Get Startedgoodbyegoodbye

Revision NotesFunctions and Graphs

Higher Maths

Functions and graphs You should know the meaning of the terms domain and range of a function;

f : x → sin (ax + b),

f : x → cos (ax + b)

f: x → ax

(a > 1 and 0 < a < 1, x ∈ R) f: x → logax (a > 1, x > 0)

Recognise the probable form of a function from

its graph

Given the graph of f(x) draw the graphs of related functions, where f(x) is a simple

polynomial or trigonometric function

Polynomial functions functions with restricted domain

Inverse of a functionComposite function

Complete the square.Complete the square.

Radian measure.Radian measure.

In both graphs• a = 360 ÷ wavelength. … the number of waves in 360˚.• b = shift to the left a.

y=sin(ax + b)

-1.5

-1

-0.5

0

0.5

1

1.5

y-sin(ax+b)

y-sin(b)

y=sin(ax)

y=sin(ax + b)

-1.5

-1

-0.5

0

0.5

1

1.5

y-sin(ax+b)

y-sin(b)

y=sin(ax)

Shift to leftShift to left

wavelengthwavelength

y=cos(ax + b)

-1.5

-1

-0.5

0

0.5

1

1.5

y= cos(ax + b)y= cos(b)y = cos(ax)

y=cos(ax + b)

-1.5

-1

-0.5

0

0.5

1

1.5

y= cos(ax + b)y= cos(b)y = cos(ax)

wavelengthwavelength

Shift to leftShift to left

-1.5

-1

-0.5

0

0.5

1

1.5

0 20 40 60 80 100 120

Imagine the red curve, the corresponding function of the form y = sin(ax).

The wavelength is 120˚The shift to the left is 60 – 40 = 20˚

a = 360 ÷ 120 = 3

b = 20 3 = 60The blue curve has equation y = sin(3x + 60)˚

Example: Find the equation of the blue curve if it is of the form y = sin(ax + b).

Test Yourself?

y = a^xy = a^x

(0, 1)(0, 1)

(1, a)(1, a)

; 0 > a > 1; 0 > a > 1y = a^xy = a^x

(0, 1)(0, 1)

(1, a)(1, a)

; a > 1; a > 1

Exponential functions

Logarithmic functions

y = logaxy = logax

(1, 0)(1, 0)

(a, 1)(a, 1)

€

y = 1a( )

x= a−x

Note:

An exponential function is the inverse of the corresponding logarithmic function.

€

loga (ax ) = x

aloga x = x

When a = e = 2·71828…

the function is called the exponential function.

Test Yourself?

polynomial functions 1

-10

-8

-6

-4

-2

0

2

4

6

8

10

-5 -4 -3 -2 -1 0 1 2 3 4 5

constant

linear

quadratic

polynomial functions 1

-10

-8

-6

-4

-2

0

2

4

6

8

10

-5 -4 -3 -2 -1 0 1 2 3 4 5

constant

linear

quadratic

polynomial functions 2

-6

-4

-2

0

2

4

6

8

10

-4 -3 -2 -1 0 1 2 3 4 5

cubic (order 3)

quartic (order 4)

polynomial functions 2

-6

-4

-2

0

2

4

6

8

10

-4 -3 -2 -1 0 1 2 3 4 5

cubic (order 3)

quartic (order 4)

Polynomial functions

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

Similar statements can be made of the other polynomials.Test

Yourself?

Constant: y = a

linear: y = ax + b

quadratic: y = ax2 + bx + c

cubic: y = ax3 + bx2 + cx + d

quartic: y = ax4 + bx3 + cx2 + dx + e

Reciprocal functions

y = a/x

-5 -4 -3 -2 -1 0 1 2 3 4 5

y = a/x

-5 -4 -3 -2 -1 0 1 2 3 4 5

Restricted domain

€

y =a

f x( ); f (x) ≠ 0

€

y =3

x + 2 then x ≠ −2

The denominator can never equal zero.

So a value of x which makes this happen is not in the domain.

For example

y = √x

0

0.5

1

1.5

2

2.5

3

3.5

0 1 2 3 4 5 6 7 8 9

y = √x

0

0.5

1

1.5

2

2.5

3

3.5

0 1 2 3 4 5 6 7 8 9

Square root function

The term within the radical sign must always be ≥ 0

So any value of x which makes this negative is not in the domain.

€

y = f (x); f (x) ≥ 0

€

y = x + 3 then x ≥ −3

For example

y = tan(x) then x ≠ 90, 270 or 90 + 180n

where n is an integer.

y = log(x) then x > 0

Others

y = sin–1(x)y = cos–1(x)

then –1 ≥ x > 1

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))

= f(3x + 4)

= 3(3x + 4) + 4

= 9x + 16

2 g(g(x))

= g(2x2 + 1)

= 2(2x2 + 1)2 + 1

= 8x4 + 8x2 + 3

3 f(g(x))

= f(2x2 + 1)

= 3(2x2 + 1) + 4

= 6x2 + 7

4 g(f(x))

= g(3x + 4)

= 2(3x + 4)2 + 4

= 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)

-15

-10

-5

0

5

10

15

-4 -3 -2 -1 0 1 2 3

y = f(x) + a

y = f(x + a)

y = af(x)

y = f(ax)

y = –f(x)

y = f(–x)

y = f–

1(x)y = f ´(x)

[The inverse]

[The derivative]

[x-translation of –a]

[y-translation of a]

[reflection in x-axis]

[reflection in y-axis]

[stretch in y-direction]

[squash in x-direction]

Test Yourself?

Completing the square

€

x + a( )2

= x 2 + 2ax + a2

⇔ x 2 + 2ax = x + a( )2

− a2

We can use this identity to simplify quadratic expressions.

Example 1

Express x2 + 6x + 1 in the form (x + a)2 + b

Given x2 + 6x + 1

By inspection a = 6 ÷ 2 = 3

So x2 + 6x + 1 = (x + 3)2 – 32 + 1

= (x + 3)2 – 8

Note: a = 3 and b = –8

Example 2(a) Express 3x2 + 12x + 1 in the form a(x +

b)2 – c(b) Find the smallest value the expression

can take.

(a) Given 3x2 + 12x + 1, Take 3 out as a common factor leaving the

coefficient of x2 as 1So 3(x2 + 4x) +1 … focus on the red text.

By inspection a = 4 ÷ 2 = 2So we get 3(x2 + 4x) +1 = 3[(x + 2)2 – 22] +

1= 3(x + 2)2 – 12 + 1= 3(x +2)2 – 11

• 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.

Test Yourself?

radians

We can measure angle size using the degree (90˚ - 1 right angle)

We can measure angle size using the grad (100 grads - 1 right angle)

1 radian R

R

R

Mathematicians find it convenient to use the radian.(π/2 radians = 1 right angle.

degreedegree π π ÷ 180÷ 180 radianradian

degreedegree÷ π÷ π 180 180radianradian

The values are often given in terms of π.

€

30° = π6 radians

60° = π3 radians

90° = π2 radians

45° = π4 radians

180° = π radians

Test Yourself?

[x-translation of –a]

y = f(x + a)

y = f(x+1)

-15

-10

-5

0

5

10

15

-4 -3 -2 -1 0 1 2 3

–1

y = f(x) + a[y-translation

of a]

y = f(x)-5

-15

-10

-5

0

5

10

15

-4 -3 -2 -1 0 1 2 3–5

y = –f(x)[reflection in

x-axis]

y = –f(x)

-15

-10

-5

0

5

10

15

-4 -3 -2 -1 0 1 2 3

y = f(–x)[reflection in

y-axis]

y = f(–x)

-15

-10

-5

0

5

10

15

-4 -3 -2 -1 0 1 2 3

y = f ´(x)

[The derivative]

y = f'(x)

-15

-10

-5

0

5

10

15

-4 -3 -2 -1 0 1 2 3

+

+

+

+

+

+

+

+

–

–

–

–

Gradient of the function is shown in red

Stationary points of the function correspond to zeros of the derived function.

Positive gradients of the function correspond to Parts below the axis on the derived function.

Negatve gradients of the function correspond to Parts above the axis on the derived function.

y = f–

1(x)[The inverse]

Inverse of a function

0

1

2

3

4

5

6

7

0 1 2 3 4 5 6 7

function

y=x

inverse

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

-5

0

5

10

15

-5 0 5 10 15

function

y=x

inverse

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(ax)[squash in x-

direction]

y = f(2x)

-15

-10

-5

0

5

10

15

-4 -3 -2 -1 0 1 2 3

y = af(x)[stretch in y-

direction]

y = 3f(x)

-15

-10

-5

0

5

10

15

-4 -3 -2 -1 0 1 2 3

Using suitable units, the distance of the tip of the rotorto the tail of a helicopter can be calculated using a formula of the formD = 3sin(ax + b) + 10. The graph is shown below.

What are the values of a and b

Rotor Distance

0123456789

1011121314

0 15 30 45 60 75 90 105 120

revealreveal

Using suitable units, the distance of the tip of the rotorto the tail of a helicopter can be calculated using a formula of the formD = 3sin(ax + b) + 10. The graph is shown below.

What are the values of a and b

Rotor Distance

0123456789

1011121314

0 15 30 45 60 75 90 105 120

Note:

The 10 translates the sine wave 10 units up.

The 3 stretches the wave by a factor of 3 in the y-direction.

The wavelength, by inspection, is 120˚.

a = 360 ÷ 120 = 3

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.

The equation is:

D = 3sin(3x + 45) + 10

Eiffel Tower

0

50

100

150

200

250

300

350

0 10 20 30 40 50 60 70

width from centre (m)

height (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.

revealreveal

Eiffel Tower

0

50

100

150

200

250

300

350

0 10 20 30 40 50 60 70

width from centre (m)

height (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.

When y = 0, x = 63

y = a ln(bx) 0 = a ln (63b) ln(63b) = 0 63b = 1 b = 1/63

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)

Eiffel tower can be modelled by

y = –92 ln(x/63)

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.

• 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).

• We know sin(π/6) = 1/2. Use this to find k as a simple fraction with a unit numerator.

-1.5

-1

-0.5

0

0.5

1

1.5

-3 -2 -1 0 1 2 3

sine

cubic

‘It fits where it touches.’

revealreveal

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.

• 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).

• We know sin(π/6) = 1/2. Use this to find k as a simple fraction with a unit numerator.

-1.5

-1

-0.5

0

0.5

1

1.5

-3 -2 -1 0 1 2 3

sine

cubic

‘It fits where it touches.’

€

y = k(x − 0) x − 6( ) x + 6( )

= kx x 2 − 6( )

(a)

(b)

€

12 = k

π

6

π 2

36− 6

⎛

⎝ ⎜

⎞

⎠ ⎟≈ −3k

⇒ k = − 16

A function is defined by f: x (x2 – x – 2)

Find the largest possible domain for the function.

revealreveal

A function is defined by f: x (x2 – x – 2)

Find the largest possible domain for the function.

√(x 2 – x –

-1

0

1

2

3

4

5

6

-3 -2 -1 0 1 2 3 4 5

The function within the radical sign must be greater than or equal to zero.

€

x 2 − x − 2 ≥ 0

⇒ x − 2( ) x +1( ) ≥ 0

y=x2 – x – 2

-3

-2

-1

0

1

2

3

-3 -2 -1 0 1 2 3 4 5

The sketch of this quadratic tells us that x ≥ 2 or x ≤ –1.

The sketch on the left shows the function in question.

f(x) = 2x – 1 and g(x) = x2 + 2.

• Find an expression for f(g(x).

• 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.

revealreveal

f(x) = 2x – 1 and g(x) = x2 + 2.

• Find an expression for f(g(x).

• 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.

(a) f(g(x) = f(x2 + 2)

= 2(x2 + 2) – 1

= 2x2 + 3

• g(f(x)) = g(2x – 1)

= (2x – 1)2 + 2

= 4x2 – 4x + 3

g(f(x)) = f(g(x)

(a) 4x2 – 4x + 3 = 2x2 + 3

(b) 2x2 – 4x = 0

(c) x(x – 2) = 0

(d) x = 0 or x = 2

y = f(x)

-5

-4

-3

-2

-1

0

1

2

3

4

5

-2 -1 0 1 2 3

revealreveal

The sketch shows part of the function y = f(x)

(a) Draw a sketch of (i) y = f(–x) (ii) y = f(1 – x)

(b) Make a sketch of y = f´(x)

(0·75, 1)

y = f(x)

-5

-4

-3

-2

-1

0

1

2

3

4

5

-2 -1 0 1 2 3

The sketch shows part of the function y = f(x)

(a) Draw a sketch of (i) y = f(–x) (ii) y = f(1 – x)

(b) Make a sketch of y = f´(x)

(0·75, 1)

y = f(-x)

-5

-4

-3

-2

-1

0

1

2

3

4

5

-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5

y = f(x)

y = f(-x)

(a (i))

y=f(1-x)

-5

-4

-3

-2

-1

0

1

2

3

4

5

-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5

y=f(x)

y=f(-x)

y=f(1-x)

(a (ii))

y=f'(x)

-5

-4

-3

-2

-1

0

1

2

3

4

5

-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5

y=f(x)

y=f'(x)

(b)

Q1 Find the maximum value of the function defined by:

€

f : x →1

x 2 + 4x + 9

Q2 Prove that y = 3x3 + 3x2 + 5x + 1

is an increasing function.

Where completing the square is useful

revealreveal

Q1 Find the maximum value of the function defined by:

€

f : x →1

x 2 + 4x + 9

Q2 Prove that y = 3x3 + 3x2 + 5x + 1

is an increasing function.

Where completing the square is useful

In Higher maths you don’t know how to differentiate this function.

Complete the square on the denominator:

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

Q1

Q2 To prove the function is always increasing , you have to prove that the derivative is always positive.

€

dy

dx= 9x 2 + 6x + 5

Complete the square:

€

9x 2 + 6x + 5 = 9 x 2 + 69 x[ ] + 5 = 9 x 2 + 2

3 x[ ] + 5

= 9 x + 13( )

2− 1

3( )2

[ ] + 5 = 9 x + 13( )

2− 1

9[ ] + 5

= 9 x + 13( )

2−1+ 5

= 9 x + 13( )

2+ 4

The minimum value of the derivative is 4.Thus its always positive and thusAlways increasing.

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.

revealreveal

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 = π6 or π − π

6 = 6π6 − π

6 = 5π6

or any number of complete revolutions more (or less) than these two solutions …

€

x = π6 or 5π

6 or 2π + π6 or 2π + 5π

6

or 4π + π6 or 4π + 5π

6

x = π6 , 5π

6 ,13π6 ,17π

6 , 25π6 , 29π

6

All other ‘answers’ are outside the desired range.