How to plot $f(x)=x^{2/3}$ I'm using Leithold's book to teach calculus. In a exercise Leithold asks how to draw $f(x)=x^{2/3}$. I don't know how to plot this function since I can't use the derivative methods he develop afterwards. Until this page of the book Leithold only covers limits, continuity, tangents and basic derivatives. He didn't talk about concavity, inflection points, absolute values, etc. yet.
So How do I plot this function using only the definitions he made until this exercise?
 A: Using continuity, you can find that f is continuous at the origin:
$$
\lim_{x \to 0} f(x) = f(0) = 0 \\
$$
Using limits, you can find what happens at the ends:
$$
\lim_{x \to -\infty} f(x) = +\infty \\
\lim_{x \to +\infty} f(x) = + \infty \\
$$
Using limits, you can find the inclination of the tagent line at the origin:
$$
\begin{eqnarray}
f'(0) &=& \lim_{\Delta x \to 0} \frac{f(0 + \Delta x) - f(0)}{\Delta x} \\
&=& \lim_{\Delta x \to 0} \frac{(\Delta x)^{2/3} - 0}{\Delta x} \\
&=& \lim_{\Delta x \to 0} \frac{1}{(\Delta x)^{1/3}} \\
&=& \infty
\end{eqnarray}
$$
Using more limits, you can find the inclination of the tangent line at any point:
$$
\begin{eqnarray}
f'(x) &=& \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} \\
&=& \lim_{\Delta x \to 0} \frac{(x + \Delta x)^{2/3} - x^{2/3}}{\Delta x} \\
&=& \lim_{\Delta x \to 0} \frac{[(x + \Delta x)^{2/3} - x^{2/3}][(x + \Delta x)^{4/3} + (x + \Delta x)^{2/3}x^{2/3} + x^{4/3}]}{\Delta x[(x + \Delta x)^{4/3} + (x + \Delta x)^{2/3}x^{2/3} + x^{4/3}]} \\
... \\
&=& \lim_{\Delta x \to 0} \frac{2x + \Delta x}{(x + \Delta x)^{4/3} + (x + \Delta x)^{2/3}x^{2/3} + x^{4/3}} \\
&=& \frac{2x}{x^{4/3} + x^{2/3}x^{2/3} + x^{4/3}} \\
&=& \frac{2}{3x^{1/3}}
\end{eqnarray}
$$
You now know the graph tends to $+\infty$ at the ends, that it is continuous at the origin but the tangent line has an inclination of $\pi/2$, and that the inclination is negative when $x < 0$ and positive when $x > 0$. That should be plenty to draw a nice graph.
A: Evaluate the function for perfect cubes, giving you perfect squares.
$$(0,0),(\pm1,1),(\pm8,4),(\pm27,9),(\pm64,16).$$
This is more than enough to draw a curve and guess how it behaves.
