Question

What is the range when f is given as ordered pairs?

Answer 1

Here's the Solution to this Question

Given A $= \{a,b,c,d\}A={a,b,c,d} and B = \{c,d,e,f,g\}B={c,d,e,f,g} .$

$R_1 = \{(a,c), (b,d), (c,e)\}, R_2 = \{(a,c), (a,g), (b,d), (c,e), (d,f)\}, \\ R_3 = \{(a,c), (b,d), (c,e), (d,f)\}$

A relation is a function when every element of set A has image in B and a element of set A can-not have more than one image in set B.

So, Relation $R_3$ is a function.

(c) Given $f$ is a real valued function defined by $f(x) = x^2 - 9$.

(I) Function is defined for all real values of $x$. Hence,

Domain of $f=R$

(ii) Now, as $x^2 \geq 0$ $\implies$ $x^2-9 \geq -9x$

Hence, Range of f=[−9,∞)

(iii) Representation of $f$ as a set of ordered pair = $\{ (x,x^2-9) : x \in R\}$