Find fog and gof if

Solution:

(i) $f(x)=e^{x}, g(x)=\log _{e} x$

$f: R \rightarrow(0, \infty) ; g:(0, \infty) \rightarrow R$

Computing fog:

Clearly, the range of g is a subset of the domain of f.

$f o g:(0, \infty) \rightarrow R$

$(f o g)(x)=f(g(x))$

$=f\left(\log _{e} x\right)$

$=\log _{e} e^{x}$

$=x$

Computing gof:

Clearly, the range of $f$ is a subset of the domain of $g$.

$\Rightarrow f o g: R \rightarrow R$

$(g \circ f)(x)=g(f(x))$

$=g\left(e^{x}\right)$

$=\log _{e} e^{x}$

$=x$

(ii) $f(x)=x^{2}, g(x)=\cos x$

$f: R \rightarrow[0, \infty) ; g: R \rightarrow[-1,1]$

Computing fog:

Clearly, the range of $g$ is not a subset of the domain of $f$.

$\Rightarrow$ Domain $(f o g)=\{x: x \in$ domain of $g$ and $g(x) \in$ domain of $f\}$

$\Rightarrow$ Domain $(f o g)=x: x \in R$ and $\cos x \in R\}$

$\Rightarrow$ Domain of $(f o g)=R$

$f o g: R \rightarrow R$

$(f o g)(x)=f(g(x))$

$=f(\cos x)$

$=\cos ^{2} x$

Computing gof:

Clearly, the range of $f$ is a subset of the domain of $g$.

$\Rightarrow f o g: R \rightarrow R$

$(g \circ f)(x)=g(f(x))$

$=g\left(x^{2}\right)$

$=\cos \left(x^{2}\right)$

(iii) $f(x)=|x|, g(x)=\sin x$

$f: R \rightarrow(0, \infty) ; g: R \rightarrow[-1,1]$

Computing fog:

Clearly, the range of $g$ is a subset of the domain of $f$.

$\Rightarrow f o g: R \rightarrow R$

$(f o g)(x)=f(g(x))$

$=f(\sin x)$

$=|\sin x|$

Computing gof:

Clearly, the range of $f$ is a subset of the domain of $g$.

$\Rightarrow f o g: R \rightarrow R$

$(g \circ f)(x)=g(f(x))$

$=g(|x|)$

$=\sin |x|$

$(i v) f(x)=x+1, g(x)=e^{x}$

$f: R \rightarrow R ; g: R \rightarrow[1, \infty)$

Computing fog:

Clearly, range of $g$ is a subset of domain of $f$.

$\Rightarrow f o g: R \rightarrow R$

$(f o g)(x)=f(g(x))$

$=f\left(e^{x}\right)$

$=e^{x}+1$

Computing gof:

Clearly, range of $f$ is a subset of domain of $g$.

$\Rightarrow f o g: R \rightarrow R$

$(g o f)(x)=g(f(x))$

$=g(x+1)$

$=e^{x+1}$

(v) $f(x)=\sin ^{-1} x, g(x)=x^{2}$

$f:[-1,1] \rightarrow\left[\frac{-\pi}{2}, \frac{\pi}{2}\right] ; g: R \rightarrow[0, \infty)$

Computing fog:

Clearly, the range of $g$ is not a subset of the domain of $f$.

Domain $(f o g)=\{x: x \in$ domain of $g$ and $g(x) \in$ domain of $f\}$

Domain $(f o g)=\left\{x: x \in R\right.$ and $\left.x^{2} \in[-1,1]\right\}$

Domain $(f o g)=\{x: x \in R$ and $x \in[-1,1]\}$

Domain of $(f o g)=[-1,1]$

$f o g:[-1,1] \rightarrow R$

$(f o g)(x)=f(g(x))$

$=f\left(x^{2}\right)$

$=\sin ^{-1}\left(x^{2}\right)$

Computing gof:

Clearly, the range of $f$ is a subset of the domain of $g$.

$f o g:[-1,1] \rightarrow R$

$(g o f)(x)=g(f(x))$

$=g\left(\sin ^{-1} x\right)$

$=\left(\sin ^{-1} x\right)^{2}$

(vi) $f(x)=x+1, g(x)=\sin x$

$f: R \rightarrow R ; g: R \rightarrow[-1,1]$

Computing fog:

Clearly, the range of $g$ is a subset of the domain of $f$.

$\Rightarrow f \circ g: R \rightarrow R$

$(f \circ g)(x)=f(g(x))$

$=f(\sin x)$

$=\sin x+1$

Computing gof:

Clearly, the range of $f$ is a subset of the domain of $g$.

$\Rightarrow f o g: R \rightarrow R$

$(g o f)(x)=g(f(x))$

$=g(x+1)$

$=\sin (x+1)$

(vii) $f(x)=x+1, g(x)=2 x+3$

$f: R \rightarrow R ; g: R \rightarrow R$

Computing fog:

Clearly, the range of $g$ is a subset of the domain of $f$.

$\Rightarrow f o g: R \rightarrow R$

$(f o g)(x)=f(g(x))$

$=f(2 x+3)$

$=2 x+3+1$

$=2 x+4$

Computing gof:

Clearly, the range of $f$ is a subset of the domain of $g$.

$\Rightarrow f o g: R \rightarrow R$

$(g o f)(x)=g(f(x))$

$=g(x+1)$

$=2(x+1)+3$

$=2 x+5$

(viii) $f(x)=c, g(x)=\sin x^{2}$

$f: R \rightarrow\{c\} ; g: R \rightarrow[0,1]$

Computing fog:

Clearly, the range of $g$ is a subset of the domain of $f$.

$f o g: R \rightarrow R$

$(f o g)(x)=f(g(x))$

$=f\left(\sin x^{2}\right)$

$=c$

Computing gof:

Clearly, the range of $f$ is a subset of the domain of $g$.

$\Rightarrow f o g: R \rightarrow R$

$(g o f)(x)=g(f(x))$

$=g(c)$

$=\sin c^{2}$

$(i x) f(x)=x^{2}+2$

$f: R \rightarrow[2, \infty)$

$g(x)=1-\frac{1}{1-x}$

For domain of $g: 1-x \neq 0$

$\Rightarrow x \neq 1$

$\Rightarrow$ Domain of $g=R-\{1\}$

$g(x)=1-\frac{1}{1-x}=\frac{1-x-1}{1-x}=\frac{-x}{1-x}$

For range of $g$ :

$y=\frac{-x}{1-x}$

$\Rightarrow y-x y=-x$

$\Rightarrow y=x y-x$

$\Rightarrow y=x(y-1)$

$\Rightarrow x=\frac{y}{y-1}$

Range of $g=R-\{1\}$

So, g: $R-\{1\} \rightarrow R-\{1\}$

Computing fog:

Clearly, the range of $g$ is a subset of the domain of $f$.

$\Rightarrow f o g: R-\{1\} \rightarrow R$

$(f o g)(x)=f(g(x))$

$=f\left(\frac{-x}{x-1}\right)$

$=\left(\frac{-x}{x-1}\right)^{2}+2$

$=\frac{x^{2}+2 x^{2}+2-4 x}{(1-x)^{2}}$

$=\frac{3 x^{2}-4 x+2}{(1-x)^{2}}$

Computing gof:

Clearly, the range of $f$ is a subset of the domain of $g$.

$\Rightarrow g o f: R \rightarrow R$

$(g o f)(x)=g(f(x))$

$=g\left(x^{2}+2\right)$

$=1-\frac{1}{1-\left(x^{2}+2\right)}$

$=1-\frac{1}{-\left(x^{2}+1\right)}$

$=\frac{x^{2}+2}{x^{2}+1}$