I have two functions, $f(x) = 2x$ and $g(x) = \frac{x^3}{3}$. I solved for $x$ where $f = g$, finding $x = \pm 6^{1/2}$, then solved for $x$ where $f > g$, $x > \pm 6^{1/2}$, and where $f < g$, $x < \pm 6^{1/2}$. However it is clear there is overlap for where I have stated $f > g$ and $f < g$. How do I show that $f > g$ only when $x \in (0, 6^{1/2})$ and $f < g$ only when $x \in (- 6^{1/2}, 0)$ without graphing?
Completely different approaches from mine are welcomed!
On
The roots, $x = 0, -\sqrt{6}, +\sqrt{6}$ show merely the crossover points. You must test the four regions defined by those three points:
On
You forgot the solution $x=0$.
By the Intermediate Value theorem the relative positions of the curves can change only at intersection points. Now, for $x\gg 1$, $f(x)<g(x)$, hence we conclude
Solve the inequality $$ 2x>\frac{x^3}{3} $$ that can be rewritten as $$ x(x-\sqrt{6})(x+\sqrt{6})<0 $$ The expression on the left changes sign only at $-\sqrt{6}$, $0$ and $\sqrt{6}$ (where it vanishes). Since its value at $1$ is $-5$ you have the diagram $$ \begin{array}{ccccccc} \_\!\_\!\_\!\_\!\_\!\_\!\_ & -\sqrt{6} & \_\!\_\!\_\!\_\!\_\!\_\!\_ & 0 & \_\!\_\!\_\!\_\!\_\!\_\!\_ & \sqrt{6} & \_\!\_\!\_\!\_\!\_\!\_\!\_ \\ & & & & <0 & \end{array} $$ and you can complete it by changing sign at the stated points $$ \begin{array}{ccccccc} \_\!\_\!\_\!\_\!\_\!\_\!\_ & -\sqrt{6} & \_\!\_\!\_\!\_\!\_\!\_\!\_ & 0 & \_\!\_\!\_\!\_\!\_\!\_\!\_ & \sqrt{6} & \_\!\_\!\_\!\_\!\_\!\_\!\_ \\ <0 & =0 & >0 & =0 & <0 & =0 & >0 \end{array} $$ So the intervals where the original inequality holds are $(-\infty,-\sqrt{6})$ and $(0,\sqrt{6})$.