Question

If two functions are given as \[f\left( x \right)=\sqrt{x}\] and \[g\left( x \right)=2x-3\], then domain of \[\left( fog \right)\left( x \right)\] is
(a) \[\left( -\infty ,-3 \right)\]
(b) \[\left( -\infty ,\dfrac{-3}{2} \right)\]
(c) \[\left[ \dfrac{-3}{2},0 \right]\]
(d) \[\left[ 0,\dfrac{3}{2} \right]\]
(e) \[\left[ \dfrac{3}{2},\left. \infty \right) \right.\]

Answer 1

Hint: Compose the given two functions \[f\left( x \right)\] and \[g\left( x \right)\] to write the function \[\left( fog \right)\left( x \right)\], by replacing x by \[g\left( x \right)\] in the function \[f\left( x \right)\]. Observe that the domain of \[g\left( x \right)\] is \[\mathbb{R}\], while \[f\left( x \right)\] is \[\left[ 0,\infty \right)\]. Use this to find the domain of \[\left( fog \right)\left( x \right)\].

Complete step-by-step solution -
We have two functions \[f\left( x \right)=\sqrt{x}\] and \[g\left( x \right)=2x-3\]. We have to find the domain of \[\left( fog \right)\left( x \right)\].
We will firstly compose the functions \[f\left( x \right)\] and \[g\left( x \right)\].
We know that \[f\left( x \right)=\sqrt{x}\]. To find the function \[\left( fog \right)\left( x \right)\], we will replace x by \[g\left( x \right)\] in the function \[f\left( x \right)\].
Thus, we have \[\left( fog \right)\left( x \right)=f\left( g\left( x \right) \right)=\sqrt{2x-3}\].
We will now find the domain of the function \[\left( fog \right)\left( x \right)\]. We know that the domain of a function is the set of all possible values of a function at which the function doesn’t blow up.
We observe that domain of \[f\left( x \right)=\sqrt{x}\] is \[\left[ 0,\infty \right)\] and \[g\left( x \right)=2x-3\] is \[\mathbb{R}\].
So, to find the domain of \[\left( fog \right)\left( x \right)=f\left( g\left( x \right) \right)\], we must have \[g\left( x \right)\ge 0\].
Thus, we have \[g\left( x \right)=2x-3\ge 0\].
Simplifying the above equation, we have \[x\ge \dfrac{3}{2}\].
Hence, the domain of the function \[\left( fog \right)\left( x \right)=f\left( g\left( x \right) \right)=\sqrt{2x-3}\] is \[x\ge \dfrac{3}{2}\], i.e., \[x\in \left[ \dfrac{3}{2},\infty \right)\], which is option (e).
One must clearly know the difference between the terms – domain, range and codomain. Domain of a function is the set of all possible input values for the function. Co domain is the set into which all of the output of the function is constrained to fall. Range of the function is the set of all possible values attained by the function.

Note: We observe that the domain of the function \[\left( fog \right)\left( x \right)\] is not the same as the domain of \[f\left( x \right)\] or \[g\left( x \right)\]. Neither it is the same as the domain of \[f\left( x \right)g\left( x \right)\]. Thus, it’s not necessary that the domain of \[\left( fog \right)\left( x \right)\] will be the same as the domain of \[f\left( x \right)\] or \[g\left( x \right)\].