Prove that

Solution:

To find: $f \circ g, g \circ f,(f \circ g)(2)$ and $(g \circ f)(-3)$

Formula used: (i) $f \circ g=f(g(x))$

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

Given: (i) $f: R \rightarrow R: f(x)=x^{2}+2$

(ii) $g: R \rightarrow R: g(x)=\frac{x}{x-1}, x \neq 1$

f o g = f(g(x))

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

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

Ans $) \Rightarrow \frac{(x)^{2}}{(x-1)^{2}}+2$

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

$=\frac{4}{1}+2$

Ans) = 6

g o f = g(f(x))

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

$\Rightarrow \frac{x^{2}+2}{x^{2}+2-1}$

Ans $) \Rightarrow \frac{x^{2}+2}{x^{2}+1}$

$(g \circ f)(-3)=\frac{-3^{2}+2}{-3^{2}+1}$

$=\frac{9+2}{9+1}$

Ans) $=\frac{11}{10}$