Core 3 Revision Sheet
Functions
The domain is the input of the function
The range is the output of the function
Modulus functions
x means the ‘magnitude’ of x and is always positive.
When sketching modulus functions, the graph will be reflected in the x axis.
When solving modulus equations, remember to find the solutions when it is positive and negative.
Example
Solve x  3 = 7
x–3=7
OR
x = 10
x – 3 = -7
x = -4
Composite Functions
When finding composite functions, apply the functions one at a time, starting with the function
closest to x.
Example
f(x) = x² + 3
Find fg(x)
g(x) = x + 5
fg(x) = f(x + 5)
= (x + 5)² + 3
= x² + 10x + 25 + 3
= x² + 10x + 28
To find the range of composite functions, find the range of the first function being applied then use
that as the domain of the next function being applied. This output will give the range of the overall
composite function.
Inverse Functions
Remember that an inverse function can only be found if the original function is one-to-one. This can
be achieved by restricting the domain.
Example
f(x) =
let y =
2x  3
x 1
2x  3
x 1
find f 1 (x)
y(x – 1) = 2x + 3
yx – y = 2x + 3
yx – 2x = 3 + y
x(y – 2) = 3 + y
x=
Therefore f 1 (x) =
3 y
y2
3 x
x2
Drawing graphs of inverse functions
Reflect the original graph in the line y = x
You may be asked to draw inverse graphs of exponential, natural logarithms and trigonometric
graphs. Make sure you are familiar with them.
Differentiation
Chain Rule
dy dy du


dx du dx
Product Rule
d (uv)
dv
du
u
v
dx
dx
dx
d u
 
dy  v 
Quotient Rule
x as a function of y
d
du
dv
u
dx
dx - this is in your formula book
v2
dy
1
=
dx
dx
dy
Integration
Integration by substitution

determine your value for u

Find

Substitute

If necessary, substitute any x values in terms of u

Now integrate your new function and substitute ‘x’ back in at the end if required

If it is a definite integral, you must either rearrange the limits for u or put the answer in
terms of x before substituting in the limits of x.
and rearrange it into the form dx =
du
du in for dx and simplify where necessary
Example
 3x

2x
dx
2
2
2x
3
u
x
1
u
u = 3x² + 2
du
=6x
dx
dx =
1
du 
6x
substitute this into the integral and cancel
1
du
6x
1
1
du
in terms of x 
=
2
3
1 2
 3u du
3x 2  2 + c
=
2 2
u +c
3
OR
2
u +c
3
Integration by parts
dv
du
 u dxdx  uv   v dx dx

this is in your formula book
Determine your value for u and your value for
dv
dx
HINT – the value for u should get simpler when differentiated
du
and v
dx

Find

Substitute the values into the formula which should leave you with a simpler integral at the
end to work out. If not, integrate by parts again.
Example
Use integration by parts to find
 xe
5x
dx
u=x
dv
= e 5x
dx
du
=1
dx
v=
1 5x
e
5
Substitute these values into the formula:
1
1 5x
xe -  e 5 x dx
5
5
=
1 5x 1 5x
xe 
e +c
5
25
Standard Integrals
-
these are in your formula book
a
2
dx
1
 x
 tan 1    c
2
a
x
a

 x
 sin 1    c
a
a2  x2
dx
Examples
1
 1  25 x
2
dx =
1
1
2

25  1 
2
  x
5

1
dx
1
2
25(  x )
25



1
1
x 
+ c
x
tan 1 
dx =
25  1 
 1  
 
  5  
5


=
1
tan 1 5 x  c
5

1
16  x 2
dx =

1
42  x 2
 x
4
dx =
= sin 1    c
Volumes of Revolution
b
About the x axis
  y 2 dx
About the y axis
  x 2 dx
a
b
a
Exponentials and Logarithms
Make sure you recognise and could sketch the exponential and natural logarithm graphs, including
transformations of them.
Differentiating exponential functions:
Integrating exponential functions:
=
Differentiating natural logarithms:
y = ln x
Integrating natural logarithms:
x
Special Functions
dy
= ke kx
dx
y = e kx
1
dx
dx = ln (f(x)) + c,
+c
dy 1
=
dx x
=
ln x + c
if f(x) > 0
Trigonometry
Reciprocal Functions
cosec  
1
sin 
sec  
1
cos 
cot  
1
tan 
You must be able to sketch the graphs of these functions
Trigonometric Identities
tan  =
sin 
cos 
cot  =
cos 
sin 
sin 2   cos 2  = 1
(÷ cos  )

sec 2  = 1 + tan 2 
sin 2   cos 2  = 1
(÷ sin  )

cosec 2  = 1 + cot 2 
You must learn all these identities.
Inverse Trigonometric Functions
You must be able to recognise and sketch the following inverse trigonometric graphs:
y = sin
1
x for -1≤ x ≤ 1
1
y = tan 1 x for -∞ ≤ x ≤ ∞
y = cos x for -1≤ x ≤ 1
Remember, the domain and range are the opposite for the inverse functions.
The domain must be restricted in to make it a one-to-one function in order to be able to find its
inverse.
Differentiating Trigonometric Functions
d
(sin x)  cos x
dx
d
(tan x)  sec 2 x
dx
d
(cos x)   sin x
dx
Integrating Trigonometric Functions
1
 cos kxdx  k sin kx  c
 sec
2
kx 
1
tan x  c
k
1
 sin kxdx   k cos kx  c
You must learn all of the differentiation and integration trigonometric functions
Numerical methods and solutions
Change of sign method
The equation must be equal to zero before the change of sign method can be used.
Example
Show that the equations y = x² + 1 and y =
1
have a solution in the interval 0.68 and 0.69.
x
First make the equations equal to each other.
x² + 1 =
1
x
x² + 1 -
1
=0
x
Make it equal to zero
Rearrange if necessary  multiply by x
x³ + x – 1 = 0
f(0.68) = 0.68³ + 0.68 – 1 = -0.005568
f(0.69) = 0.69³ + 0.69 – 1 = 0.018509
change of sign, therefore root in interval 0.68 < x < 0.69
Iterative Methods
This involves programming your calculating so that the process is repeated to find a sequence of
approximations to the required solution (converges to the solution).
You need to be able to decide if the convergence produces a cobweb or staircase diagram, and be
able to sketch it.
Mid-ordinate rule

b
a


ydx  h  y 1  y 3  .....  y 3  y 1 
n
n
2
2
2
 2
where h =
ba
n
Simpson’s rule
odds

b
a
ydx 
evens
h
 y0  y n  4( y1  y3  ...  y n1 )  2( y 2  y 4  ...  y n2 )
3
These formulae are both in your formula book.