Question

Let \[f\left( x \right) = {x^3}\] be a function with domain \[\left\{ {0,1,2,3} \right\}\]. The domain of \[{f^{ - 1}}\] is
A. \[\left\{ {3,2,1,0} \right\}\]
B. \[\left\{ {0, - 1, - 2, - 3} \right\}\]
C. \[\left\{ {0,1,8,27} \right\}\]
D. \[\left\{ {0, - 1, - 8, - 27} \right\}\]

Answer 1

Hint: First of all, find the range of \[f\]and write the function in the set builder form of range and domain. Then obtain range and domain in the set builder form of \[{f^{ - 1}}\], which will give us the required domain of \[{f^{ - 1}}\].

Complete Step-by-Step solution:
Given function is \[f\left( x \right) = {x^3}\]
Domain of the function \[f\left( x \right)\] is \[\left\{ {0,1,2,3} \right\}\]
Consider
\[
   \Rightarrow f\left( 0 \right) = {\left( 0 \right)^3} = 0 \\
   \Rightarrow f\left( 1 \right) = {\left( 1 \right)^3} = 1 \\
   \Rightarrow f\left( 2 \right) = {\left( 2 \right)^3} = 8 \\
   \Rightarrow f\left( 3 \right) = {\left( 3 \right)^3} = 27 \\
\]
Hence the range of the function \[f\left( x \right)\] is \[\left\{ {0,1,8,27} \right\}\]
So, \[f\] can be written as
\[f = \left\{ {\left( {0,0} \right),\left( {1,1} \right),\left( {2,8} \right),\left( {3,27} \right)} \right\}\]
Hence \[{f^{ - 1}}\] can be written as
\[{f^{ - 1}} = \left\{ {\left( {0,0} \right),\left( {1,1} \right),\left( {8,2} \right),\left( {27,3} \right)} \right\}\]
So, clearly the domain of \[{f^{ - 1}}\] is \[\left\{ {0,1,8,27} \right\}\]
Thus, the correct option is C. \[\left\{ {0,1,8,27} \right\}\]

Note: The domain of a function \[f\left( x \right)\] is the set of all values for which the function is defined, and the range of the function is the set of all values that \[f\] takes and the function is denoted by \[f:X \to Y\] where \[X\] is the domain and \[Y\] is the range of the function.