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