1. ## graph defined?

Hi,

I am attempting to graph $(x^2-3)^\frac{2}{3}$.

My textbook shows the graph as being defined between $\sqrt{3}$ and $-\sqrt{3}$ in fact the derivative indicates that the graph has a local maximum at $x=0$. In my textbook the graph is convex and positive between $\pm\sqrt{3}$

The problem is that $f(0)=-3^\frac{2}{3}$ is approximately $-2.08$. Which means it's negative, unless I am calculating $f(0)$ incorrectly? What has me really confused is that MuPAD does not show the graph as defined between $\sqrt{3}$ and $-\sqrt{3}$ which has got me thinking the textbook might be wrong. When calculating $-3^\frac{2}{3}$ do you square the minus three and then take a cube root? Maybe that's what's wrong. Maybe f(0) is actually positive?

Regards
Craig.

2. Re-writing f as

$(\sqrt[3]{x^2-3})^2$ should hopefully convince you that for real x this is always positive.

3. ## Calculators don't work...

Hi,

Thanks, looking at my notes that what I originally thought.
Does that mean that $f(0)=(-3)^\frac{2}{3}$ is approximately positive 2.08? My calculator gives $(\sqrt[3]{3}[\frac{1+i\sqrt{3}}{2}])^2$ which blows my pip. Why all the complex numbers if it's 2.08?

4. Cube rooting is defined all along the real line; no complex numbers should be involved at all. Yes f(0) is approximately +2.08.