Notesale: Turn your study into money

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A Level Maths - C3

Sam Robbins 13SE

C3
1

Algebraic fractions

Any polynomial F (x) can be put in the form:
F (x) = Q(x)×divisor+remainder
Where Q(x) is the quotient

2
2
...
2

Function mapping

One-to-one function: One element in the domain maps to one element in the range
Many-to-one function: To elements of the domain maps to one element in the range
Not a function: One input maps to two outputs

2
...
4

then it is not a function as values less than 0 don’t get mapped anywhere
...
5

Inverse functions

The inverse of f (x) is written as f −1 x
The domain of f (x) is the range of f −1 x
The range of f (x) is the domain of f −1 x
Example, find the inverse function of y = 2x2 − 7:
y + 7 = 2x2
y+7
= x2
2
x=

y+7
2

f −1 x =

x+7
2

When finding the graph of an inverse function, reflect f (x) in the line y=x
...
1

The exponential and log functions
Exponential functions

Exponential functions are in the form y = ax , graphs of these functions all pass through (0,1) as a0 = 1 for any value
of a
...
2

Functions including e

The function y = ex is the function where the gradient is identical to the function
...
3

−0
...
5

1

Formulas for exponential growth or decay

Example:
t
−
P = 16000e 10
Where P is the Price in £s and t is the years from new
What was the price when new?
Substitute t=0
0
P = 16000e 10
−

P = 16000 × 1

A Level Maths - C3

Sam Robbins 13SE

What is the value at 5 years old
Substitute t=5
5

P = 16000e− 10
P = £9704
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3
...
1

Numerical methods
Approximations for roots based on graphs

Approximations for roots can be found graphically by plotting the function and finding where the line crosses the x
axis
...

If trying to find a range in which a root can be found, substitute the values at the extreme of the range, and if there
is a change in sign between the two results, there will be a root in the range
...
The function
The exception to this rule is f (x) =
x
changes sign in the interval that includes x=0, but there is not a root
...
2

Iteration for finding approximations of roots

To solve an equation of the form f (x) = 0 by an iterative method, rearrange f (x) = 0 into a for x = g(x) and use the
iterative formula xn+1 = g(xn )
...

This may not work and will not converge to a root
...
1

Transforming graphs of functions
y = |f (x)| Graphs

The modulus of a number is written as |a|, this is the positive numerical value
...
5

0
...
5

−0
...
2

0

y = f (|x|) Graphs

To plot the graph of y = |x| − 2, first sketch the graph of y = x − 2 for x ≥ 0:
4

2

0

−2

−4
−4

−3

−2

−1

0

1

2

3

4

Then reflect that graph in the y axis
4

2

0

−2

−4
−4

−3

−2

−1

0

1

2

3

4

Examples:

2

2

0

0

−2

−2

−4
−4

−3

−2

−1

0

1

2

3

−4
−4

4

• f (x + a) - Horizontal translation of −a
• f (x) + a - Vertical translation of +a

• f (−x) - Reflection in the y axis

−1

0

1

2

3

4

Figure 10: y = 4|x| − |x|

Graph transformations

• f (ax) - Horizontal stretch of scale factor

−2

3

Figure 9: y = 4x − x3

5
...
1

Graphs of the new functions

sec(x)
10

5

0

−5

−10

−400 −300 −200 −100

0

100

200

300

400

0

100

200

300

400

0

100

200

300

400

csc(x)
10

5

0

−5

−10
−400 −300 −200 −100

cot(x)
10

5

0

−5

−10
−400 −300 −200 −100

6
...
3

Sam Robbins 13SE

Graphs of inverse functions

y = arcsin(x)
1
0
−1
−1

0

1

y = arccos(x)
3
2
1
0
−1

0

1

y = arctan(x)
0
...
5
−1

7

0

1

Further trigonometric identities and their applications
• sin(a ± b) ≡ sin A sin B ± cos A sin B
• cos(a ± b) ≡ cos A cos B
• tan(a ± b) ≡

7
...
2

2 tan A
1 − tan2 A

The R formula

For positive values of a and b
a sin θ ± b cos θ can be expressed in the form R sin(θ ± α), where 0 < α < 90
a cos θ ± b sin θ can be expressed in the form R cos(θ α), where 0 < α < 90
R cos√ = a, R sin α = b
α
R = a2 + b2

A Level Maths - C3

8
8
...
1
...
2

The product rule

The product rule is used to differentiate the product of two functions
...
3

The quotient rule

du
dv
v
−u
u(x)
dy
If y =
then
= dx 2 dx
v(x)
dx
v
Example
x
y=
2x + 5
u = x, v = 2x + 5
du
dv
= 1,
=2
dx
dx
dy
(2x + 5) × 1 − x × 2
=
dx
(2x + 5)2

Sam Robbins 13SE

A Level Maths - C3

dy
5
=
dx
(2x + 5)2

8
...
5

The logarithmic function

dy
1
=
dx
x
dy
f (x)
If y = ln[f (x)] then
=
dx
f (x)
Example
y = ln(6x − 1)
dy
6x − 1 = 6
dx
6
dy
ln(6x − 1) =
dx
6x − 1
If y = ln(x) then

8
...
6
...
6
...
6
...
6
...
6
...
6

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