Functions and Graphs

What you'll learn

How to read function notation, including domains and ranges.
How to work with composite and piecewise functions.
How to find inverse functions and know when they exist.
How to sketch modulus graphs and transformed graphs accurately.

1. Function notation, domain and range

A function is more than just a formula. At A-Level, you must also notice the allowed inputs.

Definition

Function, domain and range

A function is a rule that gives exactly one output for each allowed input. The domain is the set of allowed inputs; the range is the set of outputs actually produced.

If you are asked for the maximum possible domain, use the largest set of real values that makes the expression valid.

Useful restrictions:

Denominators cannot be zero.
For $expression\sqrt{\text{expression}}$ , the expression must be at least zero.
For $1expression\frac{1}{\sqrt{\text{expression}}}$ , the expression must be greater than zero.
For $ln⁡(expression)\ln(\text{expression})$ , the expression must be greater than zero.
For $arcsin⁡x\arcsin x$ , the maximum possible domain is $−1≤x≤1-1\le x\le 1$ .

To find a range, think about the graph. For a quadratic, completing the square is often the quickest route.

The completed-square form shows the vertex of the quadratic and hence its minimum value and range.

Example

Maximum domain and range

Find the maximum possible domain of $p(x)=17−2xp(x)=\frac{1}{\sqrt{7-2x}}$ . Also find the range of $f(x)=x^2-4x+6$ for $x∈Rx\in\mathbb{R}$ .

For $p (x)$ , the square root is in the denominator, so $7 - 2 x$ must be greater than zero.
Solve the inequality:
$\Rightarrow x<\frac{7}{2}$
So the maximum possible domain is $x<72x<\frac{7}{2}$ .
Complete the square for $f (x)$ :
$x^2-4x+6=(x-2)^2+2$
The square term is at least zero, so the minimum value is 2. Therefore the range is $f(x)≥2f(x)\ge 2$ .

Common Mistake

Square root in a denominator

A square root in the numerator can equal zero, but a square root in the denominator cannot. That is why $1x\frac{1}{\sqrt{x}}$ needs $x > 0$ , not $x≥0x\ge 0$ .

2. Composite and piecewise functions

Definition

Composite function

The composite $f g (x)$ means $f (g (x))$ : apply $g$ first, then apply $f$ . Similarly, $g f (x) = g (f (x))$ .

The function nearest to $x$ happens first. This matters because, in general, $fg(x)≠gf(x)fg(x)\neq gf(x)$ .

Composite functions are evaluated from the inside out, so fg means apply g first and then f.

Example

Evaluating and solving composites

Let $f (x) = 4 x - 1$ and $g(x)=3x+2g(x)=\frac{3}{x+2}$ , where $x≠−2x\neq -2$ . Find $f g (1)$ and solve $gf(x)=12gf(x)=\frac{1}{2}$ .

For $f g (1)$ , work inside out: first find $g (1)$ .
$g(1)=31+2=1g(1)=\frac{3}{1+2}=1$
Now apply $f$ to that result.
$f (1) = 4 (1) - 1 = 3$
So $f g (1) = 3$ .
For $g f (x)$ , apply $f$ first, then $g$ :
$gf(x)=g(f(x))=3(4x−1)+2=34x+1gf(x)=g(f(x))=\frac{3}{(4x-1)+2}=\frac{3}{4x+1}$
Solve $gf(x)=12gf(x)=\frac{1}{2}$ :
$34x+1=12⇒6=4x+1⇒x=54\frac{3}{4x+1}=\frac{1}{2}\Rightarrow 6=4x+1\Rightarrow x=\frac{5}{4}$
Check the denominator is not zero. Since $4x+1≠04x+1\neq 0$ , the solution $x=54x=\frac{5}{4}$ is valid.

Definition

Piecewise function

A piecewise function uses different formulas on different parts of its domain. Always choose the formula whose condition matches the current input.

A piecewise graph shows which formula applies on each part of the domain, with open and closed endpoints marking the boundary.

Example

A repeated piecewise function

For

\begin{cases} 4-2x & x<1\\ (x-1)^2+3 & x\ge 1 \end{cases}

find $pp (0)$ and solve $p (x) > 12$ .

Since $0 < 1$ , use the first branch:
$p (0) = 4 - 2 (0) = 4$
Now find $p (4)$ . Since $4≥14\ge 1$ , use the second branch:
$p(4)=(4-1)^2+3=12$
Therefore $pp (0) = 12$ .
For $p (x) > 12$ , test each branch. On $x < 1$ :
$4−2x>12⇒−2x>8⇒x<−44-2x>12\Rightarrow -2x>8\Rightarrow x<-4$
On $x≥1x\ge 1$ :
$(x−1)2+3>12⇒(x−1)2>9(x-1)^2+3>12\Rightarrow (x-1)^2>9$
Because this branch has $x≥1x\ge 1$ , we have $x−1≥0x-1\ge 0$ , so $x - 1 > 3$ . Hence $x > 4$ .
The solution is $x < - 4$ or $x > 4$ .

3. Inverse functions

Definition

Inverse function

An inverse function, written $f^{-1}$ , undoes $f$ : if $f (a) = b$ , then $f^{-1}(b)=a$ . It exists as a function only when $f$ is one-to-one, meaning no two different inputs give the same output.

A quadratic such as $x^2$ is not one-to-one on all real values, because two different inputs can give the same output. Restricting the domain can fix this.

The horizontal line test shows why x² is not one-to-one unless its domain is restricted.

Key Idea

Domain and range swap

If $f^{-1}$ exists, the domain of $f^{-1}$ is the range of $f$ , and the range of $f^{-1}$ is the domain of $f$ .

To find an inverse:

Write $y = f (x)$ .
Rearrange to make $x$ the subject.
Swap $x$ and $y$ .
State the domain of the inverse.

The graphs of $y = f (x)$ and $y=f^{-1}(x)$ are reflections in the line $y = x$ .

A function and its inverse are mirror images in the line y = x, with coordinates swapped.

Example

Finding inverse functions

Find the inverse of $u(x)=(x-2)^2+5$ , where $x≥2x\ge 2$ . Also find the inverse of $v(x)=3+ln⁡(2x+1)v(x)=3+\ln(2x+1)$ .

For $u (x)$ , the smallest value occurs at $x = 2$ , so the range is $u(x)≥5u(x)\ge 5$ .
Start with $y=(x-2)^2+5$ and rearrange:
$y-5=(x-2)^2$
Since the domain is $x≥2x\ge 2$ , choose the positive square root. Therefore $u−1(x)=2+x−5u^{-1}(x)=2+\sqrt{x-5}$ , with domain $x≥5x\ge 5$ .
For $v (x)$ , start by writing:
$y=3+ln⁡(2x+1)y=3+\ln(2x+1)$
Rearrange using exponentials to undo the logarithm:
$ey−3=2x+1⇒x=ey−3−12e^{y-3}=2x+1\Rightarrow x=\frac{e^{y-3}-1}{2}$
Swap the letters. So $v−1(x)=ex−3−12v^{-1}(x)=\frac{e^{x-3}-1}{2}$ , with domain $x∈Rx\in\mathbb{R}$ .

Common Mistake

Inverse is not reciprocal

$f^{-1}(x)$ means the inverse function, not $1f(x)\frac{1}{f(x)}$ . These are completely different unless a question specifically defines a reciprocal.

4. Modulus graphs and inequalities

Definition

Modulus

The modulus $∣ A ∣$ means the non-negative size of $A$ :

\begin{cases} A & A\ge 0\\ -A & A<0 \end{cases}

A graph like $y = ∣2 x - 1∣$ is V-shaped. The vertex occurs where the expression inside the modulus is zero.

The modulus graph has a vertex where the expression inside the modulus is zero.

For inequalities, if $k > 0$ , then $∣ A ∣ < k$ means $- k < A < k$ .

Example

A modulus sketch and inequality

Sketch $y = 2 - ∣2 x - 1∣$ , state where it meets the axes, and solve $2 - ∣2 x - 1∣ > 0$ .

The graph of y = 2 − |2x − 1| is an upside-down V, and the positive part lies between its x-intercepts.

The vertex occurs when $2 x - 1 = 0$ , so $x=12x=\frac{1}{2}$ . At this point, $y = 2$ .
The graph is an upside-down V with vertex $(12,2)\left(\frac{1}{2},2\right)$ .
For the y-intercept, set $x = 0$ : $y = 2 - ∣ - 1∣ = 1$ . So the y-intercept is $(0, 1)$ .
For the x-intercepts, solve:
$2−∣2x−1∣=0⇒∣2x−1∣=22-|2x-1|=0\Rightarrow |2x-1|=2$
This gives $2 x - 1 = 2$ or $2 x - 1 = - 2$ , so $x=32x=\frac{3}{2}$ or $x=−12x=-\frac{1}{2}$ .
For the inequality, solve $∣2 x - 1∣ < 2$ :
$−2<2x−1<2⇒−1<2x<3⇒−12<x<32-2<2x-1<2\Rightarrow -1<2x<3\Rightarrow -\frac{1}{2}<x<\frac{3}{2}$

Common Mistake

Variable right-hand sides

For an inequality like $∣2 x - 3∣ < x + 1$ , the right-hand side depends on $x$ . Remember it must be positive, or split into cases carefully.

5. Graph transformations

Definition

Graph transformation

A graph transformation moves, stretches or reflects an existing graph without needing to plot every point from scratch.

Key transformations:

A transformation summary shows how common changes to f(x) affect points on the graph.

$y = f (x) + b$ shifts the graph vertically by $b$ .
$y = f (x + a)$ shifts the graph left by $a$ .
$y = c f (x)$ stretches vertically by scale factor $c$ .
$y = f (c x)$ stretches horizontally by scale factor $1c\frac{1}{c}$ , for $c > 0$ .
$y = ∣ f (x) ∣$ reflects parts below the x-axis upwards.
$y = f (∣ x ∣)$ keeps the right-hand half and reflects it in the y-axis.

A stationary point is a point where the gradient is zero, such as a smooth maximum or minimum.

Example

Transforming key points

The graph of $y = f (x)$ has stationary points at $A (- 2, 4)$ and $B (6, - 1)$ . Find the corresponding stationary points on $y = 3 f (2 x)$ . Also find constants $a$ and $b$ so that $y = f (x + a) + b$ moves $B$ to the origin.

In $f (2 x)$ , the x-coordinates are divided by 2. So $A$ moves to $(- 1, 4)$ and $B$ moves to $(3, - 1)$ .
The outside factor 3 multiplies the y-coordinates. So the new stationary points are $(- 1, 12)$ and $(3, - 3)$ .
For $y = f (x + a) + b$ , the point $B (6, - 1)$ moves to $(6 - a, - 1 + b)$ .
To move $B$ to the origin, solve $6 - a = 0$ and $- 1 + b = 0$ .
Hence $a = 6$ and $b = 1$ .

Exam technique

In the exam

Write composites as nested functions first, for example $g f (x) = g (f (x))$ , before simplifying.
For inverses, always state the domain of the inverse; it is the range of the original function.
On sketches, mark key points: vertices, intercepts, asymptotes, stationary points and any domain endpoints.
For modulus inequalities, check whether the right-hand side is constant or variable before splitting cases.

Self review

Check yourself

If $f(x)=1x−4f(x)=\frac{1}{x-4}$ , what value can $f (x)$ never take?
Why does $x^2+1$ not have an inverse on $x∈Rx\in\mathbb{R}$ ?
What happens to the point $(6, - 2)$ under the transformation $y = 3 f (2 x)$ ?

Functions and Graphs

What you'll learn

How to read function notation, including domains and ranges.
How to work with composite and piecewise functions.
How to find inverse functions and know when they exist.
How to sketch modulus graphs and transformed graphs accurately.

1. Function notation, domain and range

A function is more than just a formula. At A-Level, you must also notice the allowed inputs.

Definition

Function, domain and range

A function is a rule that gives exactly one output for each allowed input. The domain is the set of allowed inputs; the range is the set of outputs actually produced.

If you are asked for the maximum possible domain, use the largest set of real values that makes the expression valid.

Useful restrictions:

Denominators cannot be zero.
For $expression\sqrt{\text{expression}}$ , the expression must be at least zero.
For $1expression\frac{1}{\sqrt{\text{expression}}}$ , the expression must be greater than zero.
For $ln⁡(expression)\ln(\text{expression})$ , the expression must be greater than zero.
For $arcsin⁡x\arcsin x$ , the maximum possible domain is $−1≤x≤1-1\le x\le 1$ .

To find a range, think about the graph. For a quadratic, completing the square is often the quickest route.

The completed-square form shows the vertex of the quadratic and hence its minimum value and range.

Example

Maximum domain and range

Find the maximum possible domain of $p(x)=17−2xp(x)=\frac{1}{\sqrt{7-2x}}$ . Also find the range of $f(x)=x^2-4x+6$ for $x∈Rx\in\mathbb{R}$ .

For $p (x)$ , the square root is in the denominator, so $7 - 2 x$ must be greater than zero.
Solve the inequality:
$\Rightarrow x<\frac{7}{2}$
So the maximum possible domain is $x<72x<\frac{7}{2}$ .
Complete the square for $f (x)$ :
$x^2-4x+6=(x-2)^2+2$
The square term is at least zero, so the minimum value is 2. Therefore the range is $f(x)≥2f(x)\ge 2$ .

Common Mistake

Square root in a denominator

A square root in the numerator can equal zero, but a square root in the denominator cannot. That is why $1x\frac{1}{\sqrt{x}}$ needs $x > 0$ , not $x≥0x\ge 0$ .

2. Composite and piecewise functions

Definition

Composite function

The composite $f g (x)$ means $f (g (x))$ : apply $g$ first, then apply $f$ . Similarly, $g f (x) = g (f (x))$ .

The function nearest to $x$ happens first. This matters because, in general, $fg(x)≠gf(x)fg(x)\neq gf(x)$ .

Composite functions are evaluated from the inside out, so fg means apply g first and then f.

Example

Evaluating and solving composites

Let $f (x) = 4 x - 1$ and $g(x)=3x+2g(x)=\frac{3}{x+2}$ , where $x≠−2x\neq -2$ . Find $f g (1)$ and solve $gf(x)=12gf(x)=\frac{1}{2}$ .

For $f g (1)$ , work inside out: first find $g (1)$ .
$g(1)=31+2=1g(1)=\frac{3}{1+2}=1$
Now apply $f$ to that result.
$f (1) = 4 (1) - 1 = 3$
So $f g (1) = 3$ .
For $g f (x)$ , apply $f$ first, then $g$ :
$gf(x)=g(f(x))=3(4x−1)+2=34x+1gf(x)=g(f(x))=\frac{3}{(4x-1)+2}=\frac{3}{4x+1}$
Solve $gf(x)=12gf(x)=\frac{1}{2}$ :
$34x+1=12⇒6=4x+1⇒x=54\frac{3}{4x+1}=\frac{1}{2}\Rightarrow 6=4x+1\Rightarrow x=\frac{5}{4}$
Check the denominator is not zero. Since $4x+1≠04x+1\neq 0$ , the solution $x=54x=\frac{5}{4}$ is valid.

Definition

Piecewise function

A piecewise function uses different formulas on different parts of its domain. Always choose the formula whose condition matches the current input.

A piecewise graph shows which formula applies on each part of the domain, with open and closed endpoints marking the boundary.

Example

A repeated piecewise function

For

\begin{cases} 4-2x & x<1\\ (x-1)^2+3 & x\ge 1 \end{cases}

find $pp (0)$ and solve $p (x) > 12$ .

Since $0 < 1$ , use the first branch:
$p (0) = 4 - 2 (0) = 4$
Now find $p (4)$ . Since $4≥14\ge 1$ , use the second branch:
$p(4)=(4-1)^2+3=12$
Therefore $pp (0) = 12$ .
For $p (x) > 12$ , test each branch. On $x < 1$ :
$4−2x>12⇒−2x>8⇒x<−44-2x>12\Rightarrow -2x>8\Rightarrow x<-4$
On $x≥1x\ge 1$ :
$(x−1)2+3>12⇒(x−1)2>9(x-1)^2+3>12\Rightarrow (x-1)^2>9$
Because this branch has $x≥1x\ge 1$ , we have $x−1≥0x-1\ge 0$ , so $x - 1 > 3$ . Hence $x > 4$ .
The solution is $x < - 4$ or $x > 4$ .

3. Inverse functions

Definition

Inverse function

A quadratic such as $x^2$ is not one-to-one on all real values, because two different inputs can give the same output. Restricting the domain can fix this.

The horizontal line test shows why x² is not one-to-one unless its domain is restricted.

Key Idea

Domain and range swap

If $f^{-1}$ exists, the domain of $f^{-1}$ is the range of $f$ , and the range of $f^{-1}$ is the domain of $f$ .

To find an inverse:

Write $y = f (x)$ .
Rearrange to make $x$ the subject.
Swap $x$ and $y$ .
State the domain of the inverse.

The graphs of $y = f (x)$ and $y=f^{-1}(x)$ are reflections in the line $y = x$ .

A function and its inverse are mirror images in the line y = x, with coordinates swapped.

Example

Finding inverse functions

Find the inverse of $u(x)=(x-2)^2+5$ , where $x≥2x\ge 2$ . Also find the inverse of $v(x)=3+ln⁡(2x+1)v(x)=3+\ln(2x+1)$ .

For $u (x)$ , the smallest value occurs at $x = 2$ , so the range is $u(x)≥5u(x)\ge 5$ .
Start with $y=(x-2)^2+5$ and rearrange:
$y-5=(x-2)^2$
Since the domain is $x≥2x\ge 2$ , choose the positive square root. Therefore $u−1(x)=2+x−5u^{-1}(x)=2+\sqrt{x-5}$ , with domain $x≥5x\ge 5$ .
For $v (x)$ , start by writing:
$y=3+ln⁡(2x+1)y=3+\ln(2x+1)$
Rearrange using exponentials to undo the logarithm:
$ey−3=2x+1⇒x=ey−3−12e^{y-3}=2x+1\Rightarrow x=\frac{e^{y-3}-1}{2}$
Swap the letters. So $v−1(x)=ex−3−12v^{-1}(x)=\frac{e^{x-3}-1}{2}$ , with domain $x∈Rx\in\mathbb{R}$ .

Common Mistake

Inverse is not reciprocal

$f^{-1}(x)$ means the inverse function, not $1f(x)\frac{1}{f(x)}$ . These are completely different unless a question specifically defines a reciprocal.

4. Modulus graphs and inequalities

Definition

Modulus

The modulus $∣ A ∣$ means the non-negative size of $A$ :

\begin{cases} A & A\ge 0\\ -A & A<0 \end{cases}

A graph like $y = ∣2 x - 1∣$ is V-shaped. The vertex occurs where the expression inside the modulus is zero.

The modulus graph has a vertex where the expression inside the modulus is zero.

For inequalities, if $k > 0$ , then $∣ A ∣ < k$ means $- k < A < k$ .

Example

A modulus sketch and inequality

Sketch $y = 2 - ∣2 x - 1∣$ , state where it meets the axes, and solve $2 - ∣2 x - 1∣ > 0$ .

The graph of y = 2 − |2x − 1| is an upside-down V, and the positive part lies between its x-intercepts.

The vertex occurs when $2 x - 1 = 0$ , so $x=12x=\frac{1}{2}$ . At this point, $y = 2$ .
The graph is an upside-down V with vertex $(12,2)\left(\frac{1}{2},2\right)$ .
For the y-intercept, set $x = 0$ : $y = 2 - ∣ - 1∣ = 1$ . So the y-intercept is $(0, 1)$ .
For the x-intercepts, solve:
$2−∣2x−1∣=0⇒∣2x−1∣=22-|2x-1|=0\Rightarrow |2x-1|=2$
This gives $2 x - 1 = 2$ or $2 x - 1 = - 2$ , so $x=32x=\frac{3}{2}$ or $x=−12x=-\frac{1}{2}$ .
For the inequality, solve $∣2 x - 1∣ < 2$ :
$−2<2x−1<2⇒−1<2x<3⇒−12<x<32-2<2x-1<2\Rightarrow -1<2x<3\Rightarrow -\frac{1}{2}<x<\frac{3}{2}$

Common Mistake

Variable right-hand sides

For an inequality like $∣2 x - 3∣ < x + 1$ , the right-hand side depends on $x$ . Remember it must be positive, or split into cases carefully.

5. Graph transformations

Definition

Graph transformation

A graph transformation moves, stretches or reflects an existing graph without needing to plot every point from scratch.

Key transformations:

A transformation summary shows how common changes to f(x) affect points on the graph.

$y = f (x) + b$ shifts the graph vertically by $b$ .
$y = f (x + a)$ shifts the graph left by $a$ .
$y = c f (x)$ stretches vertically by scale factor $c$ .
$y = f (c x)$ stretches horizontally by scale factor $1c\frac{1}{c}$ , for $c > 0$ .
$y = ∣ f (x) ∣$ reflects parts below the x-axis upwards.
$y = f (∣ x ∣)$ keeps the right-hand half and reflects it in the y-axis.

A stationary point is a point where the gradient is zero, such as a smooth maximum or minimum.

Example

Transforming key points

In $f (2 x)$ , the x-coordinates are divided by 2. So $A$ moves to $(- 1, 4)$ and $B$ moves to $(3, - 1)$ .
The outside factor 3 multiplies the y-coordinates. So the new stationary points are $(- 1, 12)$ and $(3, - 3)$ .
For $y = f (x + a) + b$ , the point $B (6, - 1)$ moves to $(6 - a, - 1 + b)$ .
To move $B$ to the origin, solve $6 - a = 0$ and $- 1 + b = 0$ .
Hence $a = 6$ and $b = 1$ .

Exam technique

In the exam

Write composites as nested functions first, for example $g f (x) = g (f (x))$ , before simplifying.
For inverses, always state the domain of the inverse; it is the range of the original function.
On sketches, mark key points: vertices, intercepts, asymptotes, stationary points and any domain endpoints.
For modulus inequalities, check whether the right-hand side is constant or variable before splitting cases.

Self review

Check yourself

If $f(x)=1x−4f(x)=\frac{1}{x-4}$ , what value can $f (x)$ never take?
Why does $x^2+1$ not have an inverse on $x∈Rx\in\mathbb{R}$ ?
What happens to the point $(6, - 2)$ under the transformation $y = 3 f (2 x)$ ?

Functions and Graphs

What you'll learn

1. Function notation, domain and range

2. Composite and piecewise functions

3. Inverse functions

4. Modulus graphs and inequalities

5. Graph transformations

Recap questions

You've reached the end

Functions and Graphs

What you'll learn

1. Function notation, domain and range

2. Composite and piecewise functions

3. Inverse functions

4. Modulus graphs and inequalities

5. Graph transformations

Recap questions

You've reached the end

Functions and Graphs

What you'll learn

1. Function notation, domain and range

2. Composite and piecewise functions

3. Inverse functions

4. Modulus graphs and inequalities

5. Graph transformations

Recap questions

You've reached the end

Functions and Graphs Revision Guide

Functions and Graphs

What you'll learn

1. Function notation, domain and range

2. Composite and piecewise functions

3. Inverse functions

4. Modulus graphs and inequalities

5. Graph transformations

Recap questions

You've reached the end

Functions and Graphs Revision Guide