Functions

1. 1: Functions1: Functions
© Christine Crisp
““Teach A Level Maths”Teach A Level Maths”
Vol. 2: A2 Core ModulesVol. 2: A2 Core Modules

2. Functions
Module C3
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3. Functions
e.g. and are functions.12)( −= xxf xxg sin)( =
A function is a rule , which calculates values of
for a set of values of x.)(xf
is often replaced by y.)(xf
Another Notation
12: −xxf  12)( −= xxfmeans
is called the image of x)(xf

4. Functions
=)(f 12 −xx
x )(xf−1
0
.
.
2.1
..
.
.
1.5
.
.
.
.
.
A few of
the possible
values of x
3.2.
.
.
2
.
.
−1
.
.
−3.
.
.
.
We can illustrate a function with a diagram
The rule is sometimes called a mapping.

5. Functions
We say “ real ” values because there is a branch of
mathematics which deals with numbers that are not
real.
A bit more jargon
To define a function fully, we need to know the values
of x that can be used.
The set of values of x for which the function is
defined is called the domain.
In the function any value can be
substituted for x, so the domain consists of
all real values of x
2
)( xxf =
means “ belongs to ”∈
So, means x is any real number∈x 
stands for the set of all real numbers
We write ∈x 

6. Functions
0)( ≥xf
If , the range consists of the set of y-
values, so
)(xfy =
Tip: To help remember which is the domain and
which the range, notice that d comes before r in
the alphabet and x comes before y.
domain: x-values range: y-values
e.g. Any value of x substituted into gives a
positive ( or zero ) value.
2
)( xxf =
The range of a function is the set of values
given by .
)(xf
)(xf
So the range of is
2
)( xxf =

7. Functions
Tip: To help remember which is the domain and
which the range, notice that d comes before r in
the alphabet and x comes before y.
0)( ≥xf
If , the range consists of the set of y-
values, so
)(xfy =
e.g. Any value of x substituted into gives a
positive ( or zero ) value.
2
)( xxf =
So the range of is
2
)( xxf =
The range of a function is the set of values
given by .
)(xf
)(xf
domain: x-values range: y-values

8. Functions
The range of a function is the set of values given by
the rule.
domain: x-values range: y-values
The set of values of x for which the function is
defined is called the domain.

9. Functions
Solution: The quickest way to sketch this quadratic
function is to find its vertex by completing the square.
⇒−+= 142
xxy 2
)2( += xy 4− 1−
5)2( 2
−+=⇒ xy
14)( 2
−+= xxxf
e.g. 1 Sketch the function where
and write down its domain and
range.
)(xfy =






−
−
5
2This is a translation from of
2
xy =
)5,2( −−so the vertex is .

10. Functions
so the range is 5−≥y
So, the graph of is142
−+= xxy
The x-values on the part of the graph
we’ve sketched go from −5 to +1 . . . BUT we
could have drawn the sketch for any values of x.
( y is any real number greater than, or equal to, −5 )
BUT there are no y-values less than −5, . . .
)5,2( −−x
142
−+= xxy
domain:
So, we get ( x is any real number )∈x 

11. Functions
3+= xy
domain: x-values range: y-values
3−≥x 0≥y
e.g.2 Sketch the function where .
Hence find the domain and range of .
3)( += xxf)(xfy =
)(xf





−
0
3
so the graph is:
( We could write instead of y ))(xf
Solution: is a translation from ofxy =)(xfy =

12. Functions
SUMMARY
• To define a function we need
a rule and a set of values.
)(xfy =• For ,
the x-values form the domain
2
)( xxf =2
: xxf 
• Notation:
means
the or y-values form the range)(xf
e.g. For ,
the domain is
the range is or
2
)( xxf =
0≥y0)( ≥xf
∈x 

13. Functions
(b) xy sin=3
xy =(a)
Exercise
For each function write down the domain and range
1. Sketch the functions where

xxfbxxfa sin)()()()( 3
== and
Solution:
)(xfy =
range: 11 ≤≤− y
domain: ∈x  domain: ∈x 
range: ∈y 

14. Functions
3−≥xSo, the domain is
⇒≥+ 03x 3−≥x
We can sometimes spot the domain and range of a
function without a sketch.
e.g. For we notice that we can’t
square root a negative number ( at least not if we
want a real number answer ) so,
3)( += xxf
x + 3 must be greater than or equal to zero.
3+xThe smallest value of is zero.
Other values are greater than zero.
So, the range is 0≥y

15. Functions
then, =)(f 3
Suppose and
2
)( xf =x =)(xg 3+x
Functions of a Function
x is replaced by 3

16. Functions
)(xg
and
( ) =f
2
)( 3
Suppose and
2
)( xf = =)(xg
=)(f 1−
then, =)(f 3
2
)( 1−
9=
1=
x 3+x
Functions of a Function
x is replaced by −1
x is replaced by )(xg

17. Functions
3+x)(xg
and
( ) =f
is “a function of a function” or compound
function.
( )f )(xg
2
)(= 3+x
962
++= xx
( )f
2
)( 3
Suppose and
2
)( xf = =)(xg
=)(f 1−
then, =)(f 3
2
)( 1−
9=
1=
x 3+x
We read as “f of g of x”( ))(xgf
x is “operated” on by the inner function first.
is the inner function and the outer.)(xg )(xf
So, in we do g first.( ))(xgf
Functions of a Function

18. Functions
Notation for a Function of a Function
When we meet this notation it is a good idea to
change it to the full notation.
is often written as .( )f )(xg )(xfg
does NOT mean multiply g by f.)(xfg
I’m going to write always !( )f )(xg

19. Functions
Solution:
)()()()( xggxffxgfxfg (iv)(iii)(ii)(i)
2)( 2
+= xf
x
g
1
)( =e.g. 1 Given that and findx x
( ) == fxfg )((i) )(xg

20. Functions
x
1)(xg
x
g
1
)( =
)()()()( xggxffxgfxfg (iv)(iii)(ii)(i)
2)( 2
+= xfe.g. 1 Given that and findx x
Solution:
( ) == fxfg )((i)
)()()()( xggxffxgfxfg (iv)(iii)(ii)(i)
x
=





f

21. Functions
x
1)(xg
Solution:
( ) == fxfg )((i)
)()()()( xggxffxgfxfg (iv)(iii)(ii)(i)
2)( 2
+= xf
x
g
1
)( =e.g. 1 Given that and findx x
=





f 2
2
+





x
1

22. Functions
Solution:
( ) == fxfg )((i)
)()()()( xggxffxgfxfg (iv)(iii)(ii)(i)
2)( 2
+= xf
x
g
1
)( =e.g. 1 Given that and findx x
=





f
x
1)(xg 2
2
+





x
1
( ) == gxgf )((ii) )(xf
2
1
2
+=
x

23. Functions
)(xf
Solution:
( ) == fxfg )((i)
)()()()( xggxffxgfxfg (iv)(iii)(ii)(i)
2)( 2
+= xf
x
g
1
)( =e.g. 1 Given that and findx x
=





f
x
1)(xg 2
2
+





x
1
( ) == gxgf )((ii) )(g 22
+x
2
1
2
+=
x

24. Functions
22
+x)(xf
Solution:
)()()()( xggxffxgfxfg (iv)(iii)(ii)(i)
2)( 2
+= xf
x
g
1
)( =e.g. 1 Given that and findx x
=





f
x
1)(xg 2
2
+





x
1
( ) == gxgf )((ii) )(g
( ))(xgfN.B. is not the same as ( ))(xfg
( ) == fxfg )((i)
2
1
2
+
=
x
2
1
2
+=
x

25. Functions
2
1
2
+=
x
2
1
2
+
=
x
Solution:
)()()()( xggxffxgfxfg (iv)(iii)(ii)(i)
2)( 2
+= xf
x
g
1
)( =e.g. 1 Given that and findx x
=





f
x
1)(xg 2
2
+





x
1
( ) == gxgf )((ii) )(xf )(g 22
+x
( ))(xgfN.B. is not the same as ( ))(xfg
( ) == fxfg )((i)

26. Functions
2
1
2
+
=
x
Solution:
( ) == fxfg )((i)
)()()()( xggxffxgfxfg (iv)(iii)(ii)(i)
2)( 2
+= xf
x
g
1
)( =e.g. 1 Given that and findx x
=





f
x
1)(xg 2
2
+





x
1
( ) == )()( xffxff(iii)
2
1
2
+=
x
( ) == gxgf )((ii) )(xf )(g 22
+x

27. Functions
22
+x
( ) == gxgf )((ii) )(xf )(g 22
+x
Solution:
( ) == fxfg )((i)
)()()()( xggxffxgfxfg (iv)(iii)(ii)(i)
2)( 2
+= xf
x
g
1
)( =e.g. 1 Given that and findx x
=





f
x
1)(xg 2
2
+





x
1
( ) === )()()( fxffxff(iii)
2
1
2
+=
x
2
1
2
+
=
x

28. Functions
64 24
++= xx
( ) == gxgf )((ii) )(xf )(g 22
+x
Solution:
( ) == fxfg )((i)
)()()()( xggxffxgfxfg (iv)(iii)(ii)(i)
2)( 2
+= xf
x
g
1
)( =e.g. 1 Given that and findx x
=





f
x
1)(xg 2
2
+





x
1
( ) === )()()( fxffxff(iii) 22
+x 2)( 2
+22
+x
2
1
2
+=
x
2
1
2
+
=
x

29. Functions
( ) == gxgf )((ii) )(xf )(g 22
+x
Solution:
( ) == fxfg )((i)
)()()()( xggxffxgfxfg (iv)(iii)(ii)(i)
2)( 2
+= xf
x
g
1
)( =e.g. 1 Given that and findx x
=





f
x
1)(xg 2
2
+





x
1
( ) === )()()( fxffxff(iii) 22
+x 2)( 2
+22
+x
2
1
2
+=
x
( ) == )()( xggxgg(iv)
64 24
++= xx
2
1
2
+
=
x

30. Functions
1
=
x
1
x
1
( ) == gxgf )((ii) )(xf )(g 22
+x
Solution:
( ) == fxfg )((i)
)()()()( xggxffxgfxfg (iv)(iii)(ii)(i)
2)( 2
+= xf
x
g
1
)( =e.g. 1 Given that and findx x
=





f
x
1)(xg 2
2
+





x
1
( ) === )()()( fxffxff(iii) 22
+x 2)( 2
+22
+x
2
1
2
+=
x
( ) == )()( xggxgg(iv)






g
x
g
1
)( =e.g. 1 Given that and findx
x=
64 24
++= xx
2
1
2
+
=
x

31. Functions
( ) == gxgf )((ii) )(xf )(g 22
+x
Solution:
( ) == fxfg )((i)
)()()()( xggxffxgfxfg (iv)(iii)(ii)(i)
2)( 2
+= xf
x
g
1
)( =e.g. 1 Given that and findx x
=





f
x
1)(xg 2
2
+





x
1
( ) === )()()( fxffxff(iii) 22
+x 2)( 2
+22
+x
2
1
2
+=
x
( ) == )()( xggxgg(iv)






g
x
g
1
)( =e.g. 1 Given that and find
x
1 1
=
x
1
x=
64 24
++= xx
2
1
2
+
=
x

32. Functions
SUMMARY
• A compound function is a function of a function.
• It can be written as which means)(xfg ( ).)(xgf
• is not usually the same as( ))(xgf ( ).)(xfg
• The inner function is .)(xg
• is read as “f of g of x”.( ))(xgf

33. Functions
Exercise
,1)( 2
−= xxf
1. The functions f and g are defined as follows:
(a) The range of f is
Solution:
∈x 
0≠x,
1
)(
x
xg =
(a) What is the range of f ?
(b) Find (i) and (ii))(xfg )(xgf
1−≥y
=





x
f
1
1
1
2
−





x
( ) =− 12
xg
1
1
2
−x
(b) (i) ( ) =)(xgf=)(xfg
(ii) ( ) =)(xfg=)(xgf
1
1
2
−=
x

34. Functions
Periodic Functions
Functions whose graphs have sections which repeat
are called periodic functions.
e.g.
xy cos=
This has a
period of 3.
repeats
every radians.
xcos
π2
It has a
period of π2

35. Functions
If you are studying the OCR/MEI spec you need to
know the work on the following 3 slides.
Everyone else can skip over it by clicking here:
Skip slides

36. Functions
Some functions are even
Even functions are
symmetrical about
the y - axis
e.g. 2
)( xxf =
xxf cos)( =
So, )()( xfxf −=
e.g.
)2()2( −= ff
)()( ππ −= ff
e.g.

37. Functions
Others are odd
Odd functions have 180°
rotational symmetry
about the origin
e.g.
3
)( xxf =
xxf sin)( =
)()( xfxf −−=
e.g.
)2()2( −−= ff
e.g.
( ) ( )22
ππ −−= ff

38. Functions
Many functions are neither even nor odd
e.g.
xxxf 2)( 2
−=
Try to sketch one even function, one odd and one
that is neither. Ask your partner to check.

39. Functions