Function Solved Examples

The arrow diagram represents a relation from set A to set B.

(i) Represent the relation in roster form.

(ii) Is this relation a function? Give reason for your answer.

(i) {(1, 2), (2, 4), (5, 2), (3, 10), (4, 10)}

(ii) Yes it is a function. Every first element is mapped and it is many-one. The function is many-one into.

Let f be defined by f(x) find

(i) f(2) (ii) (iii) (iv) x when .

(i) [f(2) means find the value of y when x = 2].

(ii)

(iii)

(iv)

2x + 1 = 3x

x = 1

Ans: (i) (ii) (iii) (iv) x = 1.

Which of the following relations are functions:

(i)

(ii) { for all values of y}

(iii) {(x, y) : y = 4 for all values of x}

(iv) {(x, y) : y = 5x - 6 }.

(i) Not a function: For one value of x there are many y values and for one value of y there are many x values (many-many).

(ii) Not a function: For one value of x there are many values of y (one-many).

(iii) Function: Many values of x equal to one value of y (many - one).

(iv) Function: For one value of x there is one value of y (one-one).

Given A = {0, 1, 2}, B = {5, 1, 6}, {(0, 5), (0, 1), (1, 6)} is not a function from A to B. State two reasons for the above statement.

Reasons:

(i) (0, 5), (0, 1) pairs make it one - many, one - many is not a function.

(ii) Pre-image 2 does not have an image in B.

It becomes easy to discuss a function with the help of an arrow diagram.

Given A = {-2, -1, 0, 1, 2} and B = {-3, -1, 1, 5}. list the elements of S = {(x, y): y = 2x² - 3, x Î A and y Î B}. Is S a function? If so classify it.

y = 2x² - 3 f(x) = 2x² - 3

f(-2) = 2(-2)² - 3 = 5 f(1) = 2(1)² - 3 = -1

f(-1) = 2(-1)² - 3 = -1 f(2) = 2(2)² - 3 = 5

f(0) = 2(0)² - 3 = -3

The roster form is S = {(-2, 5), (-1, -1), (0, -3), (1, -1), (2, 5)}.

Yes, S is a function. It is many-one and into function.

Domain = {-2, -1, 0, 1, 2} = A ;

Range = {-3, -1, 5}; Co-Domain = {-3, -1, 1, 5} = B.

(i) Find the domain and the range of the function.

(ii) Draw the Cartesian graph of f.

(iii) State whether f is one-one or many-one function.

(i) Domain = {-3 < x < 2, x Î R}

To find the range, f(-3) = -(-3) - 2 = 1 (-3, 1)

f(-2) = -(-2) - 2 = 0 (-2, 0)

f(-1) = -(-1) - 2 = -1 (-1, -1)

f(1) = +(1) - 2 = -1 (1, -1)

f(2) = 2 - 2 = 0 (2, 0)

Range = {-2 < y < 1, y Î R}

(ii)

(iii) It is many-one function