Edexcel IGCSE Math B 4 functions

Use this Functions hub to move from composite and inverse functions into variation, straight-line graphs, curve sketching, rates of change and kinematics.

Covered syllabus points

4.A Functions 4.B Functions as mappings 4.C Function notation 4.D Domain and range 4.E Composite functions 4.F Inverse functions 4.G Variation 4.H Cartesian coordinates 4.I Straight-line graphs 4.J Graphs of equations 4.K Graph gradients 4.L Differentiation 4.M Rates of change and stationary points 4.N Kinematics applications

[Maximum number: 9]

Figure 3

Information about the function f is shown in Figure 3.
Given that f is the mapping $\mathrm{f}: x \mapsto a x^{3}+b$ where a and b are constants,

(a)

use the information in Figure 3 to write down and simplify two equations in a and b.

[ 2 ]

(b)

Using your values of a and b, express the inverse function $f^{-1}$ in the form $f^{-1}:x\mapsto\ldots$ .

[ 2 ]

$y=2x^3+3$ $x=\sqrt[3]{\frac{y-3}{2}}$

Answer:
$f^{-1}:x\mapsto\sqrt[3]{\frac{x-3}{2}}$

Marks: M1, A1.

(c)

Find g(2 p), simplifying your answer.

[ 2 ]

$g(2p)=\frac{p(2p)}{2p-p}=2p$

Answer: 2p.
Marks: M1, A1.

(d)

Show algebraically that $\operatorname{gg}(x)=x$

Show clear algebraic working.

[ 2 ]

$g(g(x))=\frac{p\left(\frac{px}{x-p}\right)}{\frac{px}{x-p}-p}$ $=\frac{\frac{p^2x}{x-p}}{\frac{px-p(x-p)}{x-p}} =\frac{p^2x}{x-p}\times\frac{x-p}{p^2}=x$

Marks: M1, A1.

(e)

Hence write down the inverse function $\mathrm{g}^{-1}$ in the form $\mathrm{g}^{-1}(x)=\ldots$

[ 1 ]

Answer:
$g^{-1}(x)=\frac{px}{x-p}$

Marks: B1.

[Maximum number: 13]

The equation of the straight line L is
y=2-3x.

(a)

Find the exact coordinates of the points where L crosses the coordinate axes.

[ 2 ]

Coordinate-axis intersections:
$(0,2),\qquad \left(\frac23,0\right).$

Marking notes:
Equivalent unambiguous y=2 and $x=\frac23$ statements are allowed; the x-intercept must be exact.

(b)

The functions f and g are defined as
$f:x\mapsto2-3x, \qquad g:x\mapsto\frac{3+x}{1-2x},\quad x\ne\frac12.$
Find g(-2).

[ 1 ]

(c)

Find the values of x for which ff(x)+g(x)=0

Show your working clearly and give your answer in the form

\frac{a \pm \sqrt{b}}{c}

where a, b and c are integers.

[ 5 ]

ff(x)=f(2-3x)=2-3(2-3x)=9x-4.
Set
ff(x)+g(x)=0: $9x-4+\frac{3+x}{1-2x}=0.$
This gives
$18x^{2}-18x+1=0.$
Using the quadratic formula:
$x=\frac{18\pm\sqrt{(-18)^2-4(18)(1)}}{36}.$
Final answer:
$x=\frac{3\pm\sqrt7}{6}.$

(d)

Express the inverse of the composite function fg in the form $(\mathrm{fg})^{-1}: x \mapsto \ldots$

\left[\text { Solutions of } a x^{2}+b x+c=0 \text { are } x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\right]

[ 5 ]

$fg(x)=f\left(\frac{3+x}{1-2x}\right) = 2-3\left(\frac{3+x}{1-2x}\right) = \frac{7x+7}{2x-1}.$
Let y=fg(x):
y(2x-1)=7x+7.2xy-7x=7+y.x(2y-7)=7+y.
Therefore
$(fg)^{-1}:x\mapsto\frac{7+x}{2x-7}.$

Marking notes:
Equivalent form $\frac{-x-7}{7-2x}$ is allowed.

[Maximum number: 4]

Here are the functions f, g and h.
$\begin{aligned}f:x&\mapsto 2x-5,\quad x>1,\\g:x&\mapsto 7-\frac{8x}{2x+1},\\h:x&\mapsto 2x^2+4x-5,\quad x>-1.\end{aligned}$

(a)

Find the value of g(-1).

[ 1 ]

(b)

Write down the value of x for which g is not defined.

[ 1 ]

Final answer:
$x=-\frac12$

B1 oe. Allow x=-0.5 or $x\neq-0.5$ . Do not allow x<-0.5 or x>-0.5; must be a single value.

(c)

Find $h^{-1}(x)$ .

[ 2 ]

$y=2(x+1)^{2}-7$ $\frac{y+7}{2}=(x+1)^{2}$

Final answer:
$h^{-1}(x)=\sqrt{\frac{x+7}{2}}-1$

M1 for rearranging the completed-square form.
A1 ft from part (e); must be in terms of x. If $\pm$ is given, award M1 only.

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Covered syllabus points

4 Functions question 1

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)

4 Functions question 2

Question (a)

Question (b)

Question (c)

Question (d)

4 Functions question 3

Question (a)

Question (b)

Question (c)