TG TUTORIALS

TG TUTORIALS
TARUN GEHLOT

Monday, 21 April 2014

DIFFERENCE BETWEEN RELATIONS AND FUNCTIONS

There are different ways of looking at functions. We will consider a few.  But first, we need to discuss some terminology.
A "relation" is just a relationship between sets of information. Think of all the people in one of your classes, and think of their heights. The pairing of names and heights is a relation. In relations and functions, the pairs of names and heights are "ordered", which means one comes first and the other comes second. To put it another way, we could set up this pairing so that either you give me a name, and then I give you that person's height, or else you give me a height, and I give you the names of all the people who are that tall. The set of all the starting points is called "the domain" and the set of all the ending points is called "the range." The domain is what you start with; the range is what you end up with. The domain is the x's; the range is the y's. (I'll explain more on the subject of determining domains and ranges
A function is a "well-behaved" relation. Just as with members of your own family, some members of the family of pairing relationships are better behaved than other. (Warning: This means that, while all functions are relations, since they pair information, not all relations are functions. Functions are a sub-classification of relations.) When we say that a function is "a well-behaved relation", we mean that, given a starting point, we know exactly where to go; given an x, we get only and exactly one y.
Let's return to our relation of your classmates and their heights, and let's suppose that the domain is the set of everybody's heights. Let's suppose that there's a pizza-delivery guy waiting in the hallway. And all the delivery guy knows is that the pizza is for the student in your classroom who is five-foot-five. Now let the guy in. Who does he go to? What if nobody is five-foot-five? What if there are six people in the room that are five-five? Do they all have to pay? What if you are five-foot-five? And what if you're out of cash? And allergic to anchovies? Are you still on the hook? Ack! What a mess!
The relation "height indicates name" is not well-behaved. It is not a function. Given the relationship (x, y) = (five-foot-five person, name), there might be six different possibilities for = "name". For a relation to be a function, there must be only and exactly one that corresponds to a given x. Here are some pictures of this:
x=-3 goes to y=3, x=-2 goes to y=-6, x=-1 goes to y=0, x=0 goes to y=15, x=1 goes to y=-1
This is a function. You can tell by tracing from each x to each y. There is only one y for each x; there is only one arrow coming from each x.
all x-values go to y-value of -6
Ha! Bet I fooled some of you on this one! This is a function! There is only one arrow coming from each x; there is only one y for each x. It just so happens that it's always the same y for each x, but it is only that one y. So this is a function; it's just an extremely boring function!
x=-3 goes to y=-6, x=-2 goes to y=-1, x=-1 goes to y=0, x=0 goes to y=3, x=1 goes to y=15, but also x=1 goes to y=0
This one is not a function: there are two arrows coming from the number 1; the number 1 is associated with twodifferent range elements. So this is a relation, but it is not a function.
x=-3 goes to y=-6, x=-2 goes to y=-1, x=-1 goes to y=0, x=0 goes to y=3, x=1 goes to y=15, x=16 just sits there looking at you
Okay, this one's a trick question. Each element of the domain that has a pair in the range is nicely well-behaved. But what about that 16? It is in the domain, but it has no range element that corresponds to it! This won't work! So then this is not a function. Heck, it ain't even a relation!









"Vertical Line Test"
Looking at this function stuff graphically, what if we had the relation that consists of a set containing just two points: {(2, 3), (2, –2)}? We already know that this is not a function, since x = 2 goes to each ofy = 3 and y = –2.   Copyright © Elizabeth Stapel 1999-2011 All Rights Reserved
If we graph this relation, it looks like:points at (2, -2) and (2, +2)
Notice that you can draw a vertical line through the two points, like this:vertical line drawn through two points with x=2
This characteristic of non-functions was noticed by I-don't-know-who, and was codified in "The Vertical Line Test": Given the graph of a relation, if you can draw a vertical line that crosses the graph in more than one place, then the relation is not a function. Here are a couple examples:
generic upward-opening parabola-type graphThis graph shows a function, because there is no vertical line that will cross this graph twice.
ovalThis graph does not show a function, because any number of vertical lines will intersect this oval twice. For instance, the y-axis intersects (crosses) the line twice.



Determine the domain and range of the given function:
    y = (x^2 + x - 2) / (x^2 - x - 2)
The domain is all the values that x is allowed to take on. The only problem I have with this function is that I need to be careful not to divide by zero. So the only values that x can not take on are those which would cause division by zero. So I'll set the denominator equal to zero and solve; my domain will be everything else.

    • x2 – x – 2 = 0 
      (x
       – 2)(x + 1) = 0
       
      x = 2   or  x =
       –1
       
    Then the domain is "all x not equal to –1 or 2".

    The range is a bit trickier, which is why they may not ask for it. In general, though, they'll want you to graph the function and find the range from the picture. In this case:
      
    graph
    As I can see from my picture, the graph "covers" all y-values (that is, the graph will go as low as I like, and will also go as high as I like). Since the graph will eventually cover all possible values ofy, then the range is "all real numbers".



  • Determine the domain and range of the given function:
      y = -sqrt(-2x + 3)
    The domain is all values that x can take on. The only problem I have with this function is that I cannot have a negative inside the square root. So I'll set the insides greater-than-or-equal-to zero, and solve. The result will be my domain:
      2x + 3 > 0 
      2x > –3 
      2x
       < 3
       
      x < 3/2 = 1.5
    Then the domain is "all < 3/2".

      
    graph
    The graph starts at y = 0 and goes down from there. While the graph goes down very slowly, I know that, eventually, I can go as low as I like (by picking an x that is sufficiently big). Also, from my experience with graphing, I know that the graph will never start coming back up. Then the range is "< 0".
  • Determine the domain and range of the given function:
      y = –x4 + 4
    This is just a garden-variety polynomial. There are no denominators (so no division-by-zero problems) and no radicals (so no square-root-of-a-negative problems). There are no problems with a polynomial. There are no values that I can't plug in for x. When I have a polynomial, the answer is always that the domain is "all x".
      
    The range will vary from polynomial to polynomial, and they probably won't even ask, but when they do, I look at the picture:
      
    graph
    The graph goes only as high as y = 4, but it will go as low as I like. Then:
      The range is "all y < 4"







Wednesday, 19 March 2014

What is an inverse function?


The inverse function for fx), labeled f−1x) (which is read “ f inverse of x”), contains the same domain and range elements as the original function, fx). However, the sets are switched. In other words, the domain of fx) is the range of f −1x), and vice versa. In fact, for every ordered pair ( ab) belonging to fx), there is a corresponding ordered pair ( ba) that belongs to f −1x). For example, consider this function, g:


The inverse function is the set of all ordered pairs reversed:


Only one‐to‐one functions possess inverse functions. Because these functions have range elements that correspond to only one domain element each, there's no danger that their inverses will not be functions. The horizontal line test is a quick way to determine whether a graph is that of a one‐to‐one function. It works just like the vertical line test: If an arbitrary horizontal line can be drawn across the graph of fx) and it intersects f in more than one place, then f cannot be a one‐to‐one function.
Inverse functions have the unique property that, when composed with their original functions, both functions cancel out. Mathematically, this means that 

Graphs of inverse functions

Since functions and inverse functions contain the same numbers in their ordered pair, just in reverse order, their graphs will be reflections of one another across the line y = x, as shown in Figure 1.
 Figure 1 Inverse functions are symmetric about the line y = x.

Finding inverse functions

To find the inverse function for a one‐to‐one function, follow these steps:
1. Rewrite the function using y instead of fx).
2. Switch the x and y variables; leave everything else alone.
3. Solve the new equation for y.
4. Replace the y with f −1x).
5. Make sure that your resulting inverse function is one‐to‐one. If it isn't, restrict the domain to pass the horizontal line test.
Example 1: If , find f −1x).
Follow the five steps previously listed, beginning with rewriting fx) as y:


Note the restriction x ≥ 0 for f −1x). Without this restriction, f −1x) would not pass the horizontal line test. It obviously must be one‐to‐one, since it must possess an inverse of fx). You should use that portion of the graph because it is the reflection of fx) across the line y = x, unlike the portion on x < 0.

Equal Sets and Equivalent Sets

Consider two sets:
A = {−9, −3, 0, 5, 12}
B = {−2, 1, 2, 4, 7}
Did you notice any relation between the sets and B?
Let us see.
We have,
A = {−9, −3, 0, 5, 12}
B = {−2, 1, 2, 4, 7}
Therefore, we have n (A) = 5 and n (B) = 5
Observe that both the sets A and B have same number of elements. Therefore, in this case, we say that the sets A and B are equivalent sets and it can be defined as:
Two finite sets are called equivalent, if they have the same number of elements. 
Thus, two finite sets X and Y are equivalent, if n (X) = n (Y). We write it as  Y (read as “X is equivalent to Y”)
Now, consider the two sets:
X = {all letters in the word STONE}
Y = {all letters in the word NOTES}
Did you notice any relation between the sets and Y?
Let us see.
We have,
X = {S, T, O, N, E} and Y = {N, O, T, E, S}
Observe that both the sets X and Y have same elements. Therefore, in this case, we say that the sets X andY are equal sets.
Two sets are called equal, if they have same elements.
When two sets X and Y are equal, we denote it as X = Y; and if they are not equal, then we write it as X Y
Also, note that n (X) = 5 and n (Y) = 5
Therefore, we can conclude that:
If A and B are finite sets and A = B, then (A) = n (B) i.e., A and B are equivalent. However, the converse of the above statement may not be true.
For example, if A = {2, 4, 6} and B = {1, 3, 5}, then n (A) = n (B) = 3; however, A  B
Let us now look at some more examples to understand the above discussed concepts better.
Example 1:
Which of the following sets are equal?
(a) X = {xx is a letter in the word REFRESH},
Y = {A letter in the word FRESHER}
(b) X = {4}, Y = {xx∈ N, x − 4 = 0}
(d) X = {x/x is a vowel letter in the word WEIGHT},
Y = {x/x is a vowel letter in the word HEIGHT}
(c) X = {xx∈ N, 0 < x < 4}, Y = {xx∈ W}
Solution:
(a) X = {R, E, F, S, H}, Y = {F, R, E, S, H}
∴ X and Y are equal sets i.e., X = Y
(b) X = {4}, Y = {4}
∴ X and Y are equal sets i.e., X = Y
(c) X = {E, 1}, Y = {E, I}
∴ X and Y are equal sets i.e., X = Y
(d) X = {1, 2, 3}, Y = {0, 1, 2, 3, 4, 5 …}
∴ X and Y are not equal sets i.e., X  Y
Example 2:
Which of the following sets are equal?
(a) X = {xx is a vowel in the word MATRIX}
Y = {A vowel in the word SHIVANI}
(b) X = {5, 0, 1, 0, 2, 1}, Y = {3, 2, 8, 3, 3, 2}
Solution:
(a) X = {A, I}, Y = {A, I}
∴ n (X) = 2 and n (Y) = 2
This means n (X) = n (Y)
Therefore, X and Y are equivalent sets.
(b) X = {5, 0, 1, 0, 2, 1} = {0, 1, 2, 5}, Y = {3, 2, 8, 3, 3, 2} = {2, 3, 8}
∴ n (X) = 4 and n (Y) = 3
This means n (X) ≠ n (Y)
Therefore, X and Y are not equivalent sets.

Complement of a set

Let us consider a set as
X = {2, 3, 6, 8}
Let us consider its universal set ξ as
ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
Can we find the set of elements in set ξ, which are not in X?
We can find this by taking the difference of X from ξ as
ξ − X = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} − {2, 3, 6, 8} = {0, 1, 4, 5, 7, 9}
This set (consisting of all the elements ξ, which do not belong to X) is known as the complement of set Xand we denote it by or.
Let X be any set and ξ be its universal set. The complement of set X is the set consisting of all the elements of ξ, which do not belong to X. It is denoted by X′ or Xc (read as complement of set X).
Thus, X′ = {x|xξ and xX} or X′ = ξ−X
For the above sets X and ξ, we may observe that (X) = 4, (ξ) = 10, and n () = 6.
Can we find any relation among them?
We observe that:
n() = n (ξ) − n (X)
This relation holds true for a set, its complement, and a universal set.
The other properties of a set and its complement are as follows.
(a) 
(b) 
(c) 
(d) 
(e) 
(f) If X ⊆ Y then 
Apart from these properties, there are two more properties for two sets A and B. They are:
(a) 
(b) 
These are also known as De Morgan’s laws.
Let us prove the first one.
Let A = {1, 2, 3}, B = {2, 3, 4}, and ξ = {1, 2, 3, 4, 5, 6}
Now, A  B = {2, 3}
Therefore, (A  B)′ = {1, 4, 5, 6}
Also, A′ = {4, 5, 6} and B′ = {1, 5, 6}
∴ A   B′ = {1, 4, 5, 6}
Clearly, we have
Similarly, we can prove the second one.
Now, how will we represent the complement of a set A with the help of a Venn diagram?
We know that if A is a set and ξis a universal set for the set A, then the complement of the set A is Ac = ξ A.
If we represent the sets ξand A by a Venn diagram, then we can easily represent Ac on it.
For this, we represent the set A by using a circle and ξ by using a rectangle (or a square which is bigger and encloses the circle). Now, the portion outside the set A, but inside the set ξ, represents the set Ac. This can be shown as follows:
Let us look at some examples in order to understand these concepts better.
Example 1:
If A and B are two sets and ξ is their universal set such that , and
(B) = 6, then how many elements are there in the complement of set B?
Solution:
We know that,
We also know that,
Therefore, the complement of set B contains 2 elements.
Example 2:
If A = {x, 1, 2, 3, y}, B = {2, 4, 5, y}, and ξ= {xyz, 1, 2, 3, 4, 5, 6}, then show that 
Solution:
Now, Ac = ξ − A = {x, y, z, 1, 2, 3, 4, 5, 6} − {x, 1, 2, 3, y}= {z, 4, 5, 6}
Bξ − B = {x, yz, 1, 2, 3, 4, 5, 6} − {2, 4, 5, y} = {x, y, 1, 3, 6}
A  B = {x, y, 1, 2, 3, 4, 5}
(A  B)c = {z, 6}
Now, Ac  Bc = {z, 6}
Clearly, (A ∪ B)c = Ac  Bc
Example 3:
Find the following sets from the adjoining Venn-diagram.
(i) (A ∩ B)c
(ii) Ac
(iii) (A  C)c
(iv) ξ
Solution:
From the given Venn-diagram, we find that 
(i) (A ∩ B){1, 2, 3, 4, 6, 7, 9, 10, 11, 12, 13, 14, 15, 19}
(ii) A= {3, 4, 6, 7, 9, 11, 12, 13, 14, 15}
(iii) (A  C)= {3, 4, 7, 9, 11, 13}
(iv)  ξ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 19}
Example 4:
Taking the set of first ten natural numbers as the universal set, find the set
(B− A)′ ∩ B′, where A = {1, 2, 4, 9} and B = {2, 5, 7, 9, 8, 10, 1, 3}
Solution:
B − A = {5, 7, 8, 10, 3}
U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(B − A) ′ = {1, 2, 4, 6, 9}
B′ = {4, 6}
∴ (B− A)′ ∩ B′ = {4, 6}