Functions and Iteration

This lesson covers function notation, composite functions, inverse functions, and iteration — all Higher tier topics in the Edexcel GCSE (1MA1) specification.

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Key Vocabulary

Term	Meaning
Function	A rule that maps each input to exactly one output
f(x)	"f of x" — the output when x is the input to function f
Domain	The set of allowed inputs
Range	The set of possible outputs
Composite function	Applying one function then another: fg(x) means do g first, then f
Inverse function	The function that reverses f; written $f^{-1}(x)$f−1(x)
Iteration	Repeatedly applying a formula to get closer to a solution

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Function Notation

When we write f(x) = 3x + 1, this means: "the function f takes an input x and gives the output 3x + 1."

Worked Example 1

$f(x) = 2x^{2} - 3x + 1$f(x)=2x2−3x+1

Find: (a) f(4) (b) f(−2) (c) f(0)

(a) f(4) = 2(16) − 3(4) + 1 = 32 − 12 + 1 = 21

(b) f(−2) = 2(4) − 3(−2) + 1 = 8 + 6 + 1 = 15

(c) f(0) = 2(0) − 3(0) + 1 = 1

Function Machine Diagram

flowchart LR
    IN[Input x] --> OP1[Square: x squared]
    OP1 --> OP2[Multiply by 2]
    OP2 --> OP3[Subtract 3x]
    OP3 --> OP4[Add 1]
    OP4 --> OUT[Output f of x]
    INV["Inverse: undo each step<br/>in reverse order"] -.-> IN

A function machine helps visualise f(x) as a sequence of operations applied to the input. The inverse function reverses the machine: undo the last operation first, then work backwards.

Worked Example 2

f(x) = 5x − 2. Find the value of a such that f(a) = 18.

5a − 2 = 18

5a = 20

a = 4

Worked Example 3

$f(x) = x^{2} + kx - 5$f(x)=x2+kx−5 and f(3) = 1. Find k.

$(3)^{2} + 3k - 5 = 1$(3)2+3k−5=1

9 + 3k − 5 = 1

3k + 4 = 1

3k = −3

k = −1

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Composite Functions [H]

fg(x) means: apply g first, then apply f to the result.

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

Worked Example 4

f(x) = 2x + 3 and $g(x) = x^{2}$g(x)=x2.

Find: (a) fg(2) (b) gf(2) (c) fg(x) (d) gf(x)

(a) g(2) = 4, then f(4) = 2(4) + 3 = 11

(b) f(2) = 7, then g(7) = 49

$(c) fg(x) = f(x^{2}) = 2x^{2} + 3$(c)fg(x)=f(x2)=2x2+3

$(d) gf(x) = g(2x + 3) = (2x + 3)^{2}$(d)gf(x)=g(2x+3)=(2x+3)2

Worked Example 5

f(x) = x + 5 and g(x) = 3x − 1. Find fg(x) and solve fg(x) = 20.

fg(x) = f(3x − 1) = (3x − 1) + 5 = 3x + 4

$3x + 4 = 20 \to 3x = 16 \to x =$3x+4=20→3x=16→x= 16/3

Worked Example 6

f(x) = 4x − 1. Find ff(x).

ff(x) = f(4x − 1) = 4(4x − 1) − 1 = 16x − 4 − 1 = 16x − 5

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Inverse Functions [H]

The inverse function $f^{-1}$f−1 reverses the effect of f. If f(a) = b, then $f^{-1}(b) = a$f−1(b)=a.

Finding the Inverse

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

Worked Example 7

Find $f^{-1}(x)$f−1(x) for f(x) = 3x + 7.

y = 3x + 7

Swap: x = 3y + 7

x − 7 = 3y

y = (x − 7)/3

$f^{-1}(x) = (x - 7)/3$f−1(x)=(x−7)/3

Worked Example 8

Find $f^{-1}(x)$f−1(x) for f(x) = (2x − 1)/5.

y = (2x − 1)/5

Swap: x = (2y − 1)/5

5x = 2y − 1

5x + 1 = 2y

y = (5x + 1)/2

$f^{-1}(x) = (5x + 1)/2$f−1(x)=(5x+1)/2

Worked Example 9

Find $f^{-1}(x)$f−1(x) for f(x) = (x + 3)/(x − 1), $x \neq 1$x=1.

y = (x + 3)/(x − 1)

Swap: x = (y + 3)/(y − 1)

x(y − 1) = y + 3

xy − x = y + 3

xy − y = x + 3

y(x − 1) = x + 3

y = (x + 3)/(x − 1)

$f^{-1}(x) = (x + 3)/(x - 1)$f−1(x)=(x+3)/(x−1) — this function is its own inverse (a self-inverse function).

Key Properties

• $ff^{-1}(x) = f^{-1}f(x) = x ($ff−1(x)=f−1f(x)=x(the inverse undoes the function).
• The graph of $f^{-1}$f−1 is the reflection of the graph of f in the line y = x.

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Iteration [H]

Iteration means using a formula repeatedly to generate a sequence of values that converge to a solution.

The Method

If an equation can be rearranged to x = g(x), then:

1. Start with an initial value $x_{0}$x0​.
2. Calculate $x_{1} = g(x_{0})$x1​=g(x0​).
3. Calculate $x_{2} = g(x_{1})$x2​=g(x1​).
4. Continue until the values converge (settle down) to a stable value.

Worked Example 10

Show that $x^{3} + 2x - 5 = 0$x3+2x−5=0 can be rearranged to $x = \sqrt[3]{5 - 2x}$x=35−2x​. Use $x_{0} = 1$x0​=1 and iterate to find a solution to 3 decimal places.

Rearrangement: $x^{3} = 5 - 2x \to x = (5 - 2x)^{1/3}$x3=5−2x→x=(5−2x)1/3

$x_{0} = 1$x0​=1

$x_{1} = (5 - 2(1))^{1/3} = 3^{1/3} = 1.44225$x1​=(5−2(1))1/3=31/3=1.44225…

$x_{2} = (5 - 2(1.44225))^{1/3} = (2.11550)^{1/3} = 1.28396$x2​=(5−2(1.44225))1/3=(2.11550)1/3=1.28396…

$x_{3} = (5 - 2(1.28396))^{1/3} = (2.43208)^{1/3} = 1.34485$x3​=(5−2(1.28396))1/3=(2.43208)1/3=1.34485…

$x_{4} = (5 - 2(1.34485))^{1/3} = (2.31030)^{1/3} = 1.32119$x4​=(5−2(1.34485))1/3=(2.31030)1/3=1.32119…

$x_{5} = (5 - 2(1.32119))^{1/3} = (2.35762)^{1/3} = 1.33039$x5​=(5−2(1.32119))1/3=(2.35762)1/3=1.33039…

$x_{6} = (5 - 2(1.33039))^{1/3} = (2.33922)^{1/3} = 1.32682$x6​=(5−2(1.33039))1/3=(2.33922)1/3=1.32682…

$x_{7} = (5 - 2(1.32682))^{1/3} = (2.34636)^{1/3} = 1.32821$x7​=(5−2(1.32682))1/3=(2.34636)1/3=1.32821…

Values are converging to approximately x = 1.328 (3 d.p.).

Worked Example 11

The equation $x^{2} - 3x - 1 = 0$x2−3x−1=0 can be written as x = 3 + 1/x. Starting with $x_{0} = 3$x0​=3, find $x_{1}$x1​, $x_{2}$x2​, and $x_{3}$x3​ to 4 decimal places.

$x_{1} = 3 + 1/3 =$x1​=3+1/3= 3.3333

$x_{2} = 3 + 1/3.3333 = 3 + 0.3000 =$x2​=3+1/3.3333=3+0.3000= 3.3000

$x_{3} = 3 + 1/3.3000 = 3 + 0.3030 =$x3​=3+1/3.3000=3+0.3030= 3.3030

The solution converges to approximately 3.303 (this is the positive root of $x^{2} - 3x - 1 = 0)$x2−3x−1=0).

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Why Does Iteration Work?

Iteration works by rearranging an equation f(x) = 0 into the form x = g(x). The solution is where y = x and y = g(x) intersect. Starting from $x_{0}$x0​, the sequence "spirals" or "staircase" toward the intersection point — provided the iteration converges.

Not all rearrangements converge! Different rearrangements of the same equation may converge, diverge, or oscillate. The question will always tell you which rearrangement to use.

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Common Mistakes and Misconceptions