Question:
Let $f: R \rightarrow R: f(x)=x+1$ and $g: R \rightarrow R: g(x)=2 x-3 .$
Find
(i) $(f+g)(x)$
(ii) $(f-g)(x)$
(iii) (fg) (x)
(iv)(f/g) (x)
Solution:
(i) Given:
$f(x)=x+1$ and $g(x)=2 x-3$
(i) To find: $(f+g)(x)$
$(f+g)(x)=f(x)+g(x)$
$=(x+1)+(2 x-3)$
$=x+1+2 x-3$
$=3 x-2$
Therefore,
$(f+g)(x)=3 x-2$
(ii) To find: $(f-g)(x)$
$(f-g)(x)=f(x)-g(x)$
$=(x+1)-(2 x-3)$
$=x+1-2 x+3$
$=4-x$
Therefore
$(f-g)(x)=4-x$
(iii) To find: (fg)(x)
$(f g)(x)=f(x) \cdot g(x)$
$=(x+1)(2 x-3)$
$=x(2 x)-3(x)+1(2 x)-1(3)$
$=2 x^{2}-3 x+2 x-3$
$=2 x^{2}-x-3$
Therefore,
$(f g)(x)=2 x^{2}-x-3$
(iv) To find $:\left(\frac{f}{g}\right)(x)$
Sol. $\left(\frac{f}{g}\right)(x)=\left(\frac{f(x)}{g(x)}\right)$
$=\left(\frac{x+1}{2 x-3}\right)$
Therefore,
$\left(\frac{f}{g}\right)(x)=\left(\frac{x+1}{2 x-3}\right)$