Question

Let $f,g$ : $R\to R$ be defined respectively by $f\left(x\right)=x+1,g\left(x\right)=2x-3$. Find $f+g,f-g$ and $\frac{f}{g}$

Answer 1

$f,g:R\to R$ is defined as
$f\left(x\right)=x+1$
$g\left(x\right)=2x-3$
Now, $(f+g)\left(x\right)=f\left(x\right)+g\left(x\right)=(x+1)+(2x-3)=3x-2$
$\therefore (f+g)\left(x\right)=3x-2$
Now, $(f-g)\left(x\right)=f\left(x\right)-g\left(x\right)=(x+1)-(2x-3)=x+1-2x+3=-x+4$
$\therefore (f-g)\left(x\right)=-x+4$
$\left(\frac{f}{g}\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)},g\left(x\right)\ne 0,x\in R$
$\therefore \left(\frac{f}{g}\right)\left(x\right)=\frac{x+1}{2x-3},2x-3\ne 0or2x\ne 3$
$\therefore \left(\frac{f}{g}\right)\left(x\right)=\frac{x+1}{2x-3},x\ne \frac{3}{2}$