<s>[INST] What am I doing wrong when testing for concavity with $f(x)=5x^{2/3}-2x^{5/3}$? I can't figure out what I'm doing wrong when testing for concavity with $f(x)=5x^{2/3}-2x^{5/3}$. I can't find the right intervals for where the graph is concave-down and concave-up.

My Steps:
find $f'(x)$:
$$f'(x)=\frac{10}{3}x^{-1/3}-\frac{10}{3}x^{2/3}=\frac{10}{3\sqrt[3]{x}}-\frac{10\sqrt[3]{x^2}}{3}$$
find $f''(x)$:
$$\frac{d}{dx} 3\sqrt[3]{x} = 0 + \frac{1}{3\sqrt[3]{x^2}} \cdot 3 = \frac{1}{\sqrt[3]{x^2}}$$
$$\frac{d}{dx}\frac{10}{3\sqrt[3]{x}}=\frac{0-[\frac{1}{\sqrt[3]{x^2}} \cdot 10]}{(3\sqrt[3]{x})^2}=\frac{-10}{(3\sqrt[3]{x})^2(\sqrt[3]{x^2})}=\frac{-10}{9\sqrt[3]{x^4}}$$
$$\frac{d}{dx}10\sqrt[3]{x^2}= 0 +10 \cdot \frac{d}{dx}x^{2/3}=\frac{10}{3\sqrt[3]{x}}$$
$$\frac{d}{dx}\frac{10\sqrt[3]{x^2}}{3}=\frac{0 - [\frac{10}{3\sqrt[3]{x}} \cdot 3]}{9} = \frac{-30}{9(3\sqrt[3]{x})}=\frac{-10}{9\sqrt[3]{x}}$$
$$f''(x)=-\frac{10}{9\sqrt[3]{x^4}}-\frac{-10}{9\sqrt[3]{x}}= \frac{10}{9\sqrt[3]{x}}-\frac{10}{9\sqrt[3]{x^4}}$$
set $f''(x) = 0$ and solve for $x$.
$$\frac{10}{9\sqrt[3]{x}}-\frac{10}{9\sqrt[3]{x^4}}=0 \to x=1$$
find $f''(-1)$.
$$\frac{10}{9\sqrt[3]{(-1)}}-\frac{10}{9\sqrt[3]{(-1)^4}}=\frac{-20}{9}$$
find $f''(2)$
$$\frac{10}{9\sqrt[3]{(2)}}-\frac{10}{9\sqrt[3]{(2)^4}}=\frac{5\sqrt[3]{4}}{18}$$
So the graph should be concave-down on $(-\infty, 1)$ and concave-up on $(1, \infty)$.
However, this is wrong.
 [/INST] Keep going with fractional exponents, it's much easier:
$$
f'(x)=\frac{10}{3}x^{-1/3}-\frac{10}{3}x^{2/3}
$$
Therefore
$$
f''(x)=-\frac{10}{9}x^{-4/3}-\frac{20}{9}x^{-1/3}=-\frac{10}{9}x^{-1/3}(x^{-1}-2)
$$
This is undefined for $x=0$, and it should be easy study the sign elsewhere.
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