Find the value

(i) $g \circ f$

To find: g o f

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

Given: (i) $f: R \rightarrow R: f(x)=\left(x^{2}+3 x+1\right)$

(ii) g: R → R : g(x) = (2x - 3)

Solution: We have,

$g \circ f=g(f(x))=g\left(x^{2}+3 x+1\right)=\left[2\left(x^{2}+3 x+1\right)-3\right]$

$\Rightarrow 2 x^{2}+6 x+2-3$

$\Rightarrow 2 x^{2}+6 x-1$

Ans). $g \circ f(x)=2 x^{2}+6 x-1$

(ii) f o g

To find: f o g

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

Given: (i) $f: R \rightarrow R: f(x)=\left(x^{2}+3 x+1\right)$

(ii) g: R → R : g(x) = (2x - 3)

Solution: We have,

$f \circ g=f(g(x))=f(2 x-3)=\left[(2 x-3)^{2}+3(2 x-3)+1\right]$

$\Rightarrow 4 x^{2}-12 x+9+6 x-9+1$

$\Rightarrow 4 x^{2}-6 x+1$

Ans). $f \circ g(x)=4 x^{2}-6 x+1$

(iii) g o g

To find: g o g

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

Given: (i) $g: R \rightarrow R: g(x)=(2 x-3)$

Solution: We have,

$g \circ g=g(g(x))=g(2 x-3)=[2(2 x-3)-3]$

$\Rightarrow 4 x-6-3$

$\Rightarrow 4 x-9$

Ans). $\mathrm{g} \circ \mathrm{g}(\mathrm{x})=4 \mathrm{x}-9$