Verify that $f(f^{-1}(x)) = x$

infoNote

Quick verification method: If $f(a) = b$ , then $f^{-1}(b)$ should equal $a$ . Use this to check your inverse function. This is called the inverse function property and it's a reliable way to verify your answer.

bookmarkSummary

Key Points to Remember:

Finding inverses: Interchange $x$ and $y$ , make $y$ the subject, then use function notation
Domain and range swap: The domain of $f$ becomes the range of $f^{-1}$ , and vice versa
Intercepts mirror: The intercepts of $f$ and $f^{-1}$ have swapped $x$ and $y$ coordinates
Graphical reflection: $f^{-1}$ is the reflexion of $f$ about the line $y = x$
Linear inverses: If the original function is linear and invertible, the inverse is also linear
Verification: Always check that $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$

Linear Functions (Grade 12 NSC Matric Mathematics): Revision Notes