Question

If $f\left(x\right)=\sqrt{1-x}\mathrm{and}g\left(x\right)={\log }_{e}x$ are two real functions, then describe functions fog and gof.

Answer 1

$$\begin{gathered} f\left(x\right)=\sqrt{1-x} \\ \text{For domain, 1-x≥0} \\ \Rightarrow x\le 1 \\ \Rightarrow \text{domain of}f\text{=(}-\infty ,1] \\ \Rightarrow f:(-\infty ,1]\to \left(0,\infty \right) \\ g\left(x\right)={\log }_{e}x \\ \text{Clearly,}g:\left(0,\infty \right)\to R \\ \text{Computation of}fog: \\ \text{Clearly, the range of}g\text{is not a subset of the domain of}f\text{.} \\ \text{So},\text{we need to compute the domain of}fog. \\ \Rightarrow \text{Domain}\left(fog\right)\text{=}\left\{x:x\in \text{Domain}\left(g\right)\text{and}g\left(x\right)\in \text{Domain of f}\right\} \\ \Rightarrow \text{Domain}\left(fog\right)\text{=}\left\{x:x\in \left(0,\infty \right){\text{and log}}_e\text{x ∈}(-\infty ,1]\right\} \\ \Rightarrow \text{Domain}\left(fog\right)\text{=}\left\{x:x\in \left(0,\infty \right)\mathrm{and}x\in (0,e]\right\} \\ \Rightarrow \text{Domain}\left(fog\right)\text{=}\left\{x:x\in (0,e]\right\} \\ \Rightarrow \text{Domain}\left(fog\right)\text{=(0, e]} \\ \Rightarrow fog:\left(0,e\right)\to R \\ \mathrm{So},\left(fog\right)\left(x\right)=f\left(g\left(x\right)\right) \\ =f\left({\log }_{e}x\right) \\ =\sqrt{1-{\log }_{e}x} \\ \text{Computation of}gof: \\ \text{Clearly, the range of}f\text{is a subset of the domain of}g\text{.} \\ \Rightarrow gof:(-\infty ,1]\to R \\ \Rightarrow \left(gof\right)\left(x\right)=g\left(f\left(x\right)\right) \\ =g\left(\sqrt{1-x}\right) \\ ={\log }_e\sqrt{1-x} \\ ={\log }_e{\left(1-x\right)}^{\frac{1}{2}} \\ =\frac{1}{2}{\log }_e\left(1-x\right) \end{gathered}$$