 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. $\left\{ {0, - 1, - 8, - 27} \right\}$

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.