Question

If A\left\\{ { - 2, - 1,1,2} \right\\} and f = \left\\{ {\left( {x,\dfrac{1}{x}} \right):x \in A} \right\\}, write down the range of $f$. Is $f$a function from $A$ to $A$?

Answer 1

Hint: First of all, find the range of the given function by substituting the values of $A$ in the given function. The function $f$a function from $A$ to $A$ exists if both its domain and range are equal. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer:
Given A = \left\\{ { - 2, - 1,1,2} \right\\}
And f = \left\\{ {\left( {x,\dfrac{1}{x}} \right):x \in A} \right\\}
We know that function is a relation in which each element of the domain is paired with exactly one element of the range.
Now consider,

$$\Rightarrow f\left( { - 2} \right) = \dfrac{1}{{ - 2}} = \dfrac{{ - 1}}{2} \\\ \Rightarrow f\left( { - 1} \right) = \dfrac{1}{{ - 1}} = - 1 \\\ \Rightarrow f\left( 1 \right) = \dfrac{1}{1} = 1 \\\ \Rightarrow f\left( 2 \right) = \dfrac{1}{2} \\\$$

So, the range of the function $f$ is \left\\{ {\dfrac{{ - 1}}{2}, - 1,1,\dfrac{1}{2}} \right\\}.
If we consider the function $f$ from $A$ to $A$ i.e., $f:A \to A$ here it is clear that the domain and range of the function $f$ are the same.
But clearly from the above data we have different values for domain and range of the function $f$.
Hence the function $f$ from $A$ to $A$ i.e., $f:A \to A$ can`t exist.
Thus, $f$ from $A$ to $A$ is not a function.
Note: Function is a relation in which each element of the domain is paired with exactly one element of the range. Range of a function is also known as the co-domain of the function. The function $f$ from $A$ to $A$ can be related as $f:A \to A$.