How to Solve: Functions (Domain, Range, Composite, Inverse, Modulus, Transformations) – GCSE/A-Level Maths Guide

How to Solve: Functions (Domain, Range, Composite, Inverse, Modulus, Transformations) – GCSE/A-Level Maths Guide

Introduction

Mastering functions unlocks 15-20% of your GCSE/A-Level Maths exam—think graph transformations, real-world modelling, and even coding. One question on inverses or composites could be the difference between a Grade 6 and a Grade 8.

What You Need To Know First

1. Basic algebra – Solving equations, rearranging formulas.
2. Graphs – Plotting linear and quadratic functions.
3. Inequalities – Solving and representing on number lines.

Key Vocabulary

Formulas To Know

Step-by-Step Method

1. Finding the Domain

Step 1: Identify restrictions. - Denominator ≠ 0 (e.g., ( \frac{1}{x} ) → ( x \neq 0 )) - Square root ≥ 0 (e.g., ( \sqrt{x} ) → ( x \geq 0 )) - Logarithm > 0 (e.g., ( \ln(x) ) → ( x > 0 ))

Step 2: Write in interval notation or inequality form. - Example: ( f(x) = \frac{1}{x-2} ) → Domain: ( x \neq 2 ) or ( (-\infty, 2) \cup (2, \infty) ).

2. Finding the Range

Step 1: Sketch the graph (if possible). Step 2: Identify the lowest and highest y-values. - Example: ( f(x) = x^2 ) → Range: ( y \geq 0 ). - Example: ( f(x) = \sin(x) ) → Range: ( -1 \leq y \leq 1 ).

Step 3: For linear functions, range is all real numbers unless restricted.

3. Composite Functions ( f(g(x)) )

Step 1: Write ( f(g(x)) ) clearly. Step 2: Substitute ( g(x) ) into ( f ). - Example: ( f(x) = 2x + 1 ), ( g(x) = x^2 ) ( f(g(x)) = 2(x^2) + 1 = 2x^2 + 1 ).

Step 3: Simplify if needed.

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

Step 1: Write ( y = f(x) ). Step 2: Swap ( x ) and ( y ). Step 3: Solve for ( y ). - Example: ( f(x) = 3x + 2 ) ( y = 3x + 2 ) Swap: ( x = 3y + 2 ) Solve: ( y = \frac{x - 2}{3} ) So, ( f^{-1}(x) = \frac{x - 2}{3} ).

Step 4: Check by composing ( f(f^{-1}(x)) = x ).

5. Modulus Functions ( |f(x)| )

Step 1: Sketch ( f(x) ) first. Step 2: Reflect any negative parts above the x-axis. - Example: ( f(x) = x - 2 ) ( |f(x)| = |x - 2| ) (V-shape with vertex at ( x = 2 )).

6. Transformations

Step 1: Identify the transformation. - ( f(x + a) ) → Left by ( a ) - ( f(x) + a ) → Up by ( a ) - ( f(ax) ) → Horizontal stretch/compression by ( \frac{1}{a} ) - ( a f(x) ) → Vertical stretch/compression by ( a ) - ( -f(x) ) → Reflection in x-axis - ( f(-x) ) → Reflection in y-axis

Step 2: Apply one transformation at a time.

Worked Examples

Example 1 – Basic (Domain & Range)

Question: Find the domain and range of ( f(x) = \frac{1}{x-3} ).

Solution: 1. Domain: Denominator ≠ 0 → ( x - 3 \neq 0 ) → ( x \neq 3 ).
Domain: ( (-\infty, 3) \cup (3, \infty) ). 2. Range: ( \frac{1}{x-3} ) can never be 0 (numerator is 1).
Range: ( (-\infty, 0) \cup (0, \infty) ).

What we did and why: - Domain: Excluded ( x = 3 ) because division by zero is undefined. - Range: The function never touches ( y = 0 ) because ( \frac{1}{x-3} ) can’t equal zero.

Example 2 – Medium (Composite & Inverse)

Question: Given ( f(x) = 2x + 1 ) and ( g(x) = x^2 ), find: a) ( f(g(3)) ) b) ( f^{-1}(x) )

Solution: a) Composite ( f(g(3)) ): 1. First, find ( g(3) = 3^2 = 9 ). 2. Then, ( f(9) = 2(9) + 1 = 19 ).
So, ( f(g(3)) = 19 ).

b) Inverse ( f^{-1}(x) ): 1. Write ( y = 2x + 1 ). 2. Swap ( x ) and ( y ): ( x = 2y + 1 ). 3. Solve for ( y ): ( y = \frac{x - 1}{2} ).
So, ( f^{-1}(x) = \frac{x - 1}{2} ).

What we did and why: - Composite: We nested ( g ) inside ( f ) and evaluated step-by-step. - Inverse: We reversed the function by swapping ( x ) and ( y ).

Example 3 – Exam-Style (Modulus & Transformations)

Question: The function ( f(x) = x^2 - 4 ) is transformed to ( g(x) = |f(x + 1)| ). a) Sketch ( g(x) ). b) State the range of ( g(x) ).

Solution: a) Sketching ( g(x) ): 1. Start with ( f(x) = x^2 - 4 ) (U-shape, vertex at ( (0, -4) )). 2. Shift left by 1: ( f(x + 1) = (x + 1)^2 - 4 ) (vertex at ( (-1, -4) )). 3. Apply modulus: Reflect any negative parts above the x-axis.
- The graph dips to ( y = -4 ) but modulus makes it ( y = 4 ).
- Final shape: W-shape with minimum at ( y = 0 ).

b) Range of ( g(x) ): - The lowest point is ( y = 0 ) (from the modulus). - The highest point is ( y \to \infty ). So, range: ( y \geq 0 ).

What we did and why: - Transformations: We applied shift left then modulus in order. - Range: The modulus ensures no negative outputs, so the range starts at 0.

Common Mistakes

Exam Traps

1-Minute Recap

"Right, listen up—this is your last-minute functions checklist!

1. Domain: Check for denominators ≠ 0, roots ≥ 0, and logs > 0.
2. Range: Sketch the graph—what’s the lowest and highest ( y )?
3. Composite ( f(g(x)) ): Do ( g ) first, then ( f ). Never the other way round!
4. Inverse ( f^{-1}(x) ): Swap ( x ) and ( y ), then solve. Not the same as reciprocal!
5. Modulus ( |f(x)| ): Reflect the negative parts above the x-axis.
6. Transformations:
7. ( f(x + a) ) → left by ( a )
8. ( f(x) + a ) → up by ( a )
9. ( f(ax) ) → horizontal stretch by ( \frac{1}{a} )
10. ( a f(x) ) → vertical stretch by ( a )

Final tip: If stuck, draw the graph—it’s your secret weapon. Now go smash that exam!