Question:
Let $t, g: \mathrm{R} \rightarrow \mathrm{R}$ be defined, respectively by $f(x)=x+1$ and $g(x)=2 x-3 .$ Find $f+g, f-g$ and $\frac{f}{g}$.
Solution:
f, g : R → R is defined, respectively, by f(x) = x + 1 and g(x) = 2x − 3.
(f + g) (x) = f(x) + g(x)
$=(x+1)+(2 x-3)$
$=3 x-2$
$\therefore(f+g)(x)=3 x-2$
$(f-g)(x)=f(x)-g(x)$
$=(x+1)-(2 x-3)$
$=x+1-2 x+3$
$=-x+4$
$\therefore(f-g)(x)=-x+4$
$\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}, g(x) \neq 0, x \in \boldsymbol{R}$
$\Rightarrow\left(\frac{f}{g}\right)(x)=\frac{x+1}{2 x-3}, 2 x-3 \neq 0$ or $2 x \neq 3$
$\Rightarrow\left(\frac{f}{g}\right)(x)=\frac{x+1}{2 x-3}, x \neq \frac{3}{2}$