# Properties of Functions - Dornoch Mathematics Properties of Functions A 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) or f: 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 f ( x ) x. What would be a suitable domain and range? is an unsuitable domain since negative values have no image in . {0} is a suitable domain {0} is a suitable range 1 2. Consider the function f ( x ) . x1 What would be a suitable domain and range? {1} is a suitable domain {0} is a suitable range

The modulus function The modulus or absolute value of x is denoted by | x | and is defined as x when x 0 x x when x 0 y y x Properties of |x| include: 1. x x 2 2. x y x y 3. x y x y 4. x a a x a 5. x a a x or x a x Inverse functions The 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 y 2 x 1 2x y 1 x y1 2 f 1( x ) x 1 2 Note that f 1(f ( x )) f 1(2 x 1) (2 x 1) 1 2 2x 2 x 2. Find f 1( x ) when f ( x )

x . x1 x y x1 y ( x 1) x yx y x yx x y x ( y 1) y y y1 x 1 f (x) x 1 x Here f ( x ) f 1( x ) Placing restrictions on the domain and range can often ensure that a function has an inverse. 3. f ( x ) 4 x 2 1. Find f 1( x ) and state a suitable domain and range for the function. y 4 x 2 1

4 x 2 y 1 y1 4 y1 x 2 x2 f 1( x ) x 1 2 When x < 1, f -1 (x) is undefined Hence largest suitable domain is x 1 giving a range y 0. The domain and range of the inverse function give us the range and domain respectively of the function. Domain: x : x 0, x Range: y : y 1, y Inverse Trig Functions

2 2 1 1 2 2 y 2 y sin x 1 2 2 x

1 2 Reflecting the curve in y = x. We need the largest region over which the curve is always increasing (or always decreasing). y 2 y sin x 1 2

2 x 1 2 y y sin 1 x 2 Domain: x : 1 x 1, x 1.5 1 0.5 0.5

2 1 1.5 x Range: y : y , y 2 2 Odd and Even Functions Even Functions Odd Functions f ( x ) f ( x )

f ( x ) f ( x ) 1. Consider f ( x ) x 2 and f ( x ) f ( x ) x 2 f ( x ) ( x )2 x 2 f ( x ) f ( x ) Hence this is an even function. 1 x2 1 x2 1 f ( x ) ( x )2 f (x) 1 x2

f ( x ) f ( x ) Hence this is an even function. 2. Consider f ( x ) x 3 f ( x ) x 3 f ( x ) ( x )3 x 3 f ( x ) f ( x ) Hence this is an odd function. 3. Prove f ( x ) x 4 8 x 2 3 is an even function. f ( x ) x 4 8x 2 3 f ( x ) ( x )4 8( x )2 3 x 4 8x 2 3 f ( x ) f ( x ) Hence this is an even function. 4. Prove f ( x ) x 3 2 x is an odd function. f ( x ) x 3 2x f ( x ) ( x )3 2( x ) x 3 2 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), 3

Page 102 Exercise 4 Questions 1 and 2 Page 108 Exercise 8 Question 3. Vertical asymptotes g(x ) f ( x ) is a rational function if it can be expressed in the form h( x ) where g ( x ) and h( x ) are real polynomial functions and h( x ) is of degree 1 or greater. When h (a) = 0, the function is not defined at a. As x a, f ( x ) and the function is discontinuous at a. A function is said to be continuous if lim f ( x ) f (a ) x a 1. How does f ( x ) 1 behave in the neighbourhood of x 1? x 1 As x 1, the denominator, x 1 0 and f ( x ) . As x 1 from the left, x 1 0, as the numerator 0, f ( x ) 0 (to the left of the line x = 1, the curve f (x) -, negative infinity). As x 1 from the right, x 1 0, as the numerator 0, f ( x ) 0

(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. y 3 2 1 y f ( x ) 1 2 3 x 2. Consider f ( x ) 2x 3 ( x 4)( x 1)

Vertical asymptotes occur at x = -4 and x = -1. As x 4 y (-4 from the negative direction) ( ) ( )( ) As x 1 y ( ) ( )( ) As x 4 As x 1 ( ) y ( )( ) ( )

y ( )( ) Remember for f ( x ) g( x ) h( x ) Divide if the deg g ( x ) deg h( x ) (partial fractions) Page 109 Exercise 9 Question 1. TJ Exercise 1. Non vertical asymptotes m( x ) m( x ) If the function f ( x ) can be expressed as g( x ) , where is n( x ) n( x ) a proper rational function, then, m( x ) 0 and f ( x ) g ( x ). as x , n( x ) , n( 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. (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. (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. 1. Find the non vertical asymptote for the function f ( x ) f (x) 2x 3 x 2 5x 4 2x 3 2 2 x 2x x 5x 4 2 2 2

x x x 2 3 2 x x 5 4 1 2 x x 0 0 1 0 As x , f ( x ) 0 1 As x , f ( x ) Hence equation of asymptote is y = 0. 2x 3 x 2 5x 4 2 3 2 x

x f (x) 5 4 1 2 x x Let us now consider how the curve approaches y = 0. As x , f ( x ) 0 1 As x , f ( x ) 0 1 1. Find the non vertical asymptote and determine the behaviour of the curve x 2 2x 1 for the function f ( x ) 2 x 5x 4 The degree of the numerator = the degree of the denominator so we divide. 1

x 2 5 x 4 x 2 2x 1 x 2 5x 4 3x 3 x 2 2x 1 3x 3 f (x) 2 1 2 x 5x 4 x 5x 4 3 3 2 1 x x 5 4 1 2 x x As x , f ( x ) 1 0 1 1 As x , f ( x ) 1

1 1 y 1 As x , f ( x ) 1 1 1 y 1 Hence y = 1 is a horizontal asymptote. 1. Find the non vertical asymptote and determine the behaviour of the curve x 2 4x 3 for the function f ( x ) x 2 The degree of the numerator > the degree of the denominator so we divide. x +2 x 2 x 2 4x 3 f (x ) x 2

x 2 2x 2x 3 2x 4 x 2 1 x 2 1 x 1 -1 2 x As x , f ( x ) x 2 0 x 2 1 As x , f ( x ) x 2

x 2 1 y ( x 2) As x , f ( x ) x 2 x 2 1 y ( x 2) Hence y = x + 2 is an oblique asymptote. Page 110 Exercise 10 Questions 1(a), (b), (g), (f), (k), (l). TJ Exercise 2 Curve Sketching Bringing 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? (b) Where does the curve cut the x axis?

x=0 y=0 2. Extras. (a) Where is the function increasing? (b) Where is the function decreasing? f `(x) > 0 f `(x) < 0 3. Stationary Points. (a) Where is the curve stationary? (b) Are there points of inflexion? f `(x) = 0 f ``(x) = 0 4. Asymptotes. (a) Where are the vertical asymptotes? (b) Where are the non vertical asymptotes? denominator = 0 x 2x 2 x 1 1. Sketch the graph of y

x1 y (2 x 1)( x 1) x1 When x 0, y 1 (0,1) 1 When y 0, x and 1 2 1,0 , 1 ,0 2 Using the quotient rule. f ( x ) 2 x 2 x 1 f `( x ) 4 x 1 g( x ) x 1

g `( x ) 1 dy (4 x 1)( x 1) (2 x 2 x 1) dx ( x 1)2 2 x ( x 2) ( x 1)2 dy 0 for stationary points. dx 2 x ( x 2) 0 2 ( x 1) 2 x( x 2) 0 x 0 and x 2 When x 0, y 1 When x 2, y 9 Using the quotient rule for the second derivative. f ( x ) 2 x 2 4 x

f `( x ) 4 x 4 g ( x ) ( x 1)2 g `( x ) 2( x 1) d 2 y (4 x 4)( x 1)2 2( x 1)(2 x 2 4 x ) 2 dx ( x 1)4 f ``(0) 0 Max T.P. @ (0,1) f ``(2) 0 Min T.P. @ (2,9) (0,1) (2,9) Asymptotes f (x) Vertical asymptote at x = 1.

(2 x 1)( x 1) x 1 0, f ( x ) As x 1 , f ( x ) 0, f ( x ) As x 1 , f ( x ) Non vertical asymptotes The degree of the numerator > the degree of the denominator so we divide. 2x +3 x 1 2x 2 x 1 2x 2 2x 3x 1 3x 3 2 f ( x ) 2 x 3 2

x1 2 2 x 3 x 1 1 x Non vertical asymptote at y = 2x + 3 2 y 2 x 3 x 1 1 x As x , f ( x ) 2 x 3 2x 3 1 As x , f ( x ) 2 x 3 2 x 3 y (2 x 3) (below the line) 1

y (2 x 3) (above the line) Now lets put this information on a set of axis. y 30 25 20 15 10 (2,9) 5 3 (0,1) 10 5 -1 5 10

1 5 10 x You should also be able to do the following from higher. From f(x), you should be able to sketch f(x - a), f(x + a), a f(x), f(x) a, f `(x), f -1(x) etc. Page 112 Exercise 11 Questions 1(a), (c), (e), (g), (i), (k), (m) Page 114 Exercise 12A TJ Exercise 3, 4 and 5 Do the review on Page 117