Function inflection point?

Question

I have a normal function $f(x)=(x^{2}-1)^{\frac{2}{3}}$. Derivative $\frac{4x}{3(x^{2}-1)^{\frac{1}{3}}}$. Second derivative $\frac{4\left(x^2-3\right)}{9\left(x^2-1\right)^{\frac{4}{3}}}$. When I calculated it, is not normal. Because inflection point is $\sqrt{3}$ and $\sqrt{-3}$, but they are not like that when I graph it on geogebra? There is only critical point x=1,-1. How to solve this?

Answer 1

Your intuition is correct. The second derivative is $\frac{4(x^2-3)}{9(x^2-1)^{\frac{4}{3}}}$. Setting the numerator equal to $0$ yields $4(x^2-3) = 0$, or $x = \sqrt{3}, -\sqrt{3}$. These are the correct inflection points. If you graph it, it doesn't look right only because the graph looks very straight. However, rest assured, the function DOES change concavity here.

Just note it's $-\sqrt{3}$, not $\sqrt{-3}$, which is imaginary.