4 Functions

Learning objectives

4A FunctionsThe idea of a function of a variable 4B Functions as mappingsFunction as a mapping or as a correspondence between the elements of two sets 4C Function notationUse function notation in the forms f(x) = … and f: x ↦ … .4D Domain and rangeDomain and range of a function Questions will not be set on continuity, but students will be expected to recognise when parts of the domain need to be excluded (e.g. x = 0 must be excluded from the domain of the function f where f(x) = 1/x)4E Composite functionsComposite functions ‘fg’ will mean ‘do g first then f’4F Inverse functionsInverse functions Finding the inverse of a function 4G VariationVariation, direct and indirect proportion To include only the following: y ∝ x; y ∝ 1/x; y ∝ x²; y ∝ 1/x²; y ∝ x³; y ∝ 1/x³; y ∝ √x; y ∝ 1/√x 4H Cartesian coordinatesRectangular Cartesian co-ordinates 4I Straight-line graphsRecognise that equations of the form y = mx + c are straight–line graphs with gradient m and intercept on the y-axis at the point (0, c)4J Graphs of equationsGraphs and graphical treatment of the equation: y = Ax³ + Bx² + Cx + D + E/x + F/x², in which the constants are numerical and at least three of them are zero Students will be expected to draw and interpret graphs from given equations Use of the intersection of two curves (graphs) to solve equations 4K Graph gradientsThe gradients of graphs above by drawing Students will be expected to draw a reasonable tangent to the graph at a named point and to construct an appropriate right-angled triangle from which to calculate the gradient 4L DifferentiationDifferentiation of integer powers of x Use of dy/dx notation 4M Rates of change and stationary pointsDetermination of gradients, rates of change, maxima and minima, stationary points and turning points Students will either be required to differentiate or use graphical methods to arrive at solutions and relate their calculations to their graphs and vice versa 4N Kinematics applicationsApplications to linear kinematics and to other simple practical problems This includes the drawing and interpretation of distance/time and speed/time graphs, and other graphs of a similar nature Students need to be able to understand the relationship between displacement or distance, velocity and speed, and acceleration, for example: ds/dt = v and dv/dt = a

A function assigns exactly one output to each allowed input. The same input cannot produce two different outputs, although different inputs may share one output.

Part	Meaning
input	the chosen value from the domain
rule	the operation or correspondence applied
output	the single resulting value

x\mapsto x^2:\quad -3\mapsto9,;2\mapsto4

A relation is not a function if one allowed input points to more than one output. A function need not use every possible number as an input or produce every possible number as an output.

A mapping diagram shows inputs in one set and outputs in another. Every input must have one outgoing arrow for the correspondence to define a function.

Diagram feature	Function?
every input has one arrow	yes
two inputs point to one output	yes
one input points to two outputs	no
an output receives no arrow	still may be a function

A={-2,0,2},\quad f(x)=x^2\quad\Rightarrow\quad -2\mapsto4,;0\mapsto0,;2\mapsto4

The function condition is checked from inputs to outputs. Many-to-one is allowed; one-to-many is not.

The statements $f(x)=2x^2-3$ and $f:x\mapsto2x^2-3$ name the same rule. In $f(a)$ , replace every $x$ in the rule by $a$ .

f(x)=2x^2-3\quad\Rightarrow\quad f(-2)=2(-2)^2-3=5

The input may be an expression: $f(2p)=2(2p)^2-3=8p^2-3$ . Use brackets before simplifying.

$f(x)$ is the value produced by function $f$ ; it does not mean $f\times x$ . Also, $f(a+b)$ is generally not $f(a)+f(b)$ .

The domain is the set of allowed inputs; the range is the set of outputs actually produced. Restrictions come from the rule and from any stated context.

Feature in the rule	Domain consequence
denominator $x-a$	exclude $x=a$
square root $\sqrt{x-a}$	require $x\ge a$
stated finite input set	use only those listed inputs

h(x)=\frac{3}{x-2}\quad\Rightarrow\quad \text{domain: }x\ne2,\qquad \text{range: }h(x)\ne0

An excluded input is not an output restriction unless the rule implies it. This syllabus does not require continuity arguments; determine exclusions algebraically or from the stated domain.

For this course, $fg$ means do $g$ first and then $f$ : $(fg)(x)=f(g(x))$ . The function written nearest the input acts first.

f(x)=2x-1,\quad g(x)=x^2+3\quad\Rightarrow\quad fg(x)=f(x^2+3)=2x^2+5

gf(x)=g(2x-1)=(2x-1)^2+3=4x^2-4x+4

Composition is usually order-sensitive: $fg\ne gf$ . An input is valid only if it is in the domain of $g$ and the intermediate value $g(x)$ is in the domain of $f$ .

An inverse function undoes the original function. On compatible domains, $f^{-1}(f(x))=x$ and $f(f^{-1}(x))=x$ .

Write $y=f(x)$ , interchange $x$ and $y$ , then rearrange to make $y$ the subject. Finally write the result as $f^{-1}(x)$ and carry over necessary restrictions.

y=\frac{2x}{x-3}\Rightarrow x=\frac{2y}{y-3}\Rightarrow x(y-3)=2y\Rightarrow f^{-1}(x)=\frac{3x}{x-2}

$f^{-1}(x)$ is not $1/f(x)$ . A many-to-one rule needs a restricted domain before it can have an inverse function.

Statement	Equation
$y\propto x^n$	$y=kx^n$
$y\propto1/x^n$	$y=k/x^n$
$y\propto\sqrt{x}$	$y=k\sqrt{x}$
$y\propto1/\sqrt{x}$	$y=k/\sqrt{x}$

Translate the proportionality into an equation with constant $k$ , substitute one known pair to find $k$ , then use the complete equation for the new value.

y\propto\frac1{x^3},\quad y=\frac{32}{27}\text{ when }x=\frac32\Rightarrow y=\frac{k}{x^3},;k=4\Rightarrow y=\frac4{x^3}

The frozen syllabus allows only powers $1,2,3$ and square-root forms, directly or inversely. 'Inversely proportional to $x^3$ ' means $k/x^3$ , not $k/x$ cubed after substitution errors.

A point $(x,y)$ is located by moving horizontally to $x$ and vertically to $y$ . The $x$ -coordinate is always written first.

Quadrant	Sign of $(x,y)$
I	$(+,+)$
II	$(-,+)$
III	$(-,-)$
IV	$(+,-)$

A=(-3,2),\quad B=(4,-1)\quad\Rightarrow\quad \overrightarrow{AB}=(4-(-3),-1-2)=(7,-3)

A point on an axis is not in a quadrant. Plotting $(2,-5)$ as $(-5,2)$ reverses the coordinate order and gives a different point.

y=mx+c,\qquad m=\frac{y_2-y_1}{x_2-x_1},\qquad (0,c)\text{ is the }y\text{-intercept}

To find a line through two points, calculate the gradient, substitute one point into $y=mx+c$ to find $c$ , then check the other point.

(2,-10),(-3,5):\quad m=\frac{5-(-10)}{-3-2}=-3,\quad -10=-3(2)+c\Rightarrow y=-3x-4

A vertical line has equation $x=a$ and an undefined gradient, so it cannot be written as $y=mx+c$ . Parallel non-vertical lines have equal gradients.

The required family is $y=Ax^3+Bx^2+Cx+D+E/x+F/x^2$ , with numerical constants and at least three constants zero. Tables, key features and smooth curves connect an equation to its graph.

Step	Action
1	identify excluded $x$ -values and a sensible scale
2	calculate missing table values accurately
3	plot points and join with a smooth curve
4	read roots, inequalities, turning points or intersections

f(x)=g(x)\quad\Longleftrightarrow\quad \text{the graphs }y=f(x)\text{ and }y=g(x)\text{ intersect}

Do not join a nonlinear graph with straight segments or draw across an excluded value. Graphical solutions are estimates and should be quoted to precision supported by the scale.

The gradient at a point on a curve is the gradient of the tangent there. A tangent follows the curve's direction locally without being a chord through two separated curve points.

Draw a reasonable tangent at the named point. Mark two well-separated points on the tangent, construct a large right-angled triangle, and calculate vertical change divided by horizontal change.

\text{gradient}\approx\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}

Read both points from the tangent, not from the curve unless they lie on the tangent. Preserve the sign: a tangent falling from left to right has negative gradient.

\frac{d}{dx}(ax^n)=anx^{n-1}\qquad\text{for integer }n

Rewrite reciprocals as negative powers, differentiate term by term, and write a constant's derivative as zero. Use $dy/dx$ for the gradient function.

y=2x^4+\frac6x-5=2x^4+6x^{-1}-5\Rightarrow\frac{dy}{dx}=8x^3-6x^{-2}=8x^3-\frac6{x^2}

Multiply by the old power before reducing it by one. The derivative of $x^{-2}$ is $-2x^{-3}$ , not $-2x^{-1}$ .

$dy/dx$ measures instantaneous rate of change. A stationary point occurs where $dy/dx=0$ ; it may be a maximum, a minimum or a stationary point of inflection.

Gradient change through the point	Classification
positive to negative	local maximum
negative to positive	local minimum
no sign change	stationary inflection possible

y=x^3-3x^2+3\Rightarrow\frac{dy}{dx}=3x(x-2)=0\Rightarrow x=0,2\Rightarrow(0,3),(2,-1)

Solving $dy/dx=0$ finds stationary $x$ -values; substitute into the original function for coordinates and use the graph or gradient sign to classify them.

v=\frac{ds}{dt},\qquad a=\frac{dv}{dt}

Graph	Gradient means	Area means
distance–time	speed	not used here
displacement–time	velocity	not used here
speed–time	acceleration in signed-gradient contexts	distance travelled
velocity–time	acceleration	displacement

s=t^3-9t^2+15t\Rightarrow v=3t^2-18t+15,\quad a=6t-18

Distance and speed are non-negative scalars; displacement, velocity and acceleration are signed. On a distance–time graph, a steeper tangent means greater instantaneous speed; on a speed–time graph, distance is area under the graph.

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4 Functions

Learning objectives

Read a function as an input–output rule

Represent a function as a mapping between sets

Use function notation to evaluate a rule

Control the domain and range of a function

Compose functions in the stated order

Reverse a function to find its inverse

Model direct and inverse variation

Locate and interpret points in Cartesian coordinates

Read and construct straight-line equations

Draw and use graphs of equations

Estimate a graph gradient with a tangent

Differentiate integer powers of x

Connect derivatives, rates and stationary points

Relate displacement, velocity and acceleration

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