Let f, g :

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:

fg : 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 $2 x \neq 3$

$\Rightarrow\left(\frac{f}{g}\right)(x)=\frac{x+1}{2 x-3}, x \neq \frac{3}{2}$

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