Functions

Functions, domain and range Paper 1

A function is a rule that gives exactly one output for each input. In Cambridge notation

$$\mathrm{f}:x\mapsto 2x+3\text{for}x\in ℝ$$

means "f maps $x$ to $2x+3$, for every real number $x$". The same function can be written $\mathrm{f}(x)=2x+3$ for $x\in ℝ$.

• The domain is the set of inputs allowed. It is part of the definition: change the domain and you have a different function.
• The range is the set of outputs the function actually produces from that domain. It is written in terms of $\mathrm{f}(x)$, for example $\mathrm{f}(x)⩾2$, never in terms of $x$.

A rule such as $x\mapsto \pm \sqrt{x}$ is not a function, because one input gives two outputs. A rule can fail to be a function on a domain for another reason too: $x\mapsto \frac{1}{x-1}$ has no output at $x=1$, so 1 must be left out of its domain.

A function is one-one if different inputs always give different outputs. On a graph, a one-one function meets every horizontal line at most once. $\mathrm{f}(x)=x^2$ for $x\in ℝ$ is many-one, because $\mathrm{f}(3)=\mathrm{f}(-3)=9$; but $\mathrm{f}(x)=x^2$ for $x⩾0$ is one-one.

Finding the range Paper 1

To find a range, think about the graph of the function over its domain only, and ask which $y$ values it covers.

• Linear functions are increasing or decreasing, so the range runs between the values at the ends of the domain. Watch whether each end is included.
• Quadratics: complete the square to find the vertex. If the vertex is inside the domain, the least (or greatest) value is at the vertex. If it is not, the least value is at the end of the domain nearest the vertex.
• Reciprocals such as $\frac{1}{x}$ get close to 0 but never reach it.

Two of the syllabus's own illustrations:

• $\mathrm{f}:x\mapsto \frac{1}{x}$ for $x⩾1$: $\mathrm{f}(1)=1$, and as $x$ increases $\frac{1}{x}$ decreases towards 0 without reaching it. So the range is $0<\mathrm{f}(x)⩽1$.
• $\mathrm{g}:x\mapsto x^2+1$ for $x\in ℝ$: $x^2⩾0$, so the range is $\mathrm{g}(x)⩾1$.

A quick sketch is the safest method, because it shows whether the function turns round inside the domain.

Composite functions Paper 1

The composite function $\mathrm{gf}$ means "do f first, then g": $\mathrm{gf}(x)=\mathrm{g}(\mathrm{f}(x))$. The function written nearer to $x$ acts first, just as in $\mathrm{g}(\mathrm{f}(x))$.

To find $\mathrm{gf}(x)$, replace every $x$ in the rule for g by the whole expression $\mathrm{f}(x)$, in brackets. In general $\mathrm{gf}$ and $\mathrm{fg}$ are different functions. $\mathrm{ff}$ is often written ${\mathrm{f}}^2$, which means f applied twice, not f squared.

When can gf be formed? g must be able to accept every output of f. So $\mathrm{gf}$ exists only when the range of f lies within the domain of g. For example, if $\mathrm{f}(x)=x-4$ for $x\in ℝ$ and $\mathrm{g}(x)=\sqrt{x}$ for $x⩾0$, then $\mathrm{gf}$ cannot be formed: $\mathrm{f}(1)=-3$, and g cannot take $-3$. Restricting the domain of f to $x⩾4$ makes the range $\mathrm{f}(x)⩾0$, and then $\mathrm{gf}(x)=\sqrt{x-4}$ exists.

The domain of $\mathrm{gf}$ is the domain of f.

One-one functions and inverse functions Paper 1

The inverse function ${\mathrm{f}}^{-1}$ undoes f: if $\mathrm{f}(a)=b$ then ${\mathrm{f}}^{-1}(b)=a$, so ${\mathrm{f}}^{-1}\mathrm{f}(x)=x$.

Only a one-one function has an inverse. If f were many-one, say $\mathrm{f}(3)=\mathrm{f}(-3)=9$, then ${\mathrm{f}}^{-1}(9)$ would need to be both 3 and $-3$, which is not a function. This is why questions often restrict a domain, for example to $x⩾k$, "so that f has an inverse": for a quadratic, the domain must lie entirely on one side of the vertex, so the least possible $k$ is the $x$-coordinate of the vertex.

Finding an inverse:

1. Write $y=\mathrm{f}(x)$.
2. Rearrange to make $x$ the subject.
3. Swap the letters: write the result in terms of $x$ as ${\mathrm{f}}^{-1}(x)$.

When the rearranging involves a square root, the $\pm$ sign must be decided by the domain of f. The domain and range swap over:

• the domain of ${\mathrm{f}}^{-1}$ is the range of f;
• the range of ${\mathrm{f}}^{-1}$ is the domain of f.

The syllabus example, the inverse of $\mathrm{h}:x\mapsto (2x+3{)}^2-4$ for $x<-\frac{3}{2}$, needs exactly this care: for $x<-\frac{3}{2}$, $2x+3$ is negative, so the negative square root is the right one.

The graphs of a function and its inverse Paper 1

Swapping $x$ and $y$ in an equation reflects its graph in the line $y=x$. Finding an inverse swaps $x$ and $y$, so:

the graph of $y={\mathrm{f}}^{-1}(x)$ is the reflection of the graph of $y=\mathrm{f}(x)$ in the line $y=x$.

Every point $(a,b)$ on $y=\mathrm{f}(x)$ becomes $(b,a)$ on $y={\mathrm{f}}^{-1}(x)$. A sketch should show the mirror line $y=x$ (the syllabus asks for it), use the same scale on both axes, and show the two graphs as mirror images, including their end points.

Where the graph of f meets the line $y=x$, it also meets its own reflection. So the points where $y=\mathrm{f}(x)$ crosses $y={\mathrm{f}}^{-1}(x)$ on the line $y=x$ are found by solving $\mathrm{f}(x)=x$, which is usually easier than solving $\mathrm{f}(x)={\mathrm{f}}^{-1}(x)$.

$\mathrm{f}(x)=(x-2{)}^2+3$ for $x⩾2$, drawn for $2⩽x⩽5$, and ${\mathrm{f}}^{-1}(x)=2+\sqrt{x-3}$. The end point $(2,3)$ of f reflects to the end point $(3,2)$ of ${\mathrm{f}}^{-1}$.

Transformations of graphs Paper 1

Each change to the equation $y=\mathrm{f}(x)$ moves its graph in a fixed way. Changes outside the function act on $y$ and do what they look like; changes inside, to $x$, do the opposite of what they look like.

• $y=\mathrm{f}(x)+a$: a translation by $\left(\begin{array}{c}0\\ a\end{array}\right)$ (up by $a$).
• $y=\mathrm{f}(x+a)$: a translation by $\left(\begin{array}{c}-a\\ 0\end{array}\right)$ (left by $a$). For example $y=\mathrm{f}(x-2)$ moves the graph 2 to the right, because the new graph reaches the old height at $x$ values 2 larger.
• $y=a\mathrm{f}(x)$: a stretch parallel to the $y$-axis with scale factor $a$ (every $y$ multiplied by $a$).
• $y=\mathrm{f}(ax)$: a stretch parallel to the $x$-axis with scale factor $\frac{1}{a}$ (every $x$ divided by $a$).
• $y=-\mathrm{f}(x)$: a reflection in the $x$-axis.
• $y=\mathrm{f}(-x)$: a reflection in the $y$-axis.

Describe fully means give the type and all its details: "a translation by $\left(\begin{array}{c}3\\ 0\end{array}\right)$", "a stretch parallel to the $x$-axis with scale factor $\frac{1}{2}$", "a reflection in the $y$-axis". The words translation, reflection and stretch must be used; "move" or "shift" do not earn the mark.

Combinations. For $y=a\mathrm{f}(x)+b$, stretch first, then translate: the stretch multiplies $\mathrm{f}(x)$ by $a$, and then $b$ is added. Doing them the other way round would give $a(\mathrm{f}(x)+b)$. An inside change and an outside change act on different axes, so their order does not matter: $y=\mathrm{f}(x-3)+2$ is a translation by $\left(\begin{array}{c}3\\ 2\end{array}\right)$ either way.

Points and features. If $(p,q)$ is on $y=\mathrm{f}(x)$, then:

• $(p-a,q)$ is on $y=\mathrm{f}(x+a)$;
• $(p,aq)$ is on $y=a\mathrm{f}(x)$;
• $(\frac{p}{a},q)$ is on $y=\mathrm{f}(ax)$.

This works for any graph, including trigonometric ones: $y=3\sin 2x$ is $y=\sin x$ stretched parallel to the $x$-axis with factor $\frac{1}{2}$ and parallel to the $y$-axis with factor 3.

$y=(x-2{)}^2+1$ is $y=x^2$ translated by $\left(\begin{array}{c}2\\ 1\end{array}\right)$. The vertex moves from $(0,0)$ to $(2,1)$: the $-2$ inside the bracket moves the graph to the right.