Solve g(x) = -10 and g(x) > f(x) for the given equations f(x)=x^2 and g(x)=1/(x^3) and...

$g(x) = \frac{1}{x^3} = -10$

$1 = -10x^3 \qquad \rightarrow \qquad -\frac{1}{10} = x^3 \qquad \rightarrow \qquad x = \boxed{-\frac{1}{\sqrt[3]{10}}}$

$g(x) > f(x) \qquad \rightarrow \qquad \frac{1}{x^3} > x^2$

$1 > x^5 \qquad \rightarrow \qquad \boxed{x \in (-\infty, 1)}$

g(x) = 1/(x^3)

asymptote at y = 0 and x = 0

f(x) = x^2

axis of symmetry at x = 0