Functions

1. Overview

Functions are a core topic in IGCSE Additional Mathematics, appearing in many exam questions, and are fundamental to understanding calculus, coordinate geometry, and trigonometry. This topic covers the definition of a function, its domain and range, how to combine functions through composition, and how to reverse a function using inverse functions. Mastering functions is essential for success in more advanced topics. Expect questions that require you to find domains and ranges, determine if an inverse exists, calculate composite functions, and sketch modulus functions.

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Key Definitions

• Function: A rule that maps each element in a set (domain) to exactly one element in another set (range).
• Domain: The set of all possible input values ($x$-values) for which the function is defined.
• Range (Image Set): The set of all possible output values ($f(x)$ or $y$-values) resulting from the domain.
• One-one Function: A function where every output corresponds to exactly one unique input. Only one-one functions have inverses.
• Many-one Function: A function where two or more different inputs produce the same output (e.g., $f(x) = x^2$).
• Composite Function: A function formed by applying one function to the result of another (e.g., $fg(x)$).
• Inverse Function ($f^{-1}$): A function that reverses the effect of $f(x)$, mapping the range back to the domain.

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Core Content

A. Function Notation

Functions can be written as $f(x) = 2x + 3$ or using mapping notation $f: x \mapsto 2x + 3$.

• $f^2(x)$ does not mean $[f(x)]^2$; it means $f(f(x))$.
• For a composite function $gf(x)$, you apply $f$ first, then apply $g$ to the result. Order matters: $gf(x) \neq fg(x)$.

B. Domain and Range

• Domain of $f$ = Range of $f^{-1}$
• Range of $f$ = Domain of $f^{-1}$
• To find the range, consider the shape of the graph or find the maximum/minimum points.
• Notation: Always use $f(x)$ or $y$ for range (e.g., $f(x) \geq 0$) and $x$ for domain (e.g., $x \in \mathbb{R}, x \neq 2$).

C. Inverse Functions

An inverse $f^{-1}(x)$ exists if and only if $f(x)$ is a one-one function.

• Why a function may not have an inverse: If it is "many-one" (like a parabola), horizontal lines cross the graph more than once, meaning one $y$-value could lead back to multiple $x$-values, which is not allowed for a function.
• Restricting Domain: We can often make a many-one function one-one by restricting its domain (e.g., for $f(x) = x^2$, we might restrict it to $x \geq 0$).

Worked Example 1 — Finding an Inverse (Quadratic)

Question: The function $f$ is defined by $f(x) = x^2 - 6x + 5$ for $x \geq 3$. Find an expression for $f^{-1}(x)$.

1. Complete the square: $f(x) = x^2 - 6x + 5$ $f(x) = (x^2 - 6x + 9) - 9 + 5$ (Completing the square) $f(x) = (x - 3)^2 - 4$

2. Let $y = f(x)$: $y = (x - 3)^2 - 4$

3. Swap $x$ and $y$: $x = (y - 3)^2 - 4$

4. Rearrange for $y$: $x + 4 = (y - 3)^2$ $\pm \sqrt{x + 4} = y - 3$ $y = 3 \pm \sqrt{x + 4}$

5. Choose the correct sign: Since the domain of $f$ is $x \geq 3$, the range of $f^{-1}$ must be $y \geq 3$. Therefore, we take the positive root. $f^{-1}(x) = 3 + \sqrt{x + 4}$

6. State the domain of $f^{-1}(x)$: The range of $f(x)$ is $f(x) \geq -4$. Therefore the domain of $f^{-1}(x)$ is $x \geq -4$.

Answer: $f^{-1}(x) = 3 + \sqrt{x + 4}$ for $x \geq -4$

Worked Example 2 — Finding an Inverse (Rational Function)

Question: The function $g$ is defined by $g(x) = \frac{2x + 1}{x - 3}$ for $x \neq 3$. Find an expression for $g^{-1}(x)$.

1. Let $y = g(x)$: $y = \frac{2x + 1}{x - 3}$

2. Swap $x$ and $y$: $x = \frac{2y + 1}{y - 3}$

3. Rearrange for $y$: $x(y - 3) = 2y + 1$ (Multiply both sides by $y-3$) $xy - 3x = 2y + 1$ (Expand) $xy - 2y = 3x + 1$ (Collect $y$ terms on one side) $y(x - 2) = 3x + 1$ (Factor out $y$) $y = \frac{3x + 1}{x - 2}$ (Divide by $x-2$)

4. State the domain of $g^{-1}(x)$: The range of $g(x)$ can be found by considering the horizontal asymptote of $g(x)$, which is $y=2$. Therefore the domain of $g^{-1}(x)$ is $x \neq 2$.

Answer: $g^{-1}(x) = \frac{3x + 1}{x - 2}$ for $x \neq 2$

Worked Example 3 — Composite Functions

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

1. Find $fg(x)$: $fg(x) = f(g(x))$ $fg(x) = f(x^2 + 1)$ (Substitute $g(x)$ into $f$) $fg(x) = 3(x^2 + 1) - 2$ (Substitute $x^2+1$ into $f(x)$) $fg(x) = 3x^2 + 3 - 2$ (Expand) $fg(x) = 3x^2 + 1$

2. Find $gf(x)$: $gf(x) = g(f(x))$ $gf(x) = g(3x - 2)$ (Substitute $f(x)$ into $g$) $gf(x) = (3x - 2)^2 + 1$ (Substitute $3x-2$ into $g(x)$) $gf(x) = (9x^2 - 12x + 4) + 1$ (Expand) $gf(x) = 9x^2 - 12x + 5$

Answers: $fg(x) = 3x^2 + 1$ and $gf(x) = 9x^2 - 12x + 5$

D. Graphical Relationships

• The graph of $y = f^{-1}(x)$ is a reflection of $y = f(x)$ in the line $y = x$.
• Any point $(a, b)$ on $f$ becomes $(b, a)$ on $f^{-1}$.

E. Modulus Functions $y = |f(x)|$

The modulus (absolute value) makes any negative output positive.

• To sketch $y = |f(x)|$: Sketch $y = f(x)$ normally, then reflect any part of the graph that is below the $x$-axis upwards so it is above the $x$-axis.
• 📊A V-shaped graph for $y=|mx+c|$ or a "W" shape for a reflected quadratic $y=|ax^2+bx+c|$ where the vertex was originally below the x-axis.

Worked Example 4 — Modulus Equations

Question: Solve the equation $|3x + 1| = |x - 2|$.

1. Case 1: $3x + 1 = x - 2$ $2x = -3$ $x = -\frac{3}{2}$

2. Case 2: $3x + 1 = -(x - 2)$ $3x + 1 = -x + 2$ $4x = 1$ $x = \frac{1}{4}$

Answers: $x = -\frac{3}{2}, x = \frac{1}{4}$. (Always check for extraneous solutions by substituting back into the original modulus).

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Extended Content

Additional Mathematics is a single-tier syllabus — all content above applies to all students.

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Key Equations & Notation

$f: x \mapsto \dots$: Function $f$ maps $x$ to a value. $fg(x)$: Composite function: $f(g(x))$. $f^{-1}(x)$: Inverse function of $f$. $|f(x)|$: Modulus of $f(x)$ (always $\geq 0$). $f(x) \in \mathbb{R}$: The range is the set of all real numbers.

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Common Mistakes to Avoid

• ❌ Wrong Notation for Range: Writing $x > 5$ for the range. ✓ Right: Range must use $f(x)$ or $y$, e.g., $f(x) > 5$.
• ❌ Incorrect Composite Order: Calculating $gf(x)$ as $g(x) \times f(x)$. ✓ Right: $gf(x)$ means substituting the entire expression of $f(x)$ into every '$x$' in $g(x)$.
• ❌ Keeping both roots for inverses: Leaving an inverse as $f^{-1}(x) = 2 \pm \sqrt{x}$. ✓ Right: Check the original domain. If $x \geq$ vertex, use $+$. If $x \leq$ vertex, use $-$.
• ❌ Logarithm Domains: Forgetting that for $f(x) = \ln(x-k)$, the domain must be $x > k$.
• ❌ Sign Errors: Swapping $x$ and $y$ then making a sign error during algebraic rearrangement.
• ❌ Exact Value Errors: Giving a decimal approximation when the question requires an exact answer in terms of surds or $\ln$. ✓ Right: Leave your answer as $\sqrt{5}$, $\frac{1}{3}$, or $\ln 2$ unless the question explicitly asks for a decimal approximation.
• ❌ Forgetting Domain Restrictions: When finding the inverse of $f(x) = \sqrt{x-2}$, not stating the domain of $f^{-1}(x)$ as $x \geq 0$. ✓ Right: The range of $f(x)$ is $y \geq 0$, so the domain of $f^{-1}(x)$ is $x \geq 0$.
• ❌ Missing Solutions in Modulus Equations: Only considering the positive case when solving $|f(x)| = a$. ✓ Right: Remember to solve both $f(x) = a$ and $f(x) = -a$.

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Exam Tips

• Command Words: If a question says "State the range," it is usually a 1-mark question that requires no working—look for horizontal asymptotes or vertices.
• Paper 1 (Non-Calculator): Expect to handle surds and fractions. $f(x) = \frac{x+1}{x-2}$ is a common type for finding inverses.
• Formula Sheet: Note that no function formulas are provided on the IGCSE Additional Maths formula sheet. You must memorize the methods for completing the square and finding inverses.
• Show That: In "show that" questions for inverse functions, clearly state "Let $y = f(x)$" and "Swap $x$ and $y$" to earn method marks.
• Graphing: When sketching $|f(x)|$ for a quadratic, always label the $y$-intercept, $x$-intercepts, and the coordinates of the reflected maximum point.
• Existence of Inverses: If asked why $f^{-1}$ doesn't exist, the standard answer is: "Because $f$ is a many-one function (or fails the horizontal line test)."
• Paper 1 (Non-Calculator): Practice algebraic manipulation to avoid errors when finding inverses. Pay close attention to signs and the order of operations.
• Paper 2 (Calculator Allowed): Use your calculator to check your answers, especially for modulus equations. Graph both sides of the equation to visually confirm your solutions.

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0606 papers.

Exam-Style Question 1 — Paper 1 (No Calculator Allowed) [10 marks]

Question:

The function $f$ is defined by $f(x) = \frac{4x + 1}{x - 3}$ for $x > 3$.

(a) Find an expression for $f^{-1}(x)$. [5]

(b) Write down the domain and range of $f^{-1}(x)$. [3]

(c) Explain why $f(x)$ does not have an inverse if the domain is not restricted. [2]

Worked Solution:

(a)

1. Let $y = f(x)$, so $y = \frac{4x + 1}{x - 3}$. To find the inverse, swap $x$ and $y$: $x = \frac{4y + 1}{y - 3}$ [Swap x and y to begin finding the inverse]

2. Rearrange to make $y$ the subject: $x(y - 3) = 4y + 1$ $xy - 3x = 4y + 1$ $xy - 4y = 3x + 1$ $y(x - 4) = 3x + 1$ $y = \frac{3x + 1}{x - 4}$ [Isolate y]

3. Write the inverse function using correct notation: $f^{-1}(x) = \frac{3x + 1}{x - 4}$ [Express the answer in terms of f^{-1}(x)]

$\boxed{f^{-1}(x) = \frac{3x + 1}{x - 4}}$

How to earn full marks: Remember to swap $x$ and $y$ at the start, and then rearrange to make $y$ the subject, expressing your final answer as $f^{-1}(x)$.

(b)

1. The domain of $f^{-1}(x)$ is the range of $f(x)$. Since $f(x) = \frac{4x + 1}{x - 3} = \frac{4(x-3)+13}{x-3}=4+\frac{13}{x-3}$, as $x$ tends to infinity, $f(x)$ approaches 4. Also, $f(x)$ can take any value greater than 4. Therefore, the range of $f(x)$ is $f(x) > 4$ Domain of $f^{-1}(x)$: $x > 4$ [State the domain of the inverse function]

2. The range of $f^{-1}(x)$ is the domain of $f(x)$. Range of $f^{-1}(x)$: $f^{-1}(x) > 3$ [State the range of the inverse function]

$\boxed{\text{Domain: } x > 4, \text{ Range: } f^{-1}(x) > 3}$

How to earn full marks: State the domain and range clearly, remembering that the domain of $f^{-1}(x)$ is the range of $f(x)$, and vice versa.

(c)

1. For $f(x)$ to have an inverse, it must be a one-to-one function. If the domain is not restricted, then $f(x)$ is not a one-to-one function.
2. For example, $f(0) = -\frac{1}{3}$ and $f(-\frac{1}{4}) = -\frac{1}{3}$.
3. Therefore, $f(x)$ is a many-to-one function if the domain is not restricted, and does not have an inverse.

$\boxed{\text{If the domain is not restricted, the function is many-to-one, so it does not have an inverse.}}$

How to earn full marks: Explain that for a function to have an inverse, it must be one-to-one, and show with an example that $f(x)$ is many-to-one when the domain is unrestricted.

Common Pitfall: When finding the inverse, remember to swap $x$ and $y$ first before rearranging. Also, be careful with the domain and range of the inverse function – they are related to the range and domain of the original function, respectively.

Exam-Style Question 2 — Paper 1 (No Calculator Allowed) [8 marks]

Question:

The functions $g$ and $h$ are defined as follows: $g(x) = x^2 - 6x + 5$ for $x \in \mathbb{R}$ $h(x) = 3x - 2$ for $x \in \mathbb{R}$

(a) Find the range of $g(x)$. [3]

(b) Find $hg(x)$ in its simplest form. [3]

(c) Solve $h(x) = g(0)$. [2]

Worked Solution:

(a)

1. Complete the square for $g(x)$: $g(x) = (x - 3)^2 - 9 + 5 = (x - 3)^2 - 4$ [Complete the square to find the vertex]

2. The minimum value of $g(x)$ occurs when $(x - 3)^2 = 0$, which gives $g(x) = -4$. Since $(x-3)^2$ is always non-negative, the range of $g(x)$ is $g(x) \ge -4$.

$\boxed{g(x) \ge -4}$

How to earn full marks: Complete the square correctly to find the vertex of the parabola, then state the range using the correct inequality sign.

(b)

1. Substitute $g(x)$ into $h(x)$: $hg(x) = h(g(x)) = 3(g(x)) - 2 = 3(x^2 - 6x + 5) - 2$ [Substitute the function]

2. Expand and simplify: $hg(x) = 3x^2 - 18x + 15 - 2 = 3x^2 - 18x + 13$

$\boxed{hg(x) = 3x^2 - 18x + 13}$

How to earn full marks: Substitute $g(x)$ into $h(x)$ correctly, then expand and simplify the expression fully.

(c)

1. Find $g(0)$: $g(0) = (0)^2 - 6(0) + 5 = 0 - 0 + 5 = 5$ [Calculate g(0)]

2. Solve $h(x) = 5$: $3x - 2 = 5$ $3x = 7$ $x = \frac{7}{3}$

$\boxed{x = \frac{7}{3}}$

How to earn full marks: First, calculate $g(0)$, then set $h(x)$ equal to that value and solve for $x$.

Common Pitfall: When completing the square, double-check your arithmetic, especially when subtracting the squared term. Also, remember the order of operations when finding composite functions – work from the inside out!

Exam-Style Question 3 — Paper 2 (Calculator Allowed) [9 marks]

Question:

The function $f$ is defined by $f(x) = \frac{8}{3x+2}$ for $x \ge 0$.

(a) Find $f^{-1}(x)$. [4]

(b) The function $g$ is defined by $g(x) = \sqrt{x+5}$ for $x \ge -5$. Find the domain and range of $fg(x)$. Give your answers to 2 decimal places where appropriate. [5]

Worked Solution:

(a)

1. Let $y = f(x)$, so $y = \frac{8}{3x + 2}$. Swap $x$ and $y$: $x = \frac{8}{3y + 2}$ [Swap x and y to begin finding the inverse]

2. Rearrange to make $y$ the subject: $x(3y + 2) = 8$ $3xy + 2x = 8$ $3xy = 8 - 2x$ $y = \frac{8 - 2x}{3x}$ [Isolate y]

3. Write the inverse function using correct notation: $f^{-1}(x) = \frac{8 - 2x}{3x}$ [Express the answer in terms of f^{-1}(x)]

$\boxed{f^{-1}(x) = \frac{8 - 2x}{3x}}$

How to earn full marks: Remember to swap $x$ and $y$ at the beginning, and then isolate $y$ to express the inverse function correctly.

(b)

1. Find the composite function $fg(x)$: $fg(x) = f(g(x)) = f(\sqrt{x+5}) = \frac{8}{3\sqrt{x+5} + 2}$ [Substitute g(x) into f(x)]

2. Find the domain of $fg(x)$. Since $x \ge -5$ and the square root is defined, the domain is $x \ge -5$. [State the domain of the composite function, based on the domain of g(x)]

3. Find the range of $fg(x)$. When $x=-5$, $fg(x)=\frac{8}{3\sqrt{-5+5} + 2}=4$. As $x$ approaches infinity, $\sqrt{x+5}$ also approaches infinity, so $3\sqrt{x+5}+2$ approaches infinity, and thus $\frac{8}{3\sqrt{x+5} + 2}$ approaches 0. Therefore, the range is $0 < fg(x) \le 4$.

$\boxed{\text{Domain: } x \ge -5, \text{ Range: } 0 < fg(x) \le 4}$

How to earn full marks: Find the composite function, then state the domain considering the domain of $g(x)$, and determine the range by considering the minimum and maximum possible values of $fg(x)$.

Common Pitfall: When dealing with composite functions and square roots, always consider the domain restrictions of the inner function. Also, remember that the range of a function can be restricted by the denominator approaching infinity.

Exam-Style Question 4 — Paper 2 (Calculator Allowed) [11 marks]

Question:

The function $f(x)$ is defined as $f(x) = |x^2 - 3x - 4|$ for $x \in \mathbb{R}$.

(a) Sketch the graph of $y = f(x)$ for $-2 \le x \le 5$, showing the coordinates of any turning points and points where the graph intersects the $x$-axis. [5]

(b) Find the range of $f(x)$. [2]

(c) Determine the values of $k$ for which the equation $f(x) = k$ has exactly 4 solutions. [4]

Worked Solution:

(a)

1. Factorise the quadratic: $x^2 - 3x - 4 = (x+1)(x-4)$. The roots are $x=-1$ and $x=4$. [Find the roots of the quadratic]

2. Find the vertex of the quadratic before taking the modulus: $x = \frac{-(-3)}{2(1)} = 1.5$. $y = (1.5)^2 - 3(1.5) - 4 = 2.25 - 4.5 - 4 = -6.25$. [Find the x-coordinate of the vertex using -b/2a, then substitute to find the y-coordinate]

3. The vertex of $y = |x^2 - 3x - 4|$ is therefore at $(1.5, 6.25)$. The points of intersection with the x-axis are (-1,0) and (4,0).

4. Sketch the graph, reflecting the portion below the x-axis. The graph passes through (-2,6) and (5,6).

How to earn full marks: Find the roots and vertex of the quadratic, reflect the part of the graph below the x-axis, and label all key points clearly on your sketch.

(b)

1. From the graph, the minimum value of $f(x)$ is 0 and the maximum value is 6 (at $x=-2$ and $x=5$).
2. The range of $f(x)$ is $0 \le f(x) \le 6$.

$\boxed{0 \le f(x) \le 6}$

How to earn full marks: State the range using the correct inequality signs, based on the minimum and maximum y-values of the graph.

(c)

1. For $f(x) = k$ to have exactly 4 solutions, the line $y = k$ must intersect the graph of $y = f(x)$ in 4 places.
2. This occurs when $0 < k < 6.25$.

$\boxed{0 < k < 6.25}$

How to earn full marks: Determine the range of $k$ values for which a horizontal line $y=k$ intersects the graph at exactly four points, using the vertex of the reflected portion of the graph.

Common Pitfall: When sketching modulus functions, remember to reflect the part of the graph that's below the x-axis. Also, be careful when finding the vertex of the original quadratic – a small arithmetic error can throw off the entire graph.