TG TUTORIALS

TG TUTORIALS
TARUN GEHLOT

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}