Answer
On $[0,2]$: $g(x)$
On $(6,\infty)$: $g(x)$
Work Step by Step
We are given the functions:
$f(x)=x^2-\dfrac{1}{2}$
$g(x)=-3x^2-1$
Determine the function $f+g$:
$(f+g)(x)=f(x)+g(x)=x^2-\dfrac{1}{2}-3x^2-1=-2x^2-\dfrac{3}{2}$
Graph the functions $f$, $g$ and $f+g$:
On the interval $[0,2]$, $g(x)$ contributes the most to the magnitude of $f+g$.
On the interval $(6,\infty)$, $g(x)$ contributes the most to the magnitude of $f+g$.
