Question

Find fog and gof if
(i) $f\left(x\right)=e^x,g\left(x\right)={\log }_{e}x$
(ii) $f\left(x\right)=x^2,g\left(x\right)=\cos x$
(iii) $f\left(x\right)=\left|x\right|,g\left(x\right)=\sin x$
(iv) $f\left(x\right)=x+1,g\left(x\right)=e^x$
(v) $f\left(x\right)={\sin }^{-1}x,g\left(x\right)=x^2$
(vi) $f\left(x\right)=x+1,g\left(x\right)=\sin x$
(vii) $f\left(x\right)=x+1,g\left(x\right)=2x+3$
(viii) $f\left(x\right)=c,c\in R,g\left(x\right)=\sin x^2$
(ix) $f\left(x\right)=x^2+2,g\left(x\right)=1-\frac{1}{1-x}$

Answer 1

$$\begin{gathered} \left(i\right)f\left(x\right)=e^x,g\left(x\right)={\log }_{e}x \\ f:R\to \left(0,\infty \right);g:\left(0,\infty \right)\to R \\ \text{Computing}fog\text{:} \\ \text{Clearly, the range of}g\text{is a subset of the domain of}f. \\ fog:\left(0,\infty \right)\to R \\ \left(fog\right)\left(x\right)=f\left(g\left(x\right)\right) \\ =f\left({\log }_{e}x\right) \\ ={\log }_e{e}^x \\ =x \\ \text{Computing}gof\text{:} \\ \text{Clearly, the range of}f\text{is a subset of the domain of}g. \\ \Rightarrow fog:R\to R \\ \left(gof\right)\left(x\right)=g\left(f\left(x\right)\right) \\ =g\left(e^x\right) \\ ={\log }_e{e}^x \\ =x \end{gathered}$$

$$\begin{gathered} \left(ii\right)f\left(x\right)=x^2,g\left(x\right)=\cos x \\ f:R\to [0,\infty );g:R\to \left[-1,1\right] \\ \text{Computing}\mathrm{fog}\text{:} \\ \text{Clearly, the range of}g\text{is not a subset of the domain of}f. \\ \Rightarrow \text{Domain}\left(fog\right)\text{=}\left\{x:x\in \text{domain of}g\text{and}g\left(x\right)\in \text{domain of}f\right\} \\ \Rightarrow \text{Domain}\left(fog\right)\text{=}x:x\in R\text{and}\cos x\in R\} \\ \Rightarrow \text{Domain of}\left(fog\right)\text{=}R \\ fog:R\to R \\ \left(fog\right)\left(x\right)=f\left(g\left(x\right)\right) \\ =f\left(\cos x\right) \\ ={\cos }^{2}x \\ \text{Computing}gof\text{:} \\ \text{Clearly, the range of}f\text{is a subset of the domain of}g. \\ \Rightarrow fog:R\to R \\ \left(gof\right)\left(x\right)=g\left(f\left(x\right)\right) \\ =g\left(x^2\right) \\ =\cos \left(x^2\right) \end{gathered}$$


$$\begin{gathered} \left(iii\right)f\left(x\right)=\left|x\right|,g\left(x\right)=\sin x \\ f:R\to \left(0,\infty \right);g:R\to \left[-1,1\right] \\ \text{Computing}fog\text{:} \\ \text{Clearly, the range of}g\text{is a subset of the domain of}f. \\ \Rightarrow fog:R\to R \\ \left(fog\right)\left(x\right)=f\left(g\left(x\right)\right) \\ =f\left(\sin x\right) \\ =\left|\sin x\right| \\ \text{Computing}gof\text{:} \\ \text{Clearly, the range of}f\text{is a subset of the domain of}g. \\ \Rightarrow fog:R\to R \\ \left(gof\right)\left(x\right)=g\left(f\left(x\right)\right) \\ =g\left(\left|x\right|\right) \\ =\sin \left|x\right| \end{gathered}$$

$$\begin{gathered} \left(iv\right)f\left(x\right)=x+1,g\left(x\right)=e^x \\ f:R\to R;g:R\to [1,\infty ) \\ \text{Computing}fog\text{:} \\ \text{Clearly, range of}g\text{is a subset of domain of}f. \\ \Rightarrow fog:R\to R \\ \left(fog\right)\left(x\right)=f\left(g\left(x\right)\right) \\ =f\left(e^x\right) \\ =e^x+1 \\ \text{Computing}gof\text{:} \\ \text{Clearly, range of}f\text{is a subset of domain of}g. \\ \Rightarrow fog:R\to R \\ \left(gof\right)\left(x\right)=g\left(f\left(x\right)\right) \\ =g\left(x+1\right) \\ =e^{x+1} \end{gathered}$$

$$\begin{gathered} \left(v\right)f\left(x\right)={\sin }^{-1}x,g\left(x\right)=x^2 \\ f:\left[-1,1\right]\to \left[\frac{-\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right];g:R\to [0,\infty ) \\ \mathbf{\text{Computing}}\mathit{f}\mathit{o}\mathit{g}\mathbf{\text{:}} \\ \text{Clearly, the range of}g\text{is not a subset of the domain of}f. \\ \text{Domain}\left(fog\right)\text{=}\left\{x:x\in \text{domain of}g\text{and}g\left(x\right)\in \text{domain of}f\right\} \\ \text{Domain}\left(fog\right)\text{=}\left\{x:x\in R\text{and}{x}^2\in \left[-1,1\right]\right\} \\ \text{Domain}\left(fog\right)\text{=}\left\{x:x\in R\text{and}x\in \left[-1,1\right]\right\} \\ \text{Domain of}\left(fog\right)\text{=}\left[-1,1\right] \\ fog:\left[-1,1\right]\to R \\ \left(fog\right)\left(x\right)=f\left(g\left(x\right)\right) \\ =f\left(x^2\right) \\ ={\sin }^{-1}\left(x^2\right) \\ \mathbf{\text{Computing}}\mathit{g}\mathit{o}\mathit{f}\mathbf{\text{:}} \\ \text{Clearly, the range of}f\text{is a subset of the domain of}g. \\ fog:\left[-1,1\right]\to R \\ \left(gof\right)\left(x\right)=g\left(f\left(x\right)\right) \\ =g\left({\sin }^{-1}x\right) \\ ={\left({\sin }^{-1}x\right)}^2 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{vi}\right)f\left(x\right)=x+1,g\left(x\right)=\sin x \\ f:R\to R;g:R\to \left[-1,1\right] \\ \text{Computing}fog\text{:} \\ \text{Clearly, the range of}g\text{is a subset of the domain of}f. \\ \Rightarrow fog:R\to R \\ \left(fog\right)\left(x\right)=f\left(g\left(x\right)\right) \\ =f\left(\sin x\right) \\ =\sin x+1 \\ \text{Computing}gof\text{:} \\ \text{Clearly, the range of}f\text{is a subset of the domain of}g. \\ \Rightarrow fog:R\to R \\ \left(gof\right)\left(x\right)=g\left(f\left(x\right)\right) \\ =g\left(x+1\right) \\ =\sin \left(x+1\right) \end{gathered}$$

$$\begin{gathered} \left(vii\right)f\left(x\right)=x+1,g\left(x\right)=2x+3 \\ f:R\to R;g:R\to R \\ \text{Computing}fog\text{:} \\ \text{Clearly, the range of}g\text{is a subset of the domain of}f. \\ \Rightarrow fog:R\to R \\ \left(fog\right)\left(x\right)=f\left(g\left(x\right)\right) \\ =f\left(2x+3\right) \\ =2x+3+1 \\ =2x+4 \\ \text{Computing}gof\text{:} \\ \text{Clearly, the range of}f\text{is a subset of the domain of}g. \\ \Rightarrow fog:R\to R \\ \left(gof\right)\left(x\right)=g\left(f\left(x\right)\right) \\ =g\left(x+1\right) \\ =2\left(x+1\right)+3 \\ =2x+5 \end{gathered}$$

$$\begin{gathered} \left(viii\right)f\left(x\right)=c,g\left(x\right)=\sin x^2 \\ f:R\to \left\{c\right\};g:R\to \left[0,1\right] \\ \text{Computing}fog\text{:} \\ \text{Clearly, the range of}g\text{is a subset of the domain of}f. \\ fog:R\to R \\ \left(fog\right)\left(x\right)=f\left(g\left(x\right)\right) \\ =f\left(\sin x^2\right) \\ =c \\ \text{Computing}gof\text{:} \\ \text{Clearly, the range of}f\text{is a subset of the domain of}g. \\ \Rightarrow fog:R\to R \\ \left(gof\right)\left(x\right)=g\left(f\left(x\right)\right) \\ =g\left(c\right) \\ =\sin c^2 \end{gathered}$$

$$\begin{gathered} \left(ix\right)f\left(x\right)=x^2+2 \\ f:R\to [2,\infty ) \\ g\left(x\right)=1-\frac{1}{1-x} \\ \text{For domain of}g:1-x\ne 0 \\ \Rightarrow x\ne 1 \\ \Rightarrow \text{Domain of g=}R-\left\{1\right\} \\ g\left(x\right)=1-\frac{1}{1-x}=\frac{1-x-1}{1-x}=\frac{-x}{1-x} \\ \text{For range of}g: \\ y=\frac{-x}{1-x} \\ \Rightarrow y-xy=-x \\ \Rightarrow y=xy-x \\ \Rightarrow y=x\left(y-1\right) \\ \Rightarrow x=\frac{y}{y-1} \\ \text{Range of g =}R-\left\{1\right\} \\ \text{So, g:}R-\left\{1\right\}\to R-\left\{1\right\} \\ \text{Computing}fog\text{:} \\ \text{Clearly, the range of}g\text{is a subset of the domain of}f. \\ \Rightarrow fog:R-\left\{1\right\}\to R \\ \left(fog\right)\left(x\right)=f\left(g\left(x\right)\right) \\ =f\left(\frac{-x}{x-1}\right) \\ ={\left(\frac{-x}{x-1}\right)}^2+2 \\ =\frac{x^2+2x^2+2-4x}{{\left(1-x\right)}^2} \\ =\frac{3x^2-4x+2}{{\left(1-x\right)}^2} \\ \text{Computing}gof\text{:} \\ \text{Clearly, the range of}f\text{is a subset of the domain of}g. \\ \Rightarrow gof:R\to R \\ \left(gof\right)\left(x\right)=g\left(f\left(x\right)\right) \\ =g\left(x^2+2\right) \\ =1-\frac{1}{1-\left(x^2+2\right)} \\ =1-\frac{1}{-\left(x^2+1\right)} \\ =\frac{x^2+2}{x^2+1} \end{gathered}$$