1 Functions. Function is a rule but not all rules are functions.

Function is a rule but not all rules are functions. This slide will help you have a deep understanding of functions.

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

2. Functions
Module C3

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.
 14
2
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
2
This 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 is 14
2
 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
14
2
 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) xysin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,

)(f3
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
)(f1
then,

)(f3
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
96
2
 xx
 f
2
)(3
Suppose and
2
)( xf )(xg
)(f1
then,

)(f3
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 find

x 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 find

x 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 find

x 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 find

x 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 find

x x









f
x
1
)(xg 2
2









x
1
gxgf)((ii) )(g 2
2
x
2
1
2

x

24. Functions
2
2
x)(xf
Solution:
)()()()( xggxffxgfxfg (iv)(iii)(ii)(i)
2)(
2
xf
x
g
1
)(e.g. 1 Given that and find

x 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 find

x x









f
x
1
)(xg 2
2









x
1
gxgf)((ii) )(xf )(g 2
2
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 find

x x









f
x
1
)(xg 2
2









x
1
 )()( xffxff(iii)
2
1
2

x
gxgf)((ii) )(xf )(g 2
2
x

27. Functions
2
2
x
gxgf)((ii) )(xf )(g 2
2
x
Solution:
fxfg)((i)
)()()()( xggxffxgfxfg (iv)(iii)(ii)(i)
2)(
2
xf
x
g
1
)(e.g. 1 Given that and find

x 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 2
2
x
Solution:
fxfg)((i)
)()()()( xggxffxgfxfg (iv)(iii)(ii)(i)
2)(
2
xf
x
g
1
)(e.g. 1 Given that and find

x x









f
x
1
)(xg 2
2









x
1
  )()()( fxffxff(iii) 2
2
x 2)(
2
2
2
x
2
1
2

x
2
1
2


x

29. Functions
gxgf)((ii) )(xf )(g 2
2
x
Solution:
fxfg)((i)
)()()()( xggxffxgfxfg (iv)(iii)(ii)(i)
2)(
2
xf
x
g
1
)(e.g. 1 Given that and find

x x









f
x
1
)(xg 2
2









x
1
  )()()( fxffxff(iii) 2
2
x 2)(
2
2
2
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 2
2
x
Solution:
fxfg)((i)
)()()()( xggxffxgfxfg (iv)(iii)(ii)(i)
2)(
2
xf
x
g
1
)(e.g. 1 Given that and find

x x









f
x
1
)(xg 2
2









x
1
  )()()( fxffxff(iii) 2
2
x 2)(
2
2
2
x
2
1
2

x
 )()( xggxgg(iv)








g
x
g
1
)(e.g. 1 Given that and find

x
x
64
24
 xx
2
1
2


x

31. Functions
gxgf)((ii) )(xf )(g 2
2
x
Solution:
fxfg)((i)
)()()()( xggxffxgfxfg (iv)(iii)(ii)(i)
2)(
2
xf
x
g
1
)(e.g. 1 Given that and find

x x









f
x
1
)(xg 2
2









x
1
  )()()( fxffxff(iii) 2
2
x 2)(
2
2
2
x
2
1
2

x
 )()( xggxgg(iv)








g
x
g
1
)(e.g. 1 Given that and find

x
11

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
 1
2
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.
xycos
This has a
period of 3.
repeats
every radians.
xcos
2
It has a
period of 2

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:
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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
)()(  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

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

40. Functions
The following slides contain repeats of
information on earlier slides, shown without
colour, so that they can be printed and
photocopied.
For most purposes the slides can be printed
as “Handouts” with up to 6 slides per sheet.

41. 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
The set of values of x for which the function is
defined is called the domain.
is called the image of x)(xf

42. Functions
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
x
The range of a function is the set of values given by
the rule.
So, the range of is
2
)(xxf 0)(xf

43. Functions
Solution: The quickest way to sketch this quadratic
function is to find its vertex by completing the square.
 14
2
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
2
This is a translation from of
2
xy
)5,2(so the vertex is .

44. Functions
5y
So, the graph of is 14
2
 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, . . .
So, we get ( x is any real number )
)5,2(x
14
2
 xxy
so the range is
Domain:
x

45. Functions
SUMMARY
)(xfy
•For ,

the x-values form the domain
2
)(xxf
2
: xxf 
•Notation:
means
e.g. For ,
the domain is
the range is or
2
)(xxf
0y0)(xf
•To define a function we need
a rule and a set of values.
x

46. Functions
and
f )(xg
2
)( 3x
96
2
 xx
 f 3x
2
)(3
Suppose and
2
)( xf )(xg
)(f1
then,

)(f3
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
is “a function of a function” or compound
function.
f)(xg

47. Functions
x
g
1
)(
gxgf)((ii) )(xf )(g 2
2
x
Solution:
fxfg)((i)
)()()()( xggxffxgfxfg (iv)(iii)(ii)(i)
2)(
2
xfe.g. 1 Given that and find

x x









f
x
1
)(xg 2
2









x
1
  )()()( fxffxff(iii) 2
2
x 2)(
2
2
2
x
2
1
2

x
64
24
 xx
 )()( xggxgg(iv)








g
e.g. 1 Given that and find

x
x
11

x
1
x
2
1
2


x

48. 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