Functions

1. Overview

Functions are a fundamental concept in mathematics that describe relationships between inputs and outputs. A function takes an input value (usually denoted by $x$) and transforms it into a unique output value (often denoted by $f(x)$ or $y$). This topic covers function notation, evaluating functions, finding inverse functions, and working with composite functions, all of which are crucial for success in IGCSE mathematics, particularly in algebra and calculus. Understanding domains and ranges is also essential for defining the limits within which a function operates.

Key Definitions

• Function: A rule that maps each input value to exactly one output value.
• Domain: The set of all possible input values (the $x$-values) for which the function is defined.
• Range: The set of all possible output values (the $y$-values or $f(x)$ values) produced by the function.
• Mapping: A diagram or description showing how elements from the domain connect to elements in the range.
• Composite Function: A function created by applying one function to the result of another (e.g., $gf(x)$).
• Inverse Function ($f^{-1}(x)$): A function that "undoes" the original function, mapping the output back to the original input.

---

Core Content

(Note: For Topic 2.13, there are no specific Core curriculum objectives. All content for this topic falls under the Supplement/Extended curriculum.)

---

Extended Content (Extended Curriculum Only)

A. Function Notation

Functions are usually written as $f(x) = \dots$ or using mapping notation $f : x \mapsto \dots$ (read as "$f$ maps $x$ to..."). To evaluate a function, replace every $x$ in the expression with the given value.

Worked example 1 — Evaluating a Function

Question: Given $f(x) = 7x + 4$, find $f(-2)$.

1. Replace $x$ with $-2$: $f(-2) = 7(-2) + 4$ Substitute the given value into the function.
2. Multiply: $f(-2) = -14 + 4$ Perform the multiplication.
3. Add: $f(-2) = -10$ Complete the addition. Final Answer: $\bf{-10}$

B. Composite Functions

A composite function $gf(x)$ means you apply function $f$ first, and then apply function $g$ to the result. Think of it as $g(f(x))$. Important: The order matters! $gf(x)$ is usually not the same as $fg(x)$.

Worked example 2 — Forming Composite Functions

Question: Given $f(x) = 3x - 1$ and $g(x) = x^2 + 2$, find $gf(x)$.

1. Identify the functions: $f(x) = 3x - 1$ and $g(x) = x^2 + 2$ State the given functions.
2. Substitute $f(x)$ into $g(x)$: $gf(x) = g(3x - 1) = (3x - 1)^2 + 2$ Replace $x$ in $g(x)$ with the entire expression for $f(x)$.
3. Expand the brackets: $gf(x) = (9x^2 - 6x + 1) + 2$ Expand the squared term.
4. Simplify: $gf(x) = 9x^2 - 6x + 3$ Combine the constant terms. Final Answer: $\bf{9x^2 - 6x + 3}$

C. Inverse Functions ($f^{-1}(x)$ )

The inverse function reverses the process. If $f(2) = 10$, then $f^{-1}(10) = 2$.

Method to find $f^{-1}(x)$:

1. Let $y = f(x)$
2. Rearrange the equation to make $x$ the subject.
3. Replace $x$ with $f^{-1}(x)$ and replace $y$ with $x$.

Worked example 3 — Finding the Inverse

Question: Find the inverse of $f(x) = \frac{x}{4} + 3$.

1. Set $y = f(x)$: $y = \frac{x}{4} + 3$ Replace $f(x)$ with $y$.
2. Subtract 3 from both sides: $y - 3 = \frac{x}{4}$ Isolate the term with $x$.
3. Multiply both sides by 4: $4(y - 3) = x$ Solve for $x$.
4. Simplify: $4y - 12 = x$ Expand the brackets.
5. Rewrite in function notation: $f^{-1}(x) = 4x - 12$ Swap $x$ and $y$ and replace $y$ with $f^{-1}(x)$. Final Answer: $\bf{f^{-1}(x) = 4x - 12}$

Worked example 4 — Finding the Inverse with a Fractional Denominator

Question: Find the inverse of $f(x) = \frac{5}{x-2}$.

1. Set $y = f(x)$: $y = \frac{5}{x-2}$ Replace $f(x)$ with $y$.
2. Multiply both sides by $(x-2)$: $y(x-2) = 5$ Get $x$ out of the denominator.
3. Divide both sides by $y$: $x-2 = \frac{5}{y}$ Isolate the term with $x$.
4. Add 2 to both sides: $x = \frac{5}{y} + 2$ Solve for $x$.
5. Rewrite in function notation: $f^{-1}(x) = \frac{5}{x} + 2$ Swap $x$ and $y$ and replace $y$ with $f^{-1}(x)$. Final Answer: $\bf{f^{-1}(x) = \frac{5}{x} + 2}$

D. Domain and Range

• Domain: Usually given in the question (e.g., $x > 0$). If not given, it is assumed to be all real numbers ($\mathbb{R}$), unless a value makes the function "break" (like dividing by zero).
• Range: Look at the highest and lowest possible values of the output.

Example: $f(x) = x^2$. If the domain is all real numbers, the range is $f(x) \geq 0$ because a square number can never be negative.

---

Key Equations

$f(x) = y$ Output of function $f$ for input $x$. $y$ is the dependent variable.

$gf(x) = g(f(x))$ Work from the inside out.

$f^{-1}(x)$ Inverse function. Range of $f$ = Domain of $f^{-1}$.

$ff^{-1}(x) = x$ A function and its inverse cancel out.

These formulas are NOT provided on the IGCSE formula sheet; they must be memorized.

---

Common Mistakes to Avoid

• ❌ Wrong: For $f(x) = 3x^2 + 1$, calculating $f(x + 2)$ as $3x^2 + 1 + 2$. ✅ Right: Use brackets! $f(x + 2) = 3(x + 2)^2 + 1$. Expand to $3(x^2 + 4x + 4) + 1 = 3x^2 + 12x + 12 + 1 = 3x^2 + 12x + 13$.
• ❌ Wrong: Thinking $gf(x)$ means $g(x) + f(x)$. ✅ Right: $gf(x)$ means substitute the expression $f(x)$ into the function $g(x)$ wherever you see $x$.
• ❌ Wrong: Assuming that if $f(x) = 2x$, then $f^{-1}(x) = -2x$. ✅ Right: The inverse function "undoes" the operation. In this case, $f^{-1}(x) = \frac{x}{2}$.
• ❌ Mark Loss: When finding the inverse, forgetting to swap $x$ and $y$ after rearranging to make $x$ the subject.

---

Exam Tips

• Command Words:
  • "Evaluate $f(3)$": Give a numerical value.
  • "Find an expression for $gf(x)$": Give the answer in terms of $x$.
  • "Solve $f(x) = g(x)$": Set the two expressions equal and solve for $x$.
• Order of Operations: In composite functions like $g(f(2))$, calculate $f(2)$ first, then put that number into $g$.
• Algebraic Fractions: Questions often involve functions like $f(x) = \frac{1}{x+2}$. Remember that the domain cannot include $x = -2$ because you cannot divide by zero.
• Calculator Tip: When substituting negative numbers (e.g., $f(-3)$ for $f(x) = x^2$), always use brackets on your calculator: $(-3)^2$. Typing $-3^2$ without brackets will give you $-9$ instead of $9$.