Add Maths · 1 Functions

1 Functions

Before you start

Making a letter the subject of a formula, including when it appears twice: collect, factorise, divide.
Completing the square, e.g. $x^{2} - 4 x + 7 = {(x - 2)}^{2} + 3$ . The Quadratic functions lesson teaches it in full.
Inequality notation: $<$ , $\leq$ , $>$ , $\geq$ , and combined forms such as $- 1 \leq x < 4$ .
The shapes of straight lines, parabolas, cubics, and $y = \sin x$ and $y = \cos x$ for $0 ° \leq x \leq 360 °$ .

Syllabus objectives covered (8)

Understand the terms: function, domain, range (image set), one–one function, many–one function, inverse function and composition of functions.
Find the domain and range of functions.
Recognise and use function notation.
Understand the relationship between y = f(x) and y = |f(x)|, where f(x) may be linear, quadratic, cubic or trigonometric.
Explain in words why a given function does not have an inverse.
Find the inverse of a one–one function.
Form and use composite functions.
Use sketch graphs to show the relationship between a function and its inverse.

Part 1 of 7

What makes a rule a function

A function is a rule that takes each input and gives back exactly one output. The word doing the work is exactly: not two outputs, not none.
The set of inputs you are allowed to use is the domain. The set of outputs the rule actually produces is the range, which the syllabus also calls the image set.
A function always comes with its domain. ' $f (x) = 2 x + 3$ for $x \in ℝ$ ' means the rule is 'double, then add 3', and you may put in any real number ( $ℝ$ stands for the set of all real numbers). Same rule, different domain: a different function.
You need to read both ways of writing a function. The second is read 'f maps $x$ to $2 x + 3$ '.
$f (x) = 2 x + 3 f : x \mapsto 2 x + 3$
$f (4)$ means 'use 4 as the input': $f (4) = 2 (4) + 3 = 11$ . Anything can go in the bracket. For $f (a + 1)$ , replace every $x$ with the whole of $(a + 1)$ : $f (a + 1) = 2 (a + 1) + 3 = 2 a + 5$ .
To explain in words why a rule is a function, say that every value in the domain gives exactly one output. For $f (x) = 2 x + 3$ , $x \in ℝ$ : each real number, doubled and increased by 3, gives one real number.
Here is a rule that is not a function: $x \mapsto \pm \sqrt{x}$ for $x > 0$ . The input 9 gives both 3 and $- 3$ . One input with two outputs is enough to fail.
On a graph, a curve is a function of $x$ if every vertical line through the domain meets it exactly once. A circle fails, because the vertical line through its centre meets it twice.

Part 2 of 7

Domain and range, in the right notation

The domain is about inputs, so write it with $x$ : $x \geq 1$ , or $- 2 \leq x < 5$ , or $x \in ℝ$ .
The range is about outputs, so write it with $f (x)$ , or $y$ if you are describing a graph: $f (x) \geq 3$ . Writing a range as $x \geq 3$ is wrong. It describes the inputs, which is not what you were asked for.
Straight line on a restricted domain. Take $f (x) = 3 x - 1$ for $- 1 \leq x \leq 2$ . A line with positive gradient is lowest at its left end and highest at its right end, so work out both ends: $f (- 1) = - 4$ and $f (2) = 5$ .
$- 4 \leq f (x) \leq 5$
Match the signs to the domain. If the domain were $- 1 \leq x < 2$ , the value 5 would never be reached, and the range would be $- 4 \leq f (x) < 5$ .
Quadratic. End values alone are not enough, because a parabola turns. Complete the square to find the turning point first.
$x^{2} - 4 x + 7 = {(x - 2)}^{2} + 3$
${(x - 2)}^{2}$ is a square, so it is never negative. Its least value is 0, when $x = 2$ . So the least value of $f (x)$ is 3, and for $x \in ℝ$ the range is $f (x) \geq 3$ .
When the rule limits the domain. You cannot divide by zero, so $\frac{1}{x - 2}$ needs $x \neq 2$ . You cannot take the square root of a negative number, so $\sqrt{x - 3}$ needs $x - 3 \geq 0$ , which means $x \geq 3$ . A square root is never negative, so its range is $f (x) \geq 0$ .
The safest route to any range is a quick sketch of the graph over the domain only. Read off the lowest and highest $y$ -values the curve actually reaches, and whether each end is included.

Part 3 of 7

One–one, many–one, and why some functions have no inverse

A function is one–one if different inputs always give different outputs. $f (x) = 2 x + 3$ is one–one: two different numbers cannot be doubled and increased by 3 to give the same answer.
A function is many–one if at least two different inputs give the same output. $g (x) = x^{2}$ for $x \in ℝ$ is many–one, because $g (- 2) = 4$ and $g (2) = 4$ .
The graph test: a function is one–one if every horizontal line meets its graph at most once. A whole parabola fails, because a horizontal line above the vertex meets it twice.
The inverse function $f^{- 1}$ undoes $f$ . If $f$ sends $a$ to $b$ , then $f^{- 1}$ sends $b$ back to $a$ .
Now you can see why a many–one function has no inverse. For $g (x) = x^{2}$ , $x \in ℝ$ , an inverse would have to send 4 back to 2 and to $- 2$ . A rule giving one input two outputs is not a function, so $g^{- 1}$ cannot exist.
When asked to explain in words why a function has no inverse, give the reason and a specific pair: ' $g$ is many–one, since $g (- 2) = g (2) = 4$ , so an inverse would have to map 4 to two different values.' The named pair is what turns a claim into an explanation.
Restricting the domain fixes the problem. Keep only one side of the turning point: $h (x) = x^{2}$ for $x \geq 0$ is one–one, and has the inverse $h^{- 1} (x) = \sqrt{x}$ for $x \geq 0$ .
For a quadratic, cut at the turning point. $x^{2} - 4 x + 7 = {(x - 2)}^{2} + 3$ turns at $x = 2$ , so the domain $x \geq 2$ (or $x \leq 2$ ) makes it one–one. This is why completing the square keeps appearing in inverse questions.

Part 4 of 7

Composite functions: fg(x) means g first

A composite function feeds the output of one function straight into another. $f g (x)$ means $f (g (x))$ : apply $g$ to $x$ , then apply $f$ to the result.
Work from the inside out. The function written **next to the $x$ ** acts first. In $f g (x)$ that is $g$ , even though you read the $f$ first. This is the trap to watch for in every composite question.
Take $f (x) = 2 x + 1$ and $g (x) = x^{2}$ , both for $x \in ℝ$ . For $f g (x)$ , put the whole of $g (x)$ wherever $f$ has an $x$ .
$f g (x) = f (x^{2}) = 2 x^{2} + 1$
For $g f (x)$ , put the whole of $f (x)$ into $g$ .
$g f (x) = g (2 x + 1) = {(2 x + 1)}^{2}$
These are different functions: $f g (1) = 3$ but $g f (1) = 9$ . Order matters, so never swap the letters to make the algebra easier.
$f^{2} (x)$ means $f f (x) = f (f (x))$ , which is applying $f$ twice. It does not mean ${[f (x)]}^{2}$ . Here $f^{2} (x) = 2 (2 x + 1) + 1 = 4 x + 3$ . (The syllabus does not use this notation for trigonometric functions, where $\sin^{2} x$ means ${(\sin x)}^{2}$ .)
A composite only exists if every output of the first function is an allowed input of the second. For $g f$ , the range of $f$ must lie inside the domain of $g$ .
Example: $f (x) = x - 5$ for $x \in ℝ$ , and $g (x) = \sqrt{x}$ for $x \geq 0$ . Then $g f (0)$ does not exist, because $f (0) = - 5$ is not in the domain of $g$ . To form $g f (x) = \sqrt{x - 5}$ , you would have to restrict the domain of $f$ to $x \geq 5$ .
That is what the syllabus means by 'Domain $g f \subseteq$ Domain $f$ ' and 'Range $g f \subseteq$ Range $g$ '. $g f$ can never accept an input that $f$ refuses, and it can never produce an output that $g$ cannot produce.

Part 5 of 7

Finding the inverse of a one–one function

The method is the same every time: write $y = f (x)$ , rearrange to make $x$ the subject, then swap the letters so the answer is a function of $x$ .
Find $f^{- 1} (x)$ for $f (x) = \frac{x + 4}{3}$ , $x \in ℝ$ . Write $y = \frac{x + 4}{3}$ . Multiply by 3 to get $3 y = x + 4$ , then subtract 4.
$x = 3 y - 4$
Swap the letters: $f^{- 1} (x) = 3 x - 4$ , for $x \in ℝ$ . Notice the steps were undone in reverse order. $f$ adds 4 then divides by 3, so $f^{- 1}$ multiplies by 3 then subtracts 4.
Check by composing: $f f^{- 1} (x) = \frac{(3 x - 4) + 4}{3} = x$ . If $f f^{- 1} (x)$ does not simplify to exactly $x$ , the inverse is wrong. The check takes seconds and needs no calculator.
**The domain of $f^{- 1}$ is the range of $f$ **, and the range of $f^{- 1}$ is the domain of $f$ . The inverse swaps inputs with outputs, so it swaps these two sets too.
So find the range of $f$ before you state the domain of $f^{- 1}$ . A correct rule with a missing or wrong domain is an incomplete answer.
If $x$ appears twice after you clear fractions, as in $y = \frac{2 x + 1}{x - 3}$ , use the rearranging routine: multiply out, collect the $x$ terms on one side, factorise $x$ out, divide. The exam set practises this one.
If the rearrangement needs a square root, the domain decides the sign. For $h (x) = x^{2}$ , $x \geq 0$ , you get $x = \pm \sqrt{y}$ . But $x$ is never negative in this domain, so take the positive root: $h^{- 1} (x) = \sqrt{x}$ .
$f^{- 1} (x)$ is not $\frac{1}{f (x)}$ . Here the $- 1$ is notation meaning 'inverse', not a power.
The same notation covers exponentials and logarithms, which you meet in topic 6. The syllabus's own example is $f (x) = e^{2 x}$ , whose inverse is written $f^{- 1} (x) = \frac{1}{2} \ln x$ . The method is identical, using the fact that $\ln$ undoes $e$ . Leave it until you have studied logarithms.

Part 6 of 7

The graph of an inverse is a reflection in y = x

If the point $(a, b)$ lies on $y = f (x)$ , then $f (a) = b$ . So $f^{- 1} (b) = a$ , which means the point $(b, a)$ lies on $y = f^{- 1} (x)$ .
Swapping the coordinates of every point is exactly a reflection in the line $y = x$ . So the graph of $f^{- 1}$ is the reflection of the graph of $f$ in $y = x$ , and $f$ is equally the reflection of $f^{- 1}$ .
Take $f (x) = x^{2} + 1$ for $x \geq 0$ , whose inverse is $f^{- 1} (x) = \sqrt{x - 1}$ for $x \geq 1$ . The points $(0, 1)$ , $(1, 2)$ and $(2, 5)$ on $f$ become $(1, 0)$ , $(2, 1)$ and $(5, 2)$ on $f^{- 1}$ .
Every point $(a, b)$ on $y = f (x)$ has a partner $(b, a)$ on $y = f^{- 1} (x)$ , mirrored across $y = x$ .
The end points swap too. $f$ starts at $(0, 1)$ , so $f^{- 1}$ starts at $(1, 0)$ . That is the rule 'domain of $f^{- 1}$ = range of $f$ ', showing up on the graph.
When you sketch both: draw $y = x$ as a dashed line, use the same scale on both axes (otherwise the reflection looks wrong), label the end points, and make the two curves visibly mirror each other across the dashed line.
Draw $f$ only over its domain. Reflecting the whole parabola $y = x^{2} + 1$ would give a sideways parabola, which is not a function. The restricted domain is what allows the inverse to exist.
For an increasing function, any point where $y = f (x)$ and $y = f^{- 1} (x)$ meet lies on $y = x$ , so you can find those points by solving $f (x) = x$ . Here $x^{2} + 1 = x$ has discriminant $1 - 4 < 0$ , so there are no real solutions, and that is why the two curves never meet.

Part 7 of 7

The graph of y = |f(x)|

The modulus $| a |$ is the size of $a$ without its sign: $| 5 | = 5$ , $| - 5 | = 5$ , $| 0 | = 0$ . A modulus is never negative.
So to get $y = | f (x) |$ from $y = f (x)$ : any part of the graph on or above the $x$ -axis stays exactly where it is, and any part below the $x$ -axis is reflected in the $x$ -axis.
Only the negative parts move. The trap is reflecting the whole graph, or flipping the positive parts down.
Linear. $y = 2 x - 3$ is below the axis for $x < \frac{3}{2}$ . In $y = | 2 x - 3 |$ that part flips up, giving a V shape with its point at $(\frac{3}{2}, 0)$ . It now meets the $y$ -axis at $(0, 3)$ instead of $(0, - 3)$ .
Quadratic. $y = x^{2} - 2 x - 3 = (x + 1) (x - 3)$ dips below the axis between its roots $- 1$ and $3$ , with vertex $(1, - 4)$ . In $y = | x^{2} - 2 x - 3 |$ that dip becomes an arch with its top at $(1, 4)$ , and the $y$ -intercept moves from $(0, - 3)$ to $(0, 3)$ .
Only the dip below the axis flips: $(1, - 4)$ becomes $(1, 4)$ , and the curve meets the axis at sharp corners.
Where the original crosses the $x$ -axis, the modulus graph touches the axis at a sharp corner, not a smooth curve. Draw those corners sharp: they are where a reflected piece meets an unreflected one.
Cubic. Apply the same rule to each stretch below the axis. $y = (x + 2) (x - 1) (x - 3)$ is negative for $x < - 2$ and for $1 < x < 3$ , so both of those pieces flip up.
Trigonometric. The forms you will meet are $a \sin b x + c$ , $a \cos b x + c$ and $a \tan b x + c$ . When $c \neq 0$ the curve is shifted up or down, so the flipped parts are not whole half-waves. Find where the curve actually crosses the axis first; the worked example does this for $y = | 2 \sin x - 1 |$ .
Two consequences you can use now: the range of $| f (x) |$ never includes negative values, and a horizontal line $y = k$ with $k > 0$ can meet the modulus graph more times than it meets the original. Solving equations such as $| x^{2} - 2 x - 3 | = k$ is topic 4; here you only need the picture.

Worked examples

Example 1. The function $f$ is defined by $f (x) = x^{2} - 2 x + 3$ for $0 \leq x \leq 4$ . Find the range of $f$ .

Show the working

End values alone are not safe, because a quadratic turns. Complete the square to find where it turns.
$x^{2} - 2 x + 3 = {(x - 1)}^{2} + 2$
${(x - 1)}^{2} \geq 0$ , and it equals 0 when $x = 1$ . So the least value of $f (x)$ is 2, at $x = 1$ . Check that $x = 1$ is inside the domain $0 \leq x \leq 4$ . It is, so this least value is actually reached.
Now the ends: $f (0) = 3$ and $f (4) = 16 - 8 + 3 = 11$ . The larger of these, 11, is the greatest value.
Picture the curve: it starts at $(0, 3)$ , drops to $(1, 2)$ , then climbs to $(4, 11)$ . Lowest point at the vertex, highest at the right-hand end.
Both ends of the domain are included (the domain uses $\leq$ ), so both signs in the range are $\leq$ .
$2 \leq f (x) \leq 11$
The trap: using only the end values gives $3 \leq f (x) \leq 11$ , which leaves out every output between 2 and 3.

Answer: $2 \leq f (x) \leq 11$

Example 2. The function $f$ is defined by $f (x) = x^{2} + 6 x + 5$ for $x \in ℝ$ . (a) Explain in words why $f$ does not have an inverse. (b) The function $g$ is defined by $g (x) = x^{2} + 6 x + 5$ for $x \geq - 3$ . Find $g^{- 1} (x)$ , stating its domain, and describe how the graphs of $y = g (x)$ and $y = g^{- 1} (x)$ are related.

Show the working

(a) Look for two inputs with the same output. $f (x) = (x + 1) (x + 5)$ , so $f (- 1) = 0$ and $f (- 5) = 0$ .
So $f$ is many–one. An inverse would have to map 0 back to both $- 1$ and $- 5$ , and a rule giving two outputs for one input is not a function.
(b) Complete the square. It shows why $x \geq - 3$ is the right restriction.
$x^{2} + 6 x + 5 = {(x + 3)}^{2} - 4$
The vertex is $(- 3, - 4)$ . For $x \geq - 3$ the curve only rises, so $g$ is one–one. Its least output is $- 4$ , so the range of $g$ is $g (x) \geq - 4$ .
Write $y = {(x + 3)}^{2} - 4$ and make $x$ the subject.
${(x + 3)}^{2} = y + 4 \Rightarrow x + 3 = \pm \sqrt{y + 4}$
Let the domain choose the sign. $x \geq - 3$ means $x + 3 \geq 0$ , so take the positive root: $x = - 3 + \sqrt{y + 4}$ .
Swap the letters. The domain of $g^{- 1}$ is the range of $g$ .
$g^{- 1} (x) = - 3 + \sqrt{x + 4}, x \geq - 4$
Check: $g g^{- 1} (x) = {(\sqrt{x + 4})}^{2} - 4 = x$ . Correct.
The graphs: $y = g^{- 1} (x)$ is the reflection of $y = g (x)$ in the line $y = x$ . The vertex $(- 3, - 4)$ of $g$ becomes the start point $(- 4, - 3)$ of $g^{- 1}$ , and the $y$ -intercept $(0, 5)$ of $g$ becomes the $x$ -intercept $(5, 0)$ of $g^{- 1}$ .

Answer: (a) $f$ is many–one, e.g. $f (- 1) = f (- 5) = 0$ , so an inverse would have to map 0 to two values. (b) $g^{- 1} (x) = - 3 + \sqrt{x + 4}$ for $x \geq - 4$ ; the graph of $y = g^{- 1} (x)$ is the reflection of $y = g (x)$ in $y = x$ .

Example 3. The functions $f$ and $g$ are defined for $x \in ℝ$ by $f (x) = x^{2} + 2$ and $g (x) = 3 x - 1$ . (a) Find $f g (x)$ and $g f (x)$ . (b) Solve $f g (x) = 11$ . (c) Find $g^{2} (x)$ .

Show the working

(a) $f g (x)$ means $f (g (x))$ , so $g$ acts first. Put the whole of $g (x)$ , in a bracket, in place of the $x$ in $f$ .
$f g (x) = f (3 x - 1) = {(3 x - 1)}^{2} + 2$
$g f (x)$ means $g (f (x))$ : $f$ acts first, then $g$ multiplies by 3 and subtracts 1.
$g f (x) = g (x^{2} + 2) = 3 (x^{2} + 2) - 1 = 3 x^{2} + 5$
Check they really differ: $f g (1) = 2^{2} + 2 = 6$ and $g f (1) = 3 + 5 = 8$ .
(b) Set the composite equal to 11. Don't expand. Getting the bracket on its own is quicker and keeps the numbers small.
${(3 x - 1)}^{2} + 2 = 11 \Rightarrow {(3 x - 1)}^{2} = 9$
A square root gives two possibilities. Forgetting $- 3$ loses a solution.
$3 x - 1 = 3 or 3 x - 1 = - 3$
So $x = \frac{4}{3}$ or $x = - \frac{2}{3}$ . Check the first: $g (\frac{4}{3}) = 3$ , and $f (3) = 9 + 2 = 11$ . Correct.
(c) $g^{2} (x)$ means $g g (x)$ : apply $g$ twice. Don't square it.
$g^{2} (x) = g (3 x - 1) = 3 (3 x - 1) - 1 = 9 x - 4$

Answer: (a) $f g (x) = {(3 x - 1)}^{2} + 2$ and $g f (x) = 3 x^{2} + 5$ (b) $x = \frac{4}{3}$ or $x = - \frac{2}{3}$ (c) $g^{2} (x) = 9 x - 4$

Example 4. Sketch the graph of $y = | 2 \sin x - 1 |$ for $0 ° \leq x \leq 360 °$ , and state its range.

Show the working

Start with $y = 2 \sin x - 1$ itself. $\sin x$ lies between $- 1$ and $1$ , so $2 \sin x$ lies between $- 2$ and $2$ , and $2 \sin x - 1$ between $- 3$ and $1$ .
Work out the key angles, where $\sin x$ is $0$ , $1$ or $- 1$ . At $0 °$ , $180 °$ and $360 °$ the value is $- 1$ ; at $90 °$ it is $1$ ; at $270 °$ it is $- 3$ .
Find where it crosses the axis: $2 \sin x - 1 = 0$ gives $\sin x = \frac{1}{2}$ , so $x = 30 °$ or $x = 150 °$ . The curve is above the axis only between these two angles.
Apply the modulus. Between $30 °$ and $150 °$ , leave the curve alone. It rises to $(90 °, 1)$ and falls back to the axis.
Everywhere else the curve is below the axis, so reflect it. $(0 °, - 1)$ becomes $(0 °, 1)$ , $(180 °, - 1)$ becomes $(180 °, 1)$ , $(360 °, - 1)$ becomes $(360 °, 1)$ , and the lowest point $(270 °, - 3)$ becomes a peak at $(270 °, 3)$ .
The piece between $30 °$ and $150 °$ is untouched. Everything else is reflected, so the minimum $- 3$ becomes the maximum $3$ .
The curve touches the axis at $30 °$ and $150 °$ with sharp corners. On your sketch, label those two points and the peaks $(90 °, 1)$ and $(270 °, 3)$ .
Range: the least value is now 0 and the greatest is 3.
$0 \leq y \leq 3$
Read the sketch: the line $y = 2$ meets the graph twice, both times on the tall reflected hump, because the untouched piece never rises above 1.

Answer: A curve starting at $(0 °, 1)$ , touching the $x$ -axis at $(30 °, 0)$ and $(150 °, 0)$ , with peaks at $(90 °, 1)$ and $(270 °, 3)$ , passing through $(180 °, 1)$ and ending at $(360 °, 1)$ . Range $0 \leq y \leq 3$ .

Where people go wrong

Mistake: Working out $f g (x)$ by applying $f$ first.

Instead: The function next to the $x$ acts first: $f g (x) = f (g (x))$ .

Composition works from the inside out. With $f (x) = 2 x + 1$ and $g (x) = x^{2}$ , the two orders give $2 x^{2} + 1$ and ${(2 x + 1)}^{2}$ . The wrong order gives a different function, so every later mark goes with it.

Mistake: Writing the range of $f (x) = {(x - 2)}^{2} + 3$ as $x \geq 3$ .

Instead: Write $f (x) \geq 3$ .

The range describes outputs. ' $x \geq 3$ ' is a statement about inputs, and it is false here, since $x = 0$ is allowed.

Mistake: Giving $f^{- 1} (x)$ with no domain, or copying the domain of $f$ .

Instead: The domain of $f^{- 1}$ is the range of $f$ . Find that range first.

The inverse swaps inputs with outputs, so it swaps the two sets as well.

Mistake: Reflecting the whole of $y = f (x)$ in the $x$ -axis to get $y = | f (x) |$ .

Instead: Reflect only the parts below the $x$ -axis. Parts on or above it stay where they are.

$| f (x) | = f (x)$ whenever $f (x) \geq 0$ . Only negative outputs change.

Mistake: Explaining that a function has no inverse 'because it is a quadratic'.

Instead: Say it is many–one and name two inputs with the same output, e.g. $f (- 1) = f (- 5) = 0$ .

Being quadratic is not the reason: $x^{2}$ for $x \geq 0$ is a quadratic and has an inverse. Being many–one on the given domain is.

Check yourself

Explain why the rule $x \mapsto \pm \sqrt{x}$ , for $x > 0$ , is not a function.
Answer
One input gives two outputs: 4 maps to both 2 and $- 2$ . A function must give exactly one output for each input.
$f (x) = 3 x - 4$ for $x \in ℝ$ . Find $f (- 2)$ , solve $f (x) = 11$ , and simplify $f (a + 1)$ .
Answer
$f (- 2) = - 10$ . $3 x - 4 = 11$ gives $x = 5$ . $f (a + 1) = 3 (a + 1) - 4 = 3 a - 1$ .
State the range of $f (x) = 5 - x^{2}$ for $x \in ℝ$ , and the largest possible domain of $g (x) = \sqrt{2 x - 6}$ .
Answer
$x^{2} \geq 0$ , so $f (x) \leq 5$ . For $g$ you need $2 x - 6 \geq 0$ , so $x \geq 3$ .
$h (x) = {(x - 3)}^{2}$ for $x \in ℝ$ . Explain in words why $h$ has no inverse, and give a domain on which it would be one–one.
Answer
$h$ is many–one: $h (2) = h (4) = 1$ , so an inverse would have to send 1 to both 2 and 4. Restricting to $x \geq 3$ (or $x \leq 3$ ) makes it one–one.
Find $f^{- 1} (x)$ for $f (x) = \frac{4}{x - 2}$ , $x > 2$ , and state the domain of $f^{- 1}$ .
Answer
$y = \frac{4}{x - 2}$ gives $x - 2 = \frac{4}{y}$ , so $f^{- 1} (x) = 2 + \frac{4}{x}$ . For $x > 2$ , $f (x) > 0$ , so the domain of $f^{- 1}$ is $x > 0$ .
$f (x) = x + 4$ and $g (x) = x^{2}$ for $x \in ℝ$ . Find $f g (x)$ , $g f (x)$ and $f^{2} (x)$ .
Answer
$f g (x) = x^{2} + 4$ , $g f (x) = {(x + 4)}^{2}$ , $f^{2} (x) = (x + 4) + 4 = x + 8$ .
Describe the graph of $y = | (x + 2) (x - 1) (x - 3) |$ : where does it meet the axes, and which parts of $y = (x + 2) (x - 1) (x - 3)$ are reflected?
Answer
It touches the $x$ -axis at $x = - 2$ , $1$ and $3$ (sharp corners) and meets the $y$ -axis at $(0, 6)$ . The original cubic is negative for $x < - 2$ and for $1 < x < 3$ , so those two parts are reflected in the $x$ -axis.
The point $(3, 7)$ lies on $y = f (x)$ , where $f$ is one–one. Write down a point on $y = f^{- 1} (x)$ , and the line in which the two graphs are reflections of each other.
Answer
$(7, 3)$ , and the line $y = x$ .

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. The function $f$ is defined by $f (x) = \frac{2 x + 1}{x - 3}$ for $x > 3$ .

(a) Show that $f (x) = 2 + \frac{7}{x - 3}$ . 2 marks No calculator

Mark scheme

M1 writes 2x + 1 as 2(x − 3) + 7, or combines 2 + 7/(x − 3) over the common denominator x − 3
A1 given result reached with no errors seen

(b) Hence state the range of $f$ . 1 mark No calculator

Mark scheme

B1 f(x) > 2 (accept y > 2 or f > 2; x > 2 does not earn the mark)

(c) Find $f^{- 1} (x)$ and state its domain. 4 marks No calculator

Mark scheme

M1 y(x − 3) = 2x + 1 multiplied out, with the terms in x collected on one side
M1 x factorised out, e.g. x(y − 2) = 3y + 1, and divided through
A1 f⁻¹(x) = (3x + 1)/(x − 2) oe, written in terms of x
B1 domain x > 2; FT their range from (b)

Question 2. The functions $f$ and $g$ are defined for $x \in ℝ$ by $f (x) = x^{2} - 3$ and $g (x) = 2 x + 1$ .

(a) Find $g f (x)$ . 1 mark No calculator

Mark scheme

B1 2x² − 5 oe

(b) Solve $f g (x) = 6$ . 3 marks No calculator

Mark scheme

M1 fg(x) = (2x + 1)² − 3, composed in the correct order
M1 (2x + 1)² = 9 with both signs considered, or 4x² + 4x − 8 = 0 solved by a valid method
A1 x = 1 and x = −2

(c) State the range of $f g$ . 1 mark No calculator

Mark scheme

B1 fg(x) ≥ −3 (accept fg ≥ −3 or y ≥ −3)

(d) The function $h$ is defined by $h (x) = \sqrt{x - 1}$ for $x \geq 1$ . Explain why $h f (0)$ does not exist. 1 mark No calculator

Mark scheme

B1 f(0) = −3, which is not in the domain of h (h needs an input of at least 1) oe

Question 3. The function $f$ is defined by $f (x) = x^{2} - 6 x + 5$ for $x \in ℝ$ .

(a) Sketch the graph of $y = | x^{2} - 6 x + 5 |$ , stating the coordinates of the points where it meets the axes and of the turning point between $x = 1$ and $x = 5$ . 3 marks No calculator

Mark scheme

B1 correct shape: U-shaped outer arms with the section between x = 1 and x = 5 reflected above the axis, sharp corners at the x-axis
B1 (1, 0), (5, 0) and (0, 5) all stated
B1 turning point (3, 4) stated

(b) Write down the number of points at which the line $y = 3$ meets your graph. 1 mark No calculator

Mark scheme

B1 4; FT their sketch if the arch has a maximum above 3

(c) Write down the value of $k$ for which the line $y = k$ meets your graph at exactly three points. 1 mark No calculator

Mark scheme

B1 k = 4; FT the y-coordinate of their turning point from (a)

Question 4. The function $f$ is defined by $f (x) = x^{2} + 4 x + 9$ for $x \in ℝ$ .

(a) Explain in words why $f$ does not have an inverse. 1 mark No calculator

Mark scheme

B1 f is many–one, supported by a correct pair, e.g. f(−4) = f(0) = 9, or by reference to the turning point at x = −2 oe

(b) The function $g$ is defined by $g (x) = x^{2} + 4 x + 9$ for $x \geq - 2$ . Find $g^{- 1} (x)$ , stating its domain. 4 marks No calculator

Mark scheme

M1 completes the square: (x + 2)² + 5
M1 rearranges y = (x + 2)² + 5 to make x the subject, taking a square root
A1 g⁻¹(x) = −2 + √(x − 5), positive root only
B1 domain x ≥ 5

(c) On a single diagram, sketch the graphs of $y = g (x)$ and $y = g^{- 1} (x)$ , showing clearly the relationship between them. 2 marks No calculator

Mark scheme

B1 y = g(x) starting at (−2, 5) and rising; y = g⁻¹(x) starting at (5, −2) and rising more slowly; both end points labelled
B1 line y = x drawn, with the two curves symmetrical about it

How this is examined

Finding an inverse and giving its domain: Find $f^{- 1} (x)$ , often with 'state the domain of $f^{- 1}$ ' attached, frequently after an earlier part asking for the range of $f$ ; 3–4 marks. A rearrangement starting from $y = f (x)$ with $x$ made the subject, the answer written as $f^{- 1} (x) = \dots$ in terms of $x$ , and a domain equal to the range of $f$ , written as an inequality in $x$ . A square-root inverse needs the sign chosen from the domain.
Composite functions in the right order: Find $f g (x)$ , or solve an equation such as $g f (x) = k$ ; 2–4 marks. The inner function substituted, in a bracket, into the outer one (for $f g$ , $g$ goes inside $f$ ), then every solution found, including the one from the negative square root.
Range over a given domain: State or Find the range of $f$ ; 1–3 marks, often the opening part of an inverse question. The least or greatest value from a completed square or the end values, whichever applies to the domain, and an inequality written in $f (x)$ with $<$ or $\leq$ matching whether each end is reached.
Sketching y = |f(x)|: Sketch $y = | f (x) |$ for a linear, quadratic, cubic or trigonometric $f$ , showing where it meets the axes; 2–3 marks. Only the parts below the axis reflected, sharp corners where it meets the $x$ -axis, and the axis intercepts and reflected turning points labelled with coordinates.
No inverse, and sketching a function with its inverse: A short Explain on why $f$ has no inverse, or a Sketch of $y = f (x)$ and $y = f^{- 1} (x)$ on one diagram; 1–2 marks each. ' $f$ is many–one' backed by two named inputs with the same output. On a sketch: $y = x$ drawn, the curves mirror images across it, and the end points swapped.

Words to know

Function: A rule that gives exactly one output for each input in its domain.
Domain: The set of allowed inputs, written with $x$ , e.g. $x \geq 1$ .
Range (image set): The set of outputs the function actually produces, written with $f (x)$ , e.g. $f (x) \geq 3$ .
One–one function: A function in which different inputs always give different outputs. Only one–one functions have inverses.
Many–one function: A function in which at least two different inputs give the same output.
Inverse function: The function $f^{- 1}$ that undoes $f$ . Its domain is the range of $f$ , and its graph is the reflection of $y = f (x)$ in $y = x$ .
Composite function: $f g (x) = f (g (x))$ : apply $g$ first, then $f$ . $f^{2} (x)$ means $f (f (x))$ .
Modulus: $| a |$ is the size of $a$ without its sign, so it is never negative.
$ℝ$: The set of all real numbers. ' $x \in ℝ$ ' means $x$ can be any real number.

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