Question

$1)$ Functions $f,g:R\to R$ are defined, respectively, by $f\left(x\right)=x^2+3x+1,g\left(x\right)=2x–3,$ find

$\left(i\right)$ $fog$

$2)$ Functions $f,g:R\to R$ are defined, respectively, by $f\left(x\right)=x^2+3x+1,g\left(x\right)=2x–3,$ find

$\left(ii\right)$ $gof$

$3)$ Functions $f,g:R\to R$ are defined, respectively, by $f\left(x\right)=x^2+3x+1,g\left(x\right)=2x–3,$ find

$\left(iii\right)$ $fof$

$4)$ Functions $f,g:R\to R$ are defined, respectively, by $f\left(x\right)=x^2+3x+1,g\left(x\right)=2x–3,$ find

$\left(iv\right)$ $gog$

Answer 1

$1)$ Given, $f:R\to R,g:R\to R$

$f\left(x\right)=x^2+3x+1,$ and $g\left(x\right)=2x–3$

$\left(fog\right)\left(x\right)=f(2x–3)$

$(\because (fog\left)\right(x)=f(g\left(x\right))$

$=(2x–3{)}^2+3(2x–3)+1$

$=4x^2–12x+9+6x–9+1$

$(\because (a–b{)}^2=a^2–2ab+b^2)$

$=4x^2–6x+1$

Therefore, $\left(fog\right)\left(x\right)=4x^2–6x+1$

$2)$ Given, $f:R\to R,g:R\to R$

$f\left(x\right)=x^2+3x+1,$ and $g\left(x\right)=2x–3$

$\left(gof\right)\left(x\right)=g(x^2+3x+1)$

$(\because (fog\left)\right(x)=f(g\left(x\right))$

$=2(x^2+3x+1)-3$

$=2x^2+6x-1$

Therefore, $\left(gof\right)\left(x\right)=2x^2+6x-1$

$3)$ Given, $f:R\to R,g:R\to R$

$f\left(x\right)=x^2+3x+1,$ and $g\left(x\right)=2x–3$

$\left(fof\right)\left(x\right)=f(x^2+3x+1)$

$(\because (fof)x=f(f\left(x\right))$

$=(x^2+3x+1{)}^2+3(x^2+3x+1)+1$

$=x^4+9x^2+1+6x^3+6x+2x^2+3x^2$ $+9x+3+1$

$=x^4+6x^3+14x^2+15x+5$

$\left(\begin{array}{c}\because (a+b+c{)}^2=a^2+b^2+c^2\\ +2ab+2bc+2ca\end{array}\right)$

$4)$ Given, $f:R\to R,g:R\to R$

$f\left(x\right)=x^2+3x+1,$ and $g\left(x\right)=2x–3$

$\left(gog\right)\left(x\right)=g(2x–3)$

$(\because (gog\left)\right(x)=g(g\left(x\right)\left)\right)$

$=2(2x-3)-3$

$=4x-9$

$\therefore \left(gog\right)\left(x\right)=4x-9$