Let f : R → R : f(x) = x2 + 2 and \(g:R→R :g(x)=\frac{x}{x-1},x\ne

1.	Let f : R → R : f(x) = x2 + 2 and \(g:R→R :g(x)=\frac{x}{x-1},x\neq1\)find f o g and g o f and hence find (f o g) (2) and (g o f) (-3).
Answer» To find: f o g, g o f ,(f o g) (2) and (g o f) (-3) Formula used: (i) f o g = f(g(x)) (ii) g o f = g(f(x)) Given: (i) f : R → R : f(x) = x² + 2 (ii) \(g:R→R :g(x)=\frac{x}{x-1},x\neq1\) f o g = f(g(x)) \(\Rightarrow f(\frac{x}{x-1})\) \(\Rightarrow (\frac{x}{x-1})^2+2\) \(\Rightarrow\frac{(x)^2}{(x-1)^2}+2\) \(fog(2)=\frac{(2)^2}{(2-1)^2}+2\) \(=\frac{4}{1}+2\) = 6 g o f = g(f(x)) ⇒ g(x²+2) \(\Rightarrow \frac{x^2+2}{x^2+2-1}\) \(\Rightarrow \frac{x^2+2}{x^2+1}\) \((gof)(-3)=\frac{-3^2+2}{-3^2+1}\) \(=\frac{9+2}{9+1}\) \(=\frac{11}{10}\)

Let f : R → R : f(x) = x2 + 2 and \(g:R→R :g(x)=\frac{x}{x-1},x\neq1\)find f o g and g o f and hence find (f o g) (2) and (g o f) (-3).

Answer»

To find: f o g, g o f ,(f o g) (2) and (g o f) (-3)

Formula used: (i) f o g = f(g(x))

(ii) g o f = g(f(x))

Given: (i) f : R → R : f(x) = x² + 2

(ii) \(g:R→R :g(x)=\frac{x}{x-1},x\neq1\)

f o g = f(g(x))

\(\Rightarrow f(\frac{x}{x-1})\)

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

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

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

\(=\frac{4}{1}+2\)

= 6

g o f = g(f(x))

⇒ g(x²+2)

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

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

\((gof)(-3)=\frac{-3^2+2}{-3^2+1}\)

\(=\frac{9+2}{9+1}\)

\(=\frac{11}{10}\)