Higher Maths
Revision Notes
goodbye
Functions and Graphs
Get Started
Functions and graphs
You should know the meaning of the terms domain and range of a function;
Recognise the probable
form of a function from
its graph
f : x → sin (ax + b),
f: x → ax
(a > 1 and 0 < a < 1, x  R)
f : x → cos (ax + b)
f: x → logax
(a > 1, x > 0)
functions with
restricted domain
Polynomial functions
Composite function
Inverse of a function
Given the graph of f(x) draw the graphs
of related functions, where f(x) is a simple
Complete
the square.
polynomial or trigonometric function
Radian
measure.
y=cos(ax + b)
y=sin(ax + b)
1.5
1.5
Shift to left
1
1
0.5
0.5
Shift to left
y= cos(ax + b)
y= cos(b)
y = cos(ax)
y-sin(ax+b)
0
y-sin(b)
0
y=sin(ax)
-0.5
-0.5
-1
-1
-1.5
wavelengt
h
-1.5
In both graphs
•
a = 360 ÷ wavelength.
•
b = shift to the left  a.
wavelengt
h
… the number of waves in 360˚.
Example: Find the equation of the blue curve if it is of the form y = sin(ax + b).
1.5
Imagine the red curve, the corresponding function
of the form y = sin(ax).
1
0.5
0
0
20
40
60
80
100
120
-0.5
The wavelength is 120˚
The shift to the left is 60 – 40 = 20˚
a = 360 ÷ 120 = 3
b = 20  3 = 60
The blue curve has equation y = sin(3x + 60)˚
-1
-1.5
Test
Yourself?
Exponential functions
y = a^x ;
y = a^x
a>1
; 0>a>1
y  1a   ax
x
(1, a)
(0, 1)
(0, 1)

(1, a)
Logarithmic functions
y = log ax
Note:
An exponential function is the
inverse of the corresponding
logarithmic function.
loga (a x )  x
(a, 1)
(1, 0)
a loga x  x
When a = e = 2·71828…
Test
Yourself?

the function is called the exponential function.
Polynomial functions
polynomial functions 1
polynomial f unctions 2
10
10
8
8
6
6
4
4
2
constant
0
-5
-4
-3
-2
-1
-2
0
1
2
3
4
5
cubic (order 3)
2
linear
quadratic
0
-4
-4
-6
quartic (order 4)
-3
-2
-1
-2
0
1
2
3
4
5
-4
-8
-10
-6
In general a polynomial of order n will have at most n real roots and at most (n – 1) stationary points.
e.g. a cubic can have, at most, 3 real roots and 2 s.p.s
If a cubic has 3 real roots a, b and c then its equation will be of the form y = k(x – a)(x – b)(x – c) where k is a constant.
If a, b and c are known then k can be calculated if one point the curve passes through is also known.
When the brackets are expanded the cubic will have the form
y = px3 + qx2 + rx + s, where p, q, r, and s are constants and p ≠ 0
Constant: y = a
Similar statements can be made of the other polynomials.
linear: y = ax + b
quadratic: y = ax2 + bx + c
Test
Yourself?
cubic: y = ax3 + bx2 + cx + d
quartic: y = ax4 + bx3 + cx2 + dx + e
Restricted domain
Reciprocal functions
Square root function
y = a/x
y = ¦x
Others
3.5
3
y = tan(x)
then x ≠ 90, 270
or 90 + 180n
where n is an integer.
2.5
2
-5
-4
-3
-2
-1
0
1
2
3
4
5
1.5
1
0.5
0
0
1
2
3
4
5
6
7
8
9
The denominator can never equal zero.
The term within the radical sign must always be ≥ 0
So a value of x which makes this happen is not
in the domain.
So any value of x which makes this negative is not in
the domain.
y


a
; f (x)  0
f x 
For example
3
y
thenx  2
x2
y
f (x); f (x)  0
y = sin–1(x)
y = cos–1(x)
then –1 ≥ x > 1
For example
y  x  3 thenx  3


y = log(x)
then x > 0
Test
Yourself?
inverse
If f(g(x)) = x and g(f(x)) = x for all x in the domain then we say that f is the inverse of g and vice versa.
The inverse of f is denoted by f–1.
Examples [over suitable domains]
•
f(x) = x2
…
f–1(x)= √x
•
f(x) = sin(x)
…
f–1(x) = sin–1(x)
•
f(x) = 2x + 3
…
f–1(x) = (x – 3)/2
•
f(x) = loga(x)
…
f–1(x) = ax
•
f(x) = ex
…
f–1(x) = ln(x)
For the Higher exam you need not know how to find the formula for the inverse of any function.
composites
Composite functions
Example
Suppose we have two functions: f(x) = 3x + 4 and g(x) = 2x2 + 1.
We can use these definitions to create new functions:
1
f(f(x))
2
g(g(x))
3
= g(2x2 + 1)
= f(3x + 4)
2(2x2
+
1)2
= 3(3x + 4) + 4
=
+1
= 9x + 16
= 8x4 + 8x2 + 3
f(g(x))
4
g(f(x))
= f(2x2 + 1)
= g(3x + 4)
= 3(2x2 + 1) + 4
= 2(3x + 4)2 + 4
= 6x2 + 7
= 18x2 + 48x + 36
Things to note:
•
A composition can be made from more than two functions
•
Considering examples 1 and 2 leads to recurrence relations e.g. f(f(f(f(x)))))
•
In general f(g(x)) ≠ g(f(x)) … the order in which you do things are important.
•
If either f or g have restrictions on their domain, this will affect the domain of the composite function.
•
If f(g(x)) = x for all x in the domain then we say that f is the inverse of g … it can be denoted by g–1
Test
Yourself?
Related functions
y = f(x + a)
y = f(x) + a
[x-translation of –a]
[y-translation of a]
y = f (x)
15
y = af(x)
10
[stretch in ydirection]
y = –f(x)
[reflection in x-axis]
5
y = f(ax)
[squash in x-direction]
y = f(–x)
0
-4
-3
-2
-1
0
1
2
3
-5
-10
-15
y = f–1(x)
y = f ´(x)
[The inverse]
[The derivative]
Test
Yourself?
[reflection in y-axis]
Completing the square
x  a
2
 x 2  2ax  a 2
 x 2  2ax  x  a  a 2
2
We can use this identity to simplify quadratic expressions.

Example 1
Express
x2
+ 6x + 1 in the form (x +
a)2
Example 2
(a)
Express 3x2 + 12x + 1 in the form a(x + b)2 – c
(b)
Find the smallest value the expression can take.
+b
Given x2 + 6x + 1
(a)
Given 3x2 + 12x + 1,
Take 3 out as a common factor leaving the coefficient of x2 as 1
So 3(x2 + 4x) +1 … focus on the red text.
By inspection a = 6 ÷ 2 = 3
So x2 + 6x + 1 = (x + 3)2 – 32 + 1
By inspection a = 4 ÷ 2 = 2
So we get 3(x2 + 4x) +1 = 3[(x + 2)2 – 22] + 1
= 3(x + 2)2 – 12 + 1
= 3(x +2)2 – 11
= (x + 3)2 – 8
Note: a = 3 and b = –8
(b)
Test
Yourself?
The smallest a perfect square can be is zero.
So the smallest the expression can be is 0 – 11 = 11.
Ths happens when x = –2.
radians
R
1 radian
R
We can measure angle
size using the degree
(90˚ - 1 right angle)
degree
radian
We can measure angle size
using the grad
(100 grads - 1 right angle)
Mathematicians find it convenient
to use the radian.
(π/2 radians = 1 right angle.
π
÷ 180
 180
÷π
The values are often given in terms of π.
radian
degree
30  6 radians
60  3 radians
Test
Yourself?
R
90  2 radians
45  4 radians
180   radians
y = f(x + a)
[x-translation of –a]
y = f (x+1)
15
–1
10
5
0
-4
-3
-2
-1
0
-5
-10
-15
1
2
3
y = f(x) + a
[y-translation of a]
y = f (x)-5
15
10
5
0
-4
-3
-2
-1
0
-5
-10
-15
1
–5
2
3
y = –f(x)
15
10
y = –f(x)
[reflection in x-axis]
5
0
-4
-3
-2
-1
0
-5
-10
-15
1
2
3
y = f (–x)
15
10
5
y = f(–x)
0
-4
-3
-2
-1
0
-5
-10
-15
1
2
3
[reflection in y-axis]
Stationary points of the function correspond to
zeros of the derived function.
Negatve gradients of the function correspond
to
Parts above the axis on the derived function.
Positive gradients of the function correspond to
Parts below the axis on the derived function.
y = f '(x)
15
–
+
10
–
+
+
-2
+
–
0
-3
+
–
5
+
-4
+
-1
0
1
-5
-10
Gradient of the function is shown in red
-15
y = f ´(x)
[The derivative]
+
2
3
When a function has an inverse then, if (x, y) lies on the graph of the function,
(y, x) lies on the graph of the inverse function.
…one is the reflection of the other in the line y = x.
Inverse of a function
7
6
5
function
4
y=x
3
inverse
2
1
0
0
1
2
3
4
5
6
7
inverse of a f unction
15
y = f–1(x)
10
[The inverse]
function
5
y=x
inverse
0
-5
0
-5
5
10
15
Note that the example
function does not have
an inverse. The
reflection in y = x has,
for example, 3 values
corresponding to x = 5.
y = f (2x)
15
10
5
y = f(ax)
0
[squash in x-direction]
-4
-3
-2
0
-1
-5
-10
-15
1
2
3
y = 3f(x)
15
y = af(x)
10
[stretch in ydirection]
5
0
-4
-3
-2
-1
0
-5
-10
-15
1
2
3
Using suitable units, the distance of the tip of the rotor
to the tail of a helicopter can be calculated using a formula of the form
D = 3sin(ax + b) + 10. The graph is shown below.
What are the values of a and b
Rotor Distance
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
reveal
0
15
30
45
60
75
90
105
120
Using suitable units, the distance of the tip of the rotor
to the tail of a helicopter can be calculated using a formula of the form
D = 3sin(ax + b) + 10. The graph is shown below.
Note:
The 10 translates the sine wave 10 units up.
The 3 stretches the wave by a factor of 3 in
the y-direction.
What are the values of a and b
Rotor Distance
The wavelength, by inspection, is 120˚.
a = 360 ÷ 120 = 3
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
One would expect the first peak of
y = sin(3x) to occur at 90 ÷ 3 = 30.
It occurs at 15. Thus the shift to the left
is 30 – 15 = 15.
So b = shift  a = 15  3 = 45.
0
15
30
45
60
75
90
105
120
The equation is:
D = 3sin(3x + 45) + 10
Eiffel Tower
350
300
height (m)
250
200
150
reveal
100
50
0
0
10
20
30
40
50
60
70
width from centre (m)
The profile of the Eiffel tower can be modelled by the formula
y = a ln(bx)
where a and b are constants and y m is the height of a spot on the
profile and x is its distance measured horizontally from the centre.
When y = 0, x = 63. When y = 42, x = 40.
Find the values of a and b.
Eiffel Tower
height (m)
350
300
When y = 0, x = 63
250
y = a ln(bx)
 0 = a ln (63b)
 ln(63b) = 0
 63b = 1
 b = 1/63
200
150
When y = 42, x = 40
y = a ln(bx)
 42 = a ln (40 ÷ 63)
 42 = a  –0·454255 … (calculator)
a = –92 (to nearest whole number)
100
50
0
0
10
20
30
40
50
60
70
width from centre (m)
Eiffel tower can be modelled by
The profile of the Eiffel tower can be modelled by the formula
y = a ln(bx)
where a and b are constants and y m is the height of a spot on the
profile and x is its distance measured horizontally from the centre.
When y = 0, x = 63. When y = 42, x = 40.
Find the values of a and b.
y = –92 ln(x/63)
‘It fits where it touches.’
1.5
1
0.5
sine
0
-3
-2
-1
0
1
2
3
cubic
-0.5
reveal
-1
-1.5
As can be seen in the graph, there is a simple cubic function
which for –1 > x > 1, and working in radians, behaves
almost the same as the sine wave. i.e.
sin(x)  px3 + qx2 + rx + s where p, q, r, and s are constants.
(a)
The roots of this cubic are ±6 and 0. Express the cubic
in terms of its factors viz. k(x – a)(x – b)(x – c).
(b)
We know sin(π/6) = 1/2. Use this to find k as a simple
fraction with a unit numerator.
‘It fits where it touches.’
1.5
(a)

1
0.5
-2
-1
 kxx 2  6
sine
0
-3

y  k(x  0) x  6 x  6
0
1
2
3
cubic
-0.5
-1
(b)

1
2
-1.5
 k   16
As can be seen in the graph, there is a simple cubic function
which for –1 > x > 1, and working in radians, behaves
almost the same as the sine wave. i.e.
sin(x)  px3 + qx2 + rx + s where p, q, r, and s are constants.
(a)
The roots of this cubic are ±6 and 0. Express the cubic
in terms of its factors viz. k(x – a)(x – b)(x – c).
(b)
We know sin(π/6) = 1/2. Use this to find k as a simple
fraction with a unit numerator.

  2
 k   6 3k
6 36 


A function is defined by f: x  (x2 – x – 2)
Find the largest possible domain for the function.
reveal
A function is defined by f: x  (x2 – x – 2)
Find the largest possible domain for the function.
The function within the radical sign must
be greater than or equal to zero.
x2  x  2  0
 x  2x  1  0
y=x2 – x – 2
3

¦(x
2
2
–x–
1
0
6
-3
-2
-1
0
1
2
3
4
5
-1
5
-2
4
-3
3
The sketch of this quadratic tells us that
x ≥ 2 or x ≤ –1.
2
1
0
-3
-2
-1
0
-1
1
2
3
4
5
The sketch on the left shows the function
in question.
f(x) = 2x – 1 and g(x) = x2 + 2.
(a)
Find an expression for f(g(x).
(b)
In general f(g(x)) ≠g(f(x)).
However, in this case there are two values of x
for which f(g(x)) = g(f(x)).
Find these values.
reveal
f(x) = 2x – 1 and g(x) = x2 + 2.
(a)
Find an expression for f(g(x).
(b)
In general f(g(x)) ≠g(f(x)),
(a)
however, in this case there are two values of x
for which f(g(x)) = g(f(x)).
f(g(x) = f(x2 + 2)
= 2(x2 + 2) – 1
= 2x2 + 3
Find these values.
(b)
g(f(x)) = g(2x – 1)
= (2x – 1)2 + 2
= 4x2 – 4x + 3
g(f(x)) = f(g(x)

4x2 – 4x + 3 = 2x2 + 3

2x2 – 4x = 0

x(x – 2) = 0

x = 0 or x = 2
y = f(x)
5
4
3
2
(0·75, 1)
1
0
-2
-1
-1
0
1
2
3
-2
-3
-4
-5
reveal
The sketch shows part of the function y = f(x)
(a)
Draw a sketch of (i) y = f(–x)
(b)
Make a sketch of y = f´(x)
(ii) y = f(1 – x)
y = f(x)
5
y = f(-x)
4
(a (i))
3
5
4
2
(0·75, 1)
3
2
1
1
0
-2
-1
-1
0
1
2
y = f(x)
0
3
-2.5
-2
-1.5
-1
-0.5
-1
0
0.5
1
1.5
2
y = f(-x)
2.5
-2
-2
-3
-3
y=f (1-x)
-4
-4
5
-5
(a (ii))
4
-5
3
2
1
y=f(x)
y=f(-x)
0
-2.5
-2
-1.5
-1
-0.5
The sketch shows part of the function y = f(x)
-1
0
0.5
1
1.5
2
2.5
y=f(1-x)
-2
-3
-4
(a)
Draw a sketch of (i) y = f(–x)
(b)
Make a sketch of y = f´(x)
-5
(ii) y = f(1 – x)
(b)
y=f '(x)
5
4
3
2
1
y=f(x)
0
-2.5
-2
-1.5
-1
-0.5 0
-1
-2
-3
-4
-5
0.5
1
1.5
2
2.5
y=f'(x)
Where completing the square is useful
Q1
Find the maximum value of the function defined by:
1
f :x 2
x  4x  9

Q2
reveal
Prove that y = 3x3 + 3x2 + 5x + 1
is an increasing function.
Where completing the square is useful
Q1
Q1
Find the maximum value of the function defined by:
In Higher maths you don’t know how to
differentiate this function.
Complete the square on the denominator:
1
f :x 2
x  4x  9
x2 + 4x + 9 = (x + 2)2 – 22 + 9 = (x + 2)2 + 5.
The smallest this expression can be is when
the bracket takes the value zero …
When x = –2, the expression is worth 5.
This is when the function will be at its biggest.

f(–2) = 1/5
Q2
Q2
Prove that y = 3x3 + 3x2 + 5x + 1
is an increasing function.
To prove the function is always increasing , you
have to prove that the derivative is always positive.
dy
 9x 2  6x  5
dx
Complete the square:
9x 2  6x  5  9x 2  69 x  5  9x 2  23 x  5

2
2
2
 9 x  13   13   5  9 x  13   19  5

 9x 



 1 5
2
 9x  13   4

1 2
3
The minimum value of the derivative is 4.
Thus its always positive and thus
Always increasing.
Many repetitive situations can be modelled by the sine function.
Using suitable units the distance of the star from the centre of
the picture can be modelled by D = 2 sin x where x is measured
in radians.
reveal
Solve the equation 2sin x = 1 for 0 ≥ x ≥ 4π
Many repetitive situations can be modelled by the sine function.
Using suitable units the distance of the star from the centre of
the picture can be modelled by D = 2 sin x where x is measured
in radians.
2sin x  1
 sin x  12
x


6
or   6 
6
6
 6 
5
6
or any number of complete
revolutions more (or less) than these
two solutions …
x

6
or
5
6
or 2š + 6 or 2š + 56
or 4š + 6 or 4š + 56
x  6 , 56 , 13
, 17
, 25
, 29
6
6
6
6
Solve the equation 2sin x = 1 for 0 ≥ x ≥ 4π

All other ‘answers’ are outside the desired range.