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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 Verified Verified
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 \[f:A \to A\].