SQA SQA Higher 2022-style practice: The function is defined by f(x) = 5 - 2x for x R. Find a formula for the inverse function f^-1(x), and verify that f(f^-1(x)) = x.

Write y = 5 - 2x (1 mark). Swap x and y: x = 5 - 2y (1 mark). Make y the subject: 2y = 5 - x, so y = 5 - x2, giving f^-1(x) = 5 - x2 (1 mark). Verify: f(f^-1(x)) = 5 - 2(5 - x2) = 5 - (5 - x) = x, as required (1 mark). Markers reward swapping the variables, rearranging for y, and the verification by composition.

ScotlandMathsSyllabus dot point

How do functions combine, invert and transform, and how do those operations move and reshape their graphs?

Functions and their domain and range, composite functions, inverse functions, exponential and logarithmic graphs, and the graphs that result from translating, reflecting and stretching a known function.

A focused answer to the SQA Higher Mathematics functions and graphs content, covering domain and range, composite and inverse functions, the shapes of exponential and logarithmic graphs, and how translating, reflecting and stretching transforms the graph of a known function.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Domain and range
Composite functions
Inverse functions
Exponential and logarithmic graphs
Transformations
Examples in context
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What this dot point is asking

The SQA wants you to work with the domain and range of a function, form and evaluate composite functions, find inverse functions, recognise exponential and logarithmic graphs, and sketch graphs produced by translating, reflecting and stretching a known function.

Domain and range

The domain of a function is the set of inputs for which it is defined; the range is the set of outputs it produces. Two situations restrict a domain: a denominator must not be zero, so $\dfrac{1}{x - 3}$ excludes $x = 3$ , and the expression under a square root must not be negative, so $\sqrt{x - 1}$ requires $x \ge 1$ . Always quote the domain restriction when a function has one.

Composite functions

To form $f(g(x))$ , substitute $g(x)$ into $f$ . The order matters, so $f(g(x))$ is usually not the same as $g(f(x))$ . The domain of the composite must avoid any value that makes an inside function undefined.

Given

f(x) = 2x + 1

and

g(x) = x^2

, find

f(g(x))

and

g(f(x))

Form both composite functions and compare the results.

Step 1: Find $f(g(x))$ by applying $g$ first, then $f$

In $f(g(x))$ , $g$ is the inner function and acts first. Replace every $x$ in $f(x) = 2x + 1$ with $g(x) = x^2$ :

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

Step 2: Find $g(f(x))$ by applying $f$ first, then $g$

In $g(f(x))$ , $f$ is the inner function. Replace every $x$ in $g(x) = x^2$ with $f(x) = 2x + 1$ :

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

Step 3: Compare and conclude

$f(g(x)) = 2x^2 + 1$ and $g(f(x)) = (2x + 1)^2 = 4x^2 + 4x + 1$ . These are different functions, confirming that the order of composition matters: $f(g(x)) \neq g(f(x))$ in general.

Inverse functions

The inverse $f^{-1}$ reverses the effect of $f$ , so $f(f^{-1}(x)) = x$ . To find it, write $y = f(x)$ , swap $x$ and $y$ , then make $y$ the subject. The graph of $y = f^{-1}(x)$ is the reflection of $y = f(x)$ in the line $y = x$ .

Find the inverse of

f(x) = \dfrac{x + 4}{3}

Find a formula for the inverse function $f^{-1}(x)$ and verify it is correct.

Step 1: Write the function as an equation in $x$ and $y$

The inverse undoes $f$ , swapping inputs and outputs. Begin by writing $y = \dfrac{x + 4}{3}$ .

Step 2: Swap $x$ and $y$ to reflect the reversal of roles

Interchanging $x$ and $y$ gives $x = \dfrac{y + 4}{3}$ . This single step captures the idea that the output of $f$ becomes the input of $f^{-1}$ .

Step 3: Rearrange to make $y$ the subject

Multiply both sides by $3$ : $3x = y + 4$ . Subtract $4$ : $y = 3x - 4$ . Therefore $f^{-1}(x) = 3x - 4$ .

Step 4: Verify by composition

Check that $f(f^{-1}(x)) = x$ :

f(f^{-1}(x)) = \dfrac{(3x - 4) + 4}{3} = \dfrac{3x}{3} = x \checkmark

Final answer: $f^{-1}(x) = 3x - 4$ , confirmed by the composition check.

Exponential and logarithmic graphs

Transformations

Transformations split into outside changes (which affect $y$ and act vertically, in the natural direction) and inside changes (which affect $x$ and act horizontally, in the opposite direction to what they look like).

Examples in context

Graph transformations let you build a model from a known shape. A standard cooling curve $y = f(x)$ for an object can be shifted to $y = f(x) + 20$ to model the same cooling towards a room temperature of $20$ degrees rather than zero. Recognising that adding a constant outside the function raises the asymptote lets a scientist reuse one template curve for many ambient conditions without re-deriving the model.

Try this

Q1. Given $f(x) = x - 3$ and $g(x) = 4x$ , find $g(f(x))$ . [2 marks]

Cue. $g(f(x)) = 4(x - 3) = 4x - 12$ .

Q2. Describe the transformation that maps $y = f(x)$ onto $y = f(x - 2) + 5$ . [2 marks]

Cue. A translation $2$ to the right and $5$ up.

Q3. Given $f(x) = \dfrac{1}{x - 1}$ , state the value excluded from the domain and find $f^{-1}(x)$ . [3 marks]

Cue. Exclude $x = 1$ ; from $y = \dfrac{1}{x - 1}$ swap and rearrange to $f^{-1}(x) = \dfrac{1}{x} + 1$ .

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA Higher 20195 marksFunctions are defined by

f(x) = 3x - 1

and

g(x) = \dfrac{1}{x + 2}

, where

x \neq -2

. Find an expression for

h(x) = f(g(x))

in its simplest form, and state the value of

x

that must be excluded from the domain of

h

Show worked answer →

Apply $g$ first, then $f$ : $h(x) = f\big(g(x)\big) = f\!\left(\dfrac{1}{x + 2}\right)$ (1 mark).

Substitute into $f(x) = 3x - 1$ : $h(x) = 3 \cdot \dfrac{1}{x + 2} - 1 = \dfrac{3}{x + 2} - 1$ (1 mark).

Combine over a common denominator: $h(x) = \dfrac{3 - (x + 2)}{x + 2} = \dfrac{1 - x}{x + 2}$ (2 marks).

The excluded value is $x = -2$ , because $g$ (the inner function) is undefined there (1 mark). Markers reward the correct order of composition, the simplification, and identifying the domain restriction from the inner function.

SQA Higher 20224 marksThe function is defined by

f(x) = 5 - 2x

for

x \in \mathbb{R}

. Find a formula for the inverse function

f^{-1}(x)

, and verify that

f\big(f^{-1}(x)\big) = x

Show worked answer →

Write $y = 5 - 2x$ (1 mark). Swap $x$ and $y$ : $x = 5 - 2y$ (1 mark).

Make $y$ the subject: $2y = 5 - x$ , so $y = \dfrac{5 - x}{2}$ , giving $f^{-1}(x) = \dfrac{5 - x}{2}$ (1 mark).

Verify: $f\big(f^{-1}(x)\big) = 5 - 2\left(\dfrac{5 - x}{2}\right) = 5 - (5 - x) = x$ , as required (1 mark). Markers reward swapping the variables, rearranging for $y$ , and the verification by composition.

Related dot points

Sources & how we know this

SQA Higher Mathematics Course Specification — SQA (2018)