Functions and Domain/Range

Subject: Additional Mathematics
Topic: 1
Cambridge Code: 4037 / 0606

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Introduction to Functions

Function - A relationship between two variables where for each input value, there is exactly one output value

Notation

A function $f$ is denoted as: $f(x) = y$

where:

• $x$ is the input (independent variable)
• $y$ is the output (dependent variable)
• $f$ is the function name

Examples

• $f(x) = 2x + 3$
• $g(x) = x^2 - 4$
• $h(x) = \sqrt{x}$

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Domain and Range

Domain

Domain - The set of all possible input values (x-values) for which the function is defined

Range

Range - The set of all possible output values (y-values) that the function can produce

Finding Domain

Polynomial Functions

Functions like $f(x) = x^2 + 3x - 2$ have domain of all real numbers: $\mathbb{R}$ or $(-\infty, \infty)$

Rational Functions

For $f(x) = \frac{1}{x-2}$, exclude values that make denominator zero: $\text{Domain: } x \neq 2$

Square Root Functions

For $f(x) = \sqrt{x-1}$, expression under root must be non-negative: $x - 1 \geq 0 \Rightarrow x \geq 1$

Logarithmic Functions

For $f(x) = \log(x)$, argument must be positive: $x > 0$

Finding Range

Method 1: Complete the Square

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

Since $(x-2)^2 \geq 0$, minimum value is $-1$ $\text{Range: } f(x) \geq -1 \text{ or } [-1, \infty)$

Method 2: Solve for x in terms of y

For $f(x) = \frac{2x+1}{x-1}$: $y = \frac{2x+1}{x-1}$ $y(x-1) = 2x+1$ $yx - y = 2x + 1$ $yx - 2x = y + 1$ $x(y-2) = y+1$ $x = \frac{y+1}{y-2}$

Domain of this becomes: $y \neq 2$ $\text{Range: } f(x) \neq 2$

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Function Operations

Addition/Subtraction

$(f+g)(x) = f(x) + g(x)$ $(f-g)(x) = f(x) - g(x)$

Multiplication/Division

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

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Composite Functions

Composite Function - A function made by combining two or more functions

Notation

$(f \circ g)(x) = f(g(x))$ means "apply $g$ first, then $f$"

Example

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

$f(g(x)) = f(x^2) = 2(x^2) + 1 = 2x^2 + 1$

$g(f(x)) = g(2x+1) = (2x+1)^2 = 4x^2 + 4x + 1$

Note: $f(g(x)) \neq g(f(x))$ (composition is not commutative)

Using Function Machines

Think of function as transforming input:

$\boxed{\text{Input}} \rightarrow [f] \rightarrow [g] \rightarrow \boxed{\text{Output}}$

This represents $g(f(x))$

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Inverse Functions

Inverse Function - A function that reverses the effect of the original function

Notation

Inverse of $f$ is denoted $f^{-1}$

Finding Inverse Functions

Steps:

1. Write $y = f(x)$
2. Swap $x$ and $y$
3. Solve for $y$
4. This is $f^{-1}(x)$

Example: Find $f^{-1}$ where $f(x) = 2x + 3$

Step 1: $y = 2x + 3$ Step 2: $x = 2y + 3$ Step 3: $x - 3 = 2y$ $y = \frac{x-3}{2}$ Step 4: $f^{-1}(x) = \frac{x-3}{2}$

Verification

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

Example Verification

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

Domain and Range of Inverse

$\text{Domain of } f^{-1} = \text{Range of } f$ $\text{Range of } f^{-1} = \text{Domain of } f$

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Key Points to Remember

1. Function must have exactly one output for each input
2. Domain is set of possible inputs
3. Range is set of possible outputs
4. Composite functions apply in specific order: $f(g(x))$ means $g$ first
5. Inverse function reverses the original function
6. $f(f^{-1}(x)) = x$ for all $x$ in domain

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Worked Examples

Example 1: Domain and Range

Find domain and range of $f(x) = \sqrt{4-x^2}$

Domain: $4 - x^2 \geq 0$ $x^2 \leq 4$ $-2 \leq x \leq 2$ $\text{Domain: } [-2, 2]$

Range: Minimum value: $f(\pm 2) = 0$ Maximum value: $f(0) = 2$ $\text{Range: } [0, 2]$

Example 2: Composite Function

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

$(f \circ g)(2) = f(g(2)) = f(4) = 4 + 1 = 5$

Example 3: Inverse Function

Find the inverse of $f(x) = \frac{x+2}{3}$ and verify

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

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

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Practice Questions

1. Find the domain and range of:

   • $f(x) = \frac{1}{x+3}$
   • $g(x) = \sqrt{x-5}$
   • $h(x) = \frac{2x}{x^2-4}$

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

   • $(f \circ g)(x)$
   • $(g \circ f)(x)$

3. Find and verify the inverse of:

   • $f(x) = 5x - 2$
   • $g(x) = \frac{3}{x+1}$

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Revision Tips

• Domain: where function is defined
• Range: what values function can produce
• Composite: apply function on right first
• Inverse: swap $x$ and $y$, then solve
• Always check domain restrictions (denominators, roots, logs)