\[\begin{gathered}

{\text{Let us understand this with an example:}} \hfill \\

f(x) = \frac{1}{{1 + x}} \hfill \\

g(x) = \ln x \hfill \\

fog(x) = f[g(x)] = f[\ln x] = \frac{1}{{1 + (\ln x)}} \hfill \\

({\text{Put }}g(x){\text{ in place of }}x{\text{ in }}f(x){\text{, you will get }}fog(x)) \hfill \\

{\text{To find domain, you need to calculate the set of all values for which the function }} \hfill \\

{\text{is defined}}{\text{.}} \hfill \\

{\text{But in exam, just try to find out the value of x at which the function is not defined}}{\text{.}} \hfill \\

fog(x) = \frac{1}{{1 + (\ln x)}} \hfill \\

{\text{Here, }}fog(x){\text{ is not defined when denominator }} = 0{\text{, and when }}\ln x{\text{ is not defined}} \hfill \\

\ln x{\text{ is not defined for negative value of x, hence }}x{\text{ }} > 0 \hfill \\

{\text{denominator of }}fog(x){\text{ can not be zero, hence x}} \ne \frac{1}{e} \hfill \\

{\text{hence domain = all positive value except }}\frac{1}{e} \hfill \\

{\text{Similarly you can calculate the range, it is set of all possible value of outcome for }} \hfill \\

{\text{the domain as input to function}}{\text{.}} \hfill \\

\hfill \\

{\text{Tip for Exam: Use online calculator to eliminate wrong options}}{\text{.}} \hfill \\

\end{gathered} \]