Answer
a) $\dfrac{x^4+x^3+x}{x+1}$
b) $\dfrac{-x^4-x^3+x}{x+1}$
c) $\dfrac{x^4}{x+1}$
d) $\dfrac{1}{x^2(x+1)}$
Domain: $(-\infty,-1)\cup(-1,0)\cup(0,\infty)$
Work Step by Step
We are given the functions:
$f(x)=\dfrac{x}{x+1}$
$g(x)=x^3$
a) Determine $(f+g)(x)$:
$(f+g)(x)=f(x)+g)(x)=\dfrac{x}{x+1}+x^3=\dfrac{x^4+x^3+x}{x+1}$
b) Determine $(f-g)(x)$:
$(f-g)(x)=f(x)-g)(x)=\dfrac{x}{x+1}-x^3=\dfrac{-x^4-x^3+x}{x+1}$
c) Determine $(fg)(x)$:
$(fg)(x)=f(x)g)(x)=\dfrac{x}{x+1}\cdot x^3=\dfrac{x^4}{x+1}$
d) Determine $\left(\dfrac{f}{g}\right)(x)$:
$\left(\dfrac{f}{g}\right)(x)=\dfrac{\dfrac{x}{x+1}}{x^3}=\dfrac{1}{x^2(x+1)}$
Determine the domain of $\dfrac{f}{g}$:
$x^2(x+1)=0$
$x=0$ or $x=-1$
The domain is:
$(-\infty,-1)\cup(-1,0)\cup(0,\infty)$
