THE BINOMIAL THEOREM4.1
Method 2: Vertical line Test
The method is quite simple:
Step 1 Sketch the graph of the relation.
Step 2 Make a visual check of the number of times a vertical line would cut the graph.
Step 3 If the vertical line only ever cuts at one place for every value in the domain the relation is a function.
(a) Clearly, we have every first element of the ordered pairs different. This means that this relation is also a function.
(b) Using the TI–83 to provide a visual check:
From the graph shown, a vertical line drawn anywhere on the domain for which the relation is defined, will cut the graph at only one place.
This relation is therefore a function.
(c) Again we make use of a visual approach to determine if the relation is a function.
First we write the relation in a form that will enable us to enter it into the TI–83:
We can therefore define the relation and sketch both on the same set of axes.
Placing a vertical line over sections of the domain shows that the line cuts the graph in two places (except at the origin). Therefore this relation is not a function.
Algebraic proof y
x
No matter where along the graph we draw a vertical line, the graph will only ever be cut once.
Therefore this relation is a function.
Only one cut (anywhere)
y
x
The vertical line has cut the graph at more than one point for a given value of x. Therefore this relation is a not function.
Vertical has cut the graph at three different points for the same x–value.
Function Not a function
Which of the following defines a function?
(a) (b)
(c) (d)
0 2
( , ) 1 2,( , ) 2 1,( , )
{ } {(x y, ): y x= 3+1 x, ∈ } y2 = x x, ≥0 {( , ): xx y 2+y2=16}
E
XAMPLE5.7S
o l u t i o nCuts graph twice
y2 = x⇒y = ± x
Y1 = X andY2 = – X
Begin by choosing a value of x that lies in the domain. For example x = 4.
This gives the following equation:
From which we can say that when x = 4, y = 2 and y = –2, so that there are two ordered pairs, (4,2) and (4, –2). As we have two different y-values for one x-value this relation is not a function.
(d) This relation describes the equation of a circle with radius 4 units and centre at the origin.
The graph of this relation is shown alongside. The graph fails the vertical line test, and so is not a function.
We now provide a formal definition together with commonly used notation for a funcion.
The set X is called the domain of f and the set Y the co-domain. The element y is called the image of x under f and we denote this image by , the value of the function f at x (read as; f of x).
We write this mapping as:
We can also write this mapping as follows 1.
2.
3.
It is important to realise that the range of f is not necessarily the set Y. The range of f is actually a subset of Y (sometimes it could also be equal toY). Set Y, i.e., the co-domain, describes the types of numbers that will be produced when f is applied to different x-values — not necessarily which numbers will result! The range of f is given by the values of .
Translating the mathematical notation into English we have the following:
Notice that f describes not only the rule, , but also the domain, X.
y2 = 4⇒y = ± 4
y
x
two cuts
4
–4
–4 4
A function f, (or a mapping f ), from a set X to a set Y is a relation that assigns to each element x in the set X a unique element y in the set Y.
X Y
x Range
y = f x( )
maps onto (rule)
y
Domain Co-domain
(image)
f x( )
f : x f x( )
f : X Y where y, = f x( ) f : X Y y, = f x( ) y = f x( ) x X, ∈
f x( )
f : X Y where f x, ( ) = rule in terms of x
“ f is such that the set X maps onto the set Y where f of x is equal to . . .”
f x( )
(a) . Similarly, .
(b) If the image is 28, then we want the value of x for which . We then have,
[Taking the cube root of both sides]
Therefore, the element of the domain that has an image of 28 is 3.
First note that the co-domain is given by , the set of real numbers – meaning that all image values will be real. To determine the actual range of this function we sketch its graph.
From the graph, the only possible values of y are those for which y ≥ 1. In this case, because x = 0 is included in the domain, we also include the value y = 1 in the range. Therefore, we have a closed circle at the end point.
The range of f is then given by (or [1, ∞[).
For the function find
(a) . (b) the element of the domain that has an image of 28.
For the function find
(a) (b) (c)
(a)
(b) To determine the range of the function g, we first need to sketch its graph.
We begin by sketching the graph of for all real values of x. Then, as the domain is restricted to
. This means that we ‘remove’ the parts of the graph that lie outside this domain. This leaves the required part of the graph.
From our graph of g, the range of this function is given by {y: 2 ≤ y ≤ 6} (or [2, 6]).
Therefore, the solution set is .
(b) From (a) we obtained two values of x for one value of y and so is a many–to –one function.
Consider the function , where ,
(a) Find . (b) Determine the range of g.
Consider the function .
(a) Find i. ii. {x : } iii. {x : }
1. A function is defined as follows, .
(a) Find the value of .
(b) Evaluate the expressions i. ii.
(c) Find .
2. If , find (a) .
(b) .
(c) the range of .
3. For the mapping , find (a) , .
(b) a, given that .
(c) b, given that .
4. A function is defined as follows, .
(a) Find the value(s) of x such that y = 0.
(b) Sketch the graph of and determine its range.
5. The function f is defined as .
(a) Sketch the graph of i. f ii.
(b) Find i. ii.
6. Which of the following relations are also functions?
7. Use both visual tests and algebraic tests to show that the following relations are also functions.
(a) (b)
E
XERCISES5.2
f : x 2 x+3 x, ≥0 f 0( ) f 1, ( )
f x( +a) f x( +a) f x– ( ) x: f x( ) 9=
{ }
f x( ) x x+1
--- x, ∈[0 10, ]
= f 0( ) f 10, ( )
x: f x( ) 5=
{ }
f x( ) x x+1
--- x, ∈[0 10, ]
=
x 2 1 2---x2,x∈
– f x( +1) f x 1( – )
f a( ) = 1 f b( ) = 10 y = x3–x2,x∈[–2 2, ]
y = x3–x2,x∈[–2 2, ]
f :]–∞ ∞[ ,where, f x( ) = x2–4 y = x+2 x, ∈]–∞ ∞[, x: f x( ) 4=
{ } {x: f x( ) x 2= + }
(a) (b) (c)
(d) (e) (f)
x x3+2 x, ∈]0 5[, x x+1 x, ∈[0 9, )
8. Use an algebraic method to decide which of the following relations are also functions.
(b) determine its range.
10. A function is defined by and x ≥ 0.
(a) Determine the range of f.
(b) Find the value of a such that .
11. Consider the functions and .
(a) Show that .
(b) If , find the constant a.
12. Which of the following functions are identical? Explain
(a) and . (b) and .
(c) and . (d) and .
13. Find the the largest possible subset X of , so that the following relations are one to one increasing functions
(a) , where
(b) , where
(c) , where
(d) , where ,
14. An isosceles triangle ABC has two side lengths measuring 4 cm and a variable altitude.
Let the altitude be denoted by x cm.
(a) Find, in terms of x, a relation for
i. its perimeter, cm and specify its implied domain.
ii. its area, and specify its implied domain.
(b) Sketch the graph of
i. and determine its range.
ii. and determine its range.
f : x 1