Functions and Algebraic Fractions [H]

This lesson covers Higher tier content on function notation, composite functions, inverse functions, and algebraic fractions. These topics are among the most challenging in the AQA GCSE Mathematics specification and are frequently tested in the later questions on the Higher paper. A solid understanding here will also provide an excellent foundation for A-level Mathematics.

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Function Notation

A function is a rule that maps each input to exactly one output. In GCSE Mathematics, functions are written using the notation:

f(x) = ... or g(x) = ...

The letter before the bracket names the function; the letter inside the bracket is the input (usually x).

Worked Example 1

f(x) = 3x + 2

(a) Find f(4).

Replace x with 4: f(4) = 3(4) + 2 = 12 + 2 = 14

(b) Find f(-3).

f(-3) = 3(-3) + 2 = -9 + 2 = -7

(c) Find the value of x when f(x) = 20.

3x + 2 = 20

3x = 18

x = 6

Worked Example 2

g(x) = x^2 - 4x + 1

Find g(5).

g(5) = (5)^2 - 4(5) + 1 = 25 - 20 + 1 = 6

Exam Tip: f(x) = 20 does NOT mean f times x equals 20. The brackets denote a function, not multiplication. When f(x) = 20, you substitute the expression for f(x) and solve the equation.

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Composite Functions

A composite function applies one function and then another. The notation fg(x) means "apply g first, then apply f to the result."

Key Rule

fg(x) = f(g(x)) — apply g first, then f.

gf(x) = g(f(x)) — apply f first, then g.

Note: fg(x) is NOT the same as gf(x) in general.

Worked Example 3

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

(a) Find fg(3).

Step 1: g(3) = 3^2 = 9

Step 2: f(9) = 2(9) + 1 = 19

fg(3) = 19

(b) Find gf(3).

Step 1: f(3) = 2(3) + 1 = 7

Step 2: g(7) = 7^2 = 49

gf(3) = 49

(c) Find fg(x) as a single expression.

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

(d) Find gf(x) as a single expression.

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

flowchart LR
    A[Input x] --> B["Apply g: g(x) = x^2"]
    B --> C["Result: x^2"]
    C --> D["Apply f: f(x^2) = 2x^2 + 1"]
    D --> E["Output: fg(x) = 2x^2 + 1"]

Exam Tip: The order matters. fg(x) means do g first, then f. Think of it like reading right to left: in fg, the function nearest to x (which is g) is applied first. This is counter-intuitive but crucial.

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Inverse Functions

The inverse function f^(-1)(x) reverses the effect of f(x). If f maps a to b, then f^(-1) maps b back to a.

How to Find the Inverse

1. Write y = f(x).
2. Swap x and y.
3. Rearrange to make y the subject.
4. Replace y with f^(-1)(x).

Worked Example 4

Find the inverse of f(x) = 2x + 5.

Step 1: y = 2x + 5

Step 2: Swap x and y: x = 2y + 5

Step 3: Rearrange: x - 5 = 2y, so y = (x - 5)/2

f^(-1)(x) = (x - 5)/2

Check: f(3) = 2(3) + 5 = 11. f^(-1)(11) = (11 - 5)/2 = 3. Correct — the inverse takes us back.

Worked Example 5

Find the inverse of g(x) = (3x - 1)/4.

y = (3x - 1)/4

Swap: x = (3y - 1)/4

Multiply by 4: 4x = 3y - 1

Add 1: 4x + 1 = 3y

Divide by 3: y = (4x + 1)/3

g^(-1)(x) = (4x + 1)/3

Key Properties of Inverse Functions

• f(f^(-1)(x)) = x and f^(-1)(f(x)) = x
• The graph of f^(-1)(x) is a reflection of f(x) in the line y = x

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Algebraic Fractions

Algebraic fractions follow the same rules as numerical fractions, but with algebraic expressions in the numerator and/or denominator.

Simplifying Algebraic Fractions

To simplify, factorise the numerator and denominator, then cancel common factors.

Worked Example 6

Simplify: (6x^2) / (9x)

Factor: 6x^2 / 9x = (6/9) x (x^2/x) = (2/3) x x = 2x/3

Worked Example 7

Simplify: (x^2 - 9) / (x + 3)

Factorise the numerator (difference of two squares): (x + 3)(x - 3) / (x + 3)

Cancel (x + 3): x - 3

Worked Example 8

Simplify: (x^2 + 5x + 6) / (x^2 + 3x + 2)

Factorise numerator: (x + 2)(x + 3)

Factorise denominator: (x + 1)(x + 2)

Cancel (x + 2): (x + 3) / (x + 1)

Exam Tip: You can only cancel factors, never individual terms. For example, in (x + 3)/(x + 1), you cannot cancel the x's. You can only cancel if the entire bracket is a common factor of both numerator and denominator.

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Adding and Subtracting Algebraic Fractions

To add or subtract algebraic fractions:

1. Find a common denominator.
2. Rewrite each fraction with the common denominator.
3. Add or subtract the numerators.
4. Simplify if possible.

Worked Example 9

Simplify: 3/(x + 1) + 2/(x + 3)

Common denominator = (x + 1)(x + 3)

= 3(x + 3) / ((x + 1)(x + 3)) + 2(x + 1) / ((x + 1)(x + 3))

= (3(x + 3) + 2(x + 1)) / ((x + 1)(x + 3))

Expand the numerator: 3x + 9 + 2x + 2 = 5x + 11

Answer: (5x + 11) / ((x + 1)(x + 3))

Worked Example 10

Simplify: 5/(x - 2) - 3/(x + 1)

Common denominator = (x - 2)(x + 1)

= 5(x + 1) / ((x - 2)(x + 1)) - 3(x - 2) / ((x - 2)(x + 1))

Numerator: 5(x + 1) - 3(x - 2) = 5x + 5 - 3x + 6 = 2x + 11

Answer: (2x + 11) / ((x - 2)(x + 1))

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Multiplying and Dividing Algebraic Fractions

Multiplying

Multiply numerators together and denominators together, then simplify.

Worked Example 11

Simplify: (x/3) times (6/(x + 2))

= (x times 6) / (3 times (x + 2))

= 6x / (3(x + 2))

= 2x / (x + 2)

Dividing

To divide by a fraction, flip the second fraction and multiply.

Worked Example 12

Simplify: (x^2 - 4)/(x + 1) divided by (x - 2)/3

Flip and multiply: ((x^2 - 4)/(x + 1)) times (3/(x - 2))

Factorise x^2 - 4 = (x + 2)(x - 2)