Solved Examples

Example 1:

Given are two sets A {1, 2, -2, 3} and B = {1, 2, 3, 5}. Is the function f(x) = 2x - 1 defined from A to B?

Solution :

Out of all the ordered pairs, the ordered pairs which are related by the function f(x) = 2x - 1 are {(1, 1), (2, 3), (3, 5) But for (-2) in A, we do not have any value in B. So, this function does not exist from
A->B.

Example 2:

A function f is defined as f: N -> N (where N is natural number set) and f(x) = x+2. Is this function ONTO?

Solution :

        Since, N = {1, 2, 3, 4, .........} and A = B = N

        For : A->B

        When x = 1                f(x) = 3

        When x = 2                f(x) = 4

So f(x) never assume values 1 and 2. So, B have two elements which do not have any pre-image in A. So, it is not an ONTO function.

Example 3 :

Find the range and domain of the function f(x) = (2x+1)/(x-1) and also find its inverse.

Solution :

This function is not defined for x = 1. So, domain of the function is
R -{1}.

Now, for finding the range

Let,(2x+3)/(x-1) = y

        => 2x + 3 = yx - y

        => yx - 2x = y + 3

        => (y - 2)x = y + 3

        => x =(y+3)/(t-2)

So, y cannot assume value 2

Range of f(x) is R - {2}.

Inverse is y =(x+3/x-2) .

Example 4:

 Find domain and range of the function f(x) = (x²+2x+3)/(x²-3x+2)

Solution :

        This function can be written as : f(x) =(x²+2x+3)/(x-1)(x-2) .

        So, domain of f(x) is R - {1, 2}

        For range, let (x²+2x+3)/(x²-3x+2) = y

                => (1 - y)x² + (2x + 3y) x + 3 - 2y = 0

for x to be real, Discriminant of this equation must be > 0

        D > 0

       => (2 + 3y)² - 4(1 - y)(3 -2y) > 0

       => 4 + 9y² + 12y - 4(3 + 2y² - 5y) > 0

       => y² + 32y - 8 > 0

       => (y + 16)² - 264 > 0

       => y < - 16 - √264  or  y  > -  16 + √264.