Answer
a) $\dfrac{2x^2+x-2}{x(x^2-1)}$
b) $\dfrac{2x^2-x-2}{x(x^2-1)}$
c) $\dfrac{2}{x(x^2-1)}$
d) $\dfrac{2(x^2-1)}{x}$
Domain: $(-\infty,-1)\cup(-1,0)\cup(0,1)\cup(1,\infty)$
Work Step by Step
We are given the functions:
$f(x)=\dfrac{2}{x}$
$g(x)=\dfrac{1}{x^2-1}$
a) Determine $(f+g)(x)$:
$(f+g)(x)=f(x)+g)(x)=\dfrac{2}{x}+\dfrac{1}{x^2-1}=\dfrac{2x^2+x-2}{x(x^2-1)}$
b) Determine $(f-g)(x)$:
$(f-g)(x)=f(x)-g)(x)=\dfrac{2}{x}-\dfrac{1}{x^2-1}=\dfrac{2x^2-x-2}{x(x^2-1)}$
c) Determine $(fg)(x)$:
$(fg)(x)=f(x)g)(x)=\dfrac{2}{x}\cdot\dfrac{1}{x^2-1}=\dfrac{2}{x(x^2-1)}$
d) Determine $\left(\dfrac{f}{g}\right)(x)$:
$\left(\dfrac{f}{g}\right)(x)=\dfrac{\dfrac{2}{x}}{\dfrac{1}{x^2-1}}=\dfrac{2(x^2-1)}{x}$
Determine the domain of $\dfrac{f}{g}$:
$x^2-1=0\Rightarrow x=\pm 1$
$x=0$
The domain is:
$(-\infty,-1)\cup(-1,0)\cup(0,1)\cup(1,\infty)$
