Question

If $f:\left(-\frac{\pi }{2},\frac{\pi }{2}\right)\to R\mathrm{and}g:\left[-1,1\right]\to R$ be defined as
$f\left(x\right)=\tan x\mathrm{and}g\left(x\right)=\sqrt{1-x^2}$ respectively, describe fog and gof.

Answer 1

$$\begin{gathered} g\left(x\right)=\sqrt{1-x^2} \\ \Rightarrow x^2\ge 0,\forall x\in \left[-1,1\right] \\ \Rightarrow -x^2\le 0,\forall x\in \left[-1,1\right] \\ \Rightarrow 1-x^2\le 1,\forall x\in \left[-1,1\right] \\ \text{We know that 1-}{\mathit{\text{x}}}^2\text{≥0} \\ \text{⇒0≤1-}{\mathit{\text{x}}}^2\text{≤1} \\ \text{⇒Range of}g\left(x\right)=\left[0,1\right] \\ \mathrm{So},f\mathit{:}\left(\frac{\mathit{-}\pi }{\mathit{2}}\mathit{,}\mathit{}\frac{\pi }{\mathit{2}}\right)\mathit{\to }R\mathit{}\text{and}g\mathit{:}\left[\mathit{-}\mathit{1}\mathit{,}\mathit{}\mathit{1}\right]\mathit{\to }\mathit{}\left[\mathit{0}\mathit{,}\mathit{}\mathit{1}\right] \\ \text{Computation of}\mathit{\text{fog:}} \\ \text{Clearly, the range of}g\mathit{}\text{is a subset of the domain of}\mathit{\text{f.}} \\ \text{So,}\mathit{\text{fog:}}\left[-1,1\right]\to R \\ \left(fog\right)\mathit{}\left(x\right)=f\mathit{}\left(g\mathit{}\left(x\right)\right) \\ =f\mathit{}\left(\sqrt{1-{\mathrm{x}}^2}\right) \\ =\mathit{tan}\mathit{}\sqrt{1-{\mathrm{x}}^2} \\ \text{Computation of}\mathit{\text{gof:}} \\ \text{Clearly, the range of}\mathit{\text{f}}\text{is not a subset of the domain of}g\mathit{.} \\ \Rightarrow \mathrm{Domain}\left(gof\right)=\left\{x\in \mathrm{domain}\mathrm{of}f\mathit{}\mathrm{and}\mathrm{f}\left(\mathrm{x}\right)\in \mathrm{domain}\mathrm{of}g\right\} \\ \Rightarrow \mathrm{Domain}\left(\mathrm{g}\mathrm{o}\mathrm{f}\right)=\left\{x\in \left(\frac{\mathit{-}\mathrm{\pi }}{\mathit{2}}\mathit{,}\mathit{}\frac{\mathrm{\pi }}{\mathit{2}}\right)\mathrm{and}\tan x\mathit{}\mathit{\in }\left[\mathit{-}\mathit{1}\mathit{,}\mathit{1}\right]\right\} \\ \Rightarrow \mathrm{Domain}\left(\mathrm{g}\mathrm{o}\mathrm{f}\right)=\left\{\mathrm{x}\in \left(\frac{\mathit{-}\mathrm{\pi }}{\mathit{2}}\mathit{,}\mathit{}\frac{\mathrm{\pi }}{\mathit{2}}\right)\mathrm{and}\mathrm{x}\mathit{}\mathit{\in }\left(\frac{\mathit{-}\pi }{\mathit{4}}\mathit{,}\mathit{}\frac{\pi }{\mathit{4}}\right)\right\} \\ \Rightarrow \mathrm{Domain}\left(\mathrm{g}\mathrm{o}\mathrm{f}\right)=\left\{\mathrm{x}\mathit{\in }\left(\frac{\mathit{-}\mathrm{\pi }}{\mathit{4}}\mathit{,}\mathit{}\frac{\mathrm{\pi }}{\mathit{4}}\right)\right\} \\ \mathrm{Now},gof\mathit{:}\left(\frac{\mathit{-}\pi }{\mathit{4}}\mathit{,}\mathit{}\frac{\pi }{\mathit{4}}\right)\mathit{\to }R \\ \mathrm{So},\left(gof\right)\mathit{}\left(x\right)=g\mathit{}\left(f\mathit{}\left(x\right)\right) \\ \mathit{=}g\mathit{}\left(\tan \mathit{}x\right) \\ \mathit{=}\sqrt{1-{\tan }^{2}x}\mathit{} \end{gathered}$$