Functions

What a function is Extended

A function is a rule that takes an input and gives exactly one output. The rule "multiply by 3, then subtract 5" is a function, and we write it as

$$\mathrm{f}(x)=3x-5$$

Read $\mathrm{f}(x)$ as "f of $x$". The letter f names the function and $x$ is the input. The notation tells you what to do with any input:

• $\mathrm{f}(4)$ means put $x=4$ into the rule: $\mathrm{f}(4)=3(4)-5=7$.
• $\mathrm{f}(-2)=3(-2)-5=-11$.
• $\mathrm{f}(a)$ is $3a-5$: the input can be a letter, or even another expression.

Other letters work the same way. Cambridge often uses several functions in one question, such as $\mathrm{g}(x)=\frac{3(x+4)}{5}$ and $\mathrm{h}(x)=2x^2+3$. Put brackets round every input you substitute, especially a negative one: $\mathrm{h}(-2)=2(-2{)}^2+3=2(4)+3=11$, whereas $-2^2$ without brackets would wrongly give $-4$.

Solving $\mathrm{f}(x)=13$ is not the same as finding $\mathrm{f}(13)$. $\mathrm{f}(x)=13$ asks which input gives the output 13: $3x-5=13$, so $x=6$.

Mapping diagrams Extended

A mapping diagram shows a function as arrows from each input to its output; the table above lists the same pairs. Each input has exactly one arrow leaving it. That is what makes it a function. Two inputs may share an output (for $x^2$, both $2$ and $-2$ map to $4$), but one input can never have two outputs.

Domain and range Extended

The domain of a function is the set of inputs it is allowed to take. The range is the set of outputs it produces from that domain.

Inputs that must be left out. You cannot divide by zero, so for $\frac{5}{x-2}$ the value $x=2$ is not in the domain. You cannot take the square root of a negative number, so for $\sqrt{x+1}$ the domain is $x⩾-1$.

Finding a range.

• If the domain is a short list, work out every output. For $\mathrm{g}(x)=4x-1$ with domain $\{0,1,2,3\}$, the range is $\{-1,3,7,11\}$.
• If the domain is an interval, think about the outputs at the ends. $\mathrm{g}(x)=4x-1$ for $0⩽x⩽3$ has range $-1⩽\mathrm{g}(x)⩽11$, because a straight line goes steadily from one end to the other.
• A square is never negative. For $\mathrm{h}(x)=2x^2+3$, the smallest value of $2x^2$ is 0, so the range is $\mathrm{h}(x)⩾3$.

Write a range in terms of the output, $\mathrm{h}(x)⩾3$ or $y⩾3$, not in terms of $x$. You are not expected to find the domain or range of a composite function.

Composite functions Extended

A composite function applies one function and then another. $\mathrm{g}\mathrm{f}(x)$ means $\mathrm{g}(\mathrm{f}(x))$: apply f first, then apply g to the result.

The function nearest to $x$ acts first. In $\mathrm{g}\mathrm{f}(x)$, f is next to $x$, so f goes first. This is the opposite of the order you read the letters in, and it is the commonest source of lost marks.

To find $\mathrm{g}\mathrm{f}(x)$ as an expression, take the rule for g and replace every $x$ in it by the whole expression $\mathrm{f}(x)$, in brackets. To find a value such as $\mathrm{g}\mathrm{f}(3)$, it is usually quicker to work inside out with numbers: find $\mathrm{f}(3)$, then put that answer into g.

In general $\mathrm{f}\mathrm{g}(x)$ and $\mathrm{g}\mathrm{f}(x)$ are different. $\mathrm{f}\mathrm{f}(x)$ means apply f twice.

Inverse functions Extended

The inverse function ${\mathrm{f}}^{-1}$ undoes f. If f takes 4 to 7, then ${\mathrm{f}}^{-1}$ takes 7 back to 4. Applying a function and then its inverse returns you to where you started: ${\mathrm{f}}^{-1}\mathrm{f}(x)=x$ and $\mathrm{f}{\mathrm{f}}^{-1}(x)=x$.

Method.

1. Write $y=\mathrm{f}(x)$.
2. Rearrange to make $x$ the subject.
3. Swap the letters: replace $y$ by $x$, and you have ${\mathrm{f}}^{-1}(x)$.

The rearranging in step 2 undoes the operations of f in reverse order, which is exactly what an inverse must do. For $\mathrm{f}(x)=3x-5$ (multiply by 3, then subtract 5), the inverse adds 5 and then divides by 3: ${\mathrm{f}}^{-1}(x)=\frac{x+5}{3}$.

${\mathrm{f}}^{-1}(x)$ is not $\frac{1}{\mathrm{f}(x)}$. The $-1$ is a label for "inverse", not a power.

A function only has an inverse if each output comes from exactly one input. On the graph, $y={\mathrm{f}}^{-1}(x)$ is the reflection of $y=\mathrm{f}(x)$ in the line $y=x$.

$\mathrm{f}(x)=2x+1$ and ${\mathrm{f}}^{-1}(x)=\frac{x-1}{2}$. Folding the page along $y=x$ would place one line exactly on the other.

Solving equations with functions Extended

Many questions end with an equation. Turn the function notation into ordinary algebra first, then solve as usual.

• $\mathrm{f}(x)=k$: write out the rule and set it equal to $k$.
• $\mathrm{f}\mathrm{f}(x)=k$ or $\mathrm{f}\mathrm{g}(x)=k$: find the composite expression first, then solve.
• ${\mathrm{f}}^{-1}(x)=k$: there is a shortcut. If the inverse takes $x$ to $k$, then f takes $k$ to $x$, so $x=\mathrm{f}(k)$.
• ${\mathrm{f}}^{-1}\mathrm{f}(a)=a$ for any input $a$ in the domain, with no working needed.