properties of functions a function, f, is defined as a rule which assigns each member of a set...
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Properties of FunctionsA function, f, is defined as a rule which assigns each member of a set ‘A’ uniquely to a member of a set ‘B’.
A function f assigns exactly one value y to each x. We write y = f (x) orf: x y (‘f maps x to y’).
y is referred to as the image of x in f.
Set ‘A’ is referred to as the Domain of the function and the set ‘B’ as the co-domain.
The subset of ‘B’ which is the set of all images of the function is called the range of the function.
It is possible for f (a) = f (b) and yet a ≠ b.
Often restrictions have to be placed on these sets to provide a suitable domain and range.
1. Consider the function ( ) .
What would be a suitable domain and range?
f x x
is an unsuitable domain since negative values have no image in .
{0} is a suitable domain
{0} is a suitable range
12. Consider the function ( ) .
1What would be a suitable domain and range?
f xx
{1} is a suitable domain
{0} is a suitable range
The modulus functionThe modulus or absolute value of x is denoted by | x | and is defined as
when 0
when 0
x xx
x x
Properties of |x| include:
21. x x
2. x y x y
3. x y x y
4. x a a x a
5. or x a a x x a
y
x
y x
Inverse functionsThe graph of and inverse function f -1 (x) is obtained by reflecting f (x) in the line y = x
f -1 (x) is obtained by making x the subject of the formula.
For example find the inverse function of f (x) = 2x +1
2 1y x
2 1x y
12
yx
1 1( )
2x
f x
(2 1) 12
x
22x
x
1 1Note that ( ( )) (2 1)f f x f x
12. Find ( ) when ( ) .1
xf x f x
x
1x
yx
( 1)y x x
yx y x
yx x y
( 1)x y y
1y
xy
1( )1
xf x
x
1Here ( ) ( )f x f x
Placing restrictions on the domain and range can often ensure that a function has an inverse.
2 13. ( ) 4 1. Find ( ) and state a suitable domain and range for the function.f x x f x
24 1y x 24 1x y
2 14
yx
12
yx
1 1( )
2x
f x When x < 1, f -1 (x) is undefined
Hence largest suitable domain is x ≥ 1 giving a range y ≥ 0.
Domain: : 0,x x x
The domain and range of the inverse function give us the range and domain respectively of the function.
Range: : 1,y y y
Inverse Trig Functionsy
x
2
2
–
–
– 2
– 2
1
1
2
2
– 1
– 1
– 2
– 2
siny x
Reflecting the curve in y = x.
We need the largest region over which the curve is always increasing (or always decreasing).
y
x
0.5
0.5
1
1
1.5
1.5
– 0.5
– 0.5
– 1
– 1
– 1.5
– 1.5
2
2
–2
–2
1siny x
y
x
2
2
–
–
– 2
– 2
1
1
2
2
– 1
– 1
– 2
– 2
siny x
Domain: : 1 1,x x x
Range: : ,2 2
y y y
Odd and Even Functions
Even Functions
( ) ( )f x f x
Odd Functions
( ) ( )f x f x
22
11. Consider ( ) and ( )f x x f x
x
2( )f x x2( ) ( )f x x
2x
( ) ( )f x f x
Hence this is an even function.
2
1( )f x
x
2
1( )
( )f x
x
2
1x
( ) ( )f x f x
Hence this is an even function.
4 23. Prove ( ) 8 3 is an even function.f x x x
4 2( ) 8 3f x x x
4 2( ) ( ) 8( ) 3f x x x
4 28 3x x
( ) ( )f x f x
Hence this is an even function.
34. Prove ( ) 2 is an odd function.f x x x 3( ) 2f x x x
3( ) ( ) 2( )f x x x 3 2x x
( ) ( )f x f x
Hence this is an odd function.
Page 100 Exercise 3 Questions 1(a), (e), (g), (h), 2(e), (c), (g), 3Page 102 Exercise 4 Questions 1 and 2Page 108 Exercise 8 Question 3.
Vertical asymptotes
( )( ) is a rational function if it can be expressed in the form
( )g x
f xh x
where ( ) and ( ) are real polynomial functions and ( ) is of
degree 1 or greater.
g x h x h x
When h (a) = 0, the function is not defined at a.
As , ( ) and the function is discontinuous at .x a f x a
A function is said to be continuous if lim ( ) ( )x a
f x f a
11. How does ( ) behave in the neighbourhood of 1?
1f x x
x
As 1, the denominator, 1 0 and ( ) .x x f x
As 1 from the left, 1 0, as the numerator 0, ( ) 0x x f x
(to the left of the line x = 1, the curve f (x) ∞-, negative infinity).
As 1 from the right, 1 0, as the numerator 0, ( ) 0x x f x
(to the right of the line x = 1, the curve f (x) ∞+, positive infinity).
The line x = 1, is called a vertical asymptote of the function. The function is said to approach the line x = 1 asymptotically.
2 32. Consider ( )
( 4)( 1)x
f xx x
Vertical asymptotes occur at x = -4 and x = -1.
As 4x (-4 from the negative direction)
( )( )( )
y
As 4x
( )( )( )
y
As 1x
( )( )( )
y
As 1x
( )( )( )
y
( )Remember for ( )
( )g x
f xh x
Divide if the deg ( ) deg ( ) (partial fractions)g x h x
Page 109 Exercise 9 Question 1.
TJ Exercise 1.
Non vertical asymptotes
(i) If the degree of the numerator < degree of the denominator, divide each term of the numerator and denominator by the highest power of x.
( ) ( )If the function ( ) can be expressed as ( ) , where is
( ) ( )
a proper rational function, then,
m x m xf x g x
n x n x
as , ( ) ,x n x ( )
0( )
m xn x
and ( ) ( ).f x g x
If g(x) is a linear function, it is known as either a horizontal asymptote or an oblique asymptote depending on the gradient of the line y = g(x).
The behaviour of the function as x ∞+ and as x ∞- can be considered for each particular case.
(ii) If the degree of the numerator = degree of the denominator or the degree of the numerator > degree of the denominator, divide the numerator by the denominator.
2
2 31. Find the non vertical asymptote for the function ( )
5 4x
f xx x
2
2 3( )
5 4x
f xx x
2 2
2
2 2 2
2 3
5 4
xx x
x xx x x
2
2
2 3
5 41
x x
x x
0As , ( ) 0
1x f x
0As , ( ) 0
1x f x
Hence equation of asymptote is y = 0.
2
2
2 3
( )5 4
1
x xf x
x x
Let us now consider how the curve approaches y = 0.
As , ( ) 01
x f x
As , ( ) 01
x f x
2
2
1. Find the non vertical asymptote and determine the behaviour of the curve
2 1for the function ( )
5 4x x
f xx x
The degree of the numerator = the degree of the denominator so we divide.
2 25 4 2 1x x x x 1
2 5 4x x 3 3x
2
2 2
2 1 3 3( ) 1
5 4 5 4x x x
f xx x x x
2
2
3 3
15 4
1
x x
x x
0As , ( ) 1 1
1x f x Hence y = 1 is a horizontal asymptote.
As , ( ) 1 11
x f x
As , ( ) 1 11
x f x
1y
1y
2
1. Find the non vertical asymptote and determine the behaviour of the curve
4 3for the function ( )
2x x
f xx
The degree of the numerator > the degree of the denominator so we divide.
22 4 3x x x x
2 2x x2 3x
1( ) 2
2f x x
x
1
22
1
xx
x
0As , ( ) 2 2
1x f x x x Hence y = x + 2 is an oblique asymptote.
As , ( ) 2 21
x f x x x
As , ( ) 2 21
x f x x x
( 2)y x
( 2)y x
+2
2 4x -1
Curve SketchingBringing it all together
When sketching curves, gather as much information as possible form the list below.
1. Intercepts.
(a) Where does the curve cut the y axis? x = 0(b) Where does the curve cut the x axis? y = 0
2. Extras.
(a) Where is the function increasing? f `(x) > 0(b) Where is the function decreasing? f `(x) < 0
3. Stationary Points.
(a) Where is the curve stationary? f `(x) = 0(b) Are there points of inflexion? f ``(x) = 0
4. Asymptotes.
(a) Where are the vertical asymptotes? denominator = 0
(b) Where are the non vertical asymptotes? x ± ∞
22 11. Sketch the graph of
1x x
yx
(2 1)( 1)1
x xy
x
When 0, 1 (0,1)x y
1 1When y 0, and 1 1,0 , ,0
2 2x
Using the quotient rule.
2( ) 2 1 ( ) 1
`( ) 4 1 `( ) 1
f x x x g x x
f x x g x
2
2
(4 1)( 1) (2 1)( 1)
dy x x x xdx x
2
2 ( 2)( 1)x xx
0 for stationary points.dydx
2
2 ( 2)0
( 1)x xx
2 ( 2) 0x x
0 and 2x x
When 0, 1 (0,1)
When 2, 9 (2,9)
x y
x y
Using the quotient rule for the second derivative.
2 2( ) 2 4 ( ) ( 1)
`( ) 4 4 `( ) 2( 1)
f x x x g x x
f x x g x x
2 2 2
2 4
(4 4)( 1) 2( 1)(2 4 )( 1)
d y x x x x xdx x
``(0) 0 Max T.P. @ (0,1)f
``(2) 0 Min T.P. @ (2,9)f
Asymptotes
Vertical asymptote at x = 1.
As 1 , ( ) 0,x f x
(2 1)( 1)( )
1x x
f xx
( )f x
As 1 , ( ) 0,x f x
( )f x
Non vertical asymptotes
The degree of the numerator > the degree of the denominator so we divide.
21 2 1x x x 2x
22 2x x3 1x
+3
3 3x 2
2( ) 2 3
1f x x
x
2
2 31
1
xx
x
Non vertical asymptote at y = 2x + 3
2
2 31
1
xy x
x
As , ( ) 2 3 2 31
x f x x x
As , ( ) 2 3 2 31
x f x x x
(2 3) (above the line)y x
(2 3) (below the line)y x
Now lets put this information on a set of axis.
y
x
5
5
10
10
– 5
– 5
– 10
– 10
5
5
10
10
15
15
20
20
25
25
30
30
– 5
– 5
– 10
– 10
-1 ½ 1
3
(2,9)
(0,1)