Answer
a) $\sqrt{x^2-4}+\dfrac{x^2}{x^2+1}$
b) $\sqrt{x^2-4}-\dfrac{x^2}{x^2+1}$
c) $\dfrac{x^2\sqrt{x^2-4}}{1+x^2}$
d) $\dfrac{(x^2+1)\sqrt{x^2-4}}{x^2}$
Domain: $(-\infty,-2)\cup(2,\infty)$
Work Step by Step
We are given the functions:
$f(x)=\sqrt{x^2-4}$
$g(x)=\dfrac{x^2}{x^2+1}$
a) Determine $(f+g)(x)$:
$(f+g)(x)=f(x)+g)(x)=\sqrt{x^2-4}+\dfrac{x^2}{x^2+1}$
b) Determine $(f-g)(x)$:
$(f-g)(x)=f(x)-g)(x)=\sqrt{x^2-4}-\dfrac{x^2}{x^2+1}$
c) Determine $(fg)(x)$:
$(fg)(x)=f(x)g)(x)=\sqrt{x^2-4}\left(\dfrac{x^2}{x^2+1}\right)=\dfrac{x^2\sqrt{x^2-4}}{1+x^2}$
d) Determine $\left(\dfrac{f}{g}\right)(x)$:
$\left(\dfrac{f}{g}\right)(x)=\dfrac{\sqrt{x^2-4}}{\dfrac{x^2}{x^2+1}}=\dfrac{(x^2+1)\sqrt{x^2-4}}{x^2}$
Determine the domain of $\dfrac{f}{g}$:
$\begin{cases}
x^2-4\geq 0\\
x^2\not=0
\end{cases}$
The domain is:
$(-\infty,-2)\cup(2,\infty)$
