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.
It is a bit easier if you factor out the smallest fractional power of $x$ at each step.
\begin{eqnarray} f(x)&=&5x^{2/3}-2x^{5/3}=x^{2/3}(5-2x)\\ f^\prime(x)&=&\frac{10}{3}x^{-1/3}-\frac{10}{3}x^{2/3}=\frac{10}{3}x^{-1/3}(1-x)\\ f^{\prime\prime}(x)&=&-\frac{10}{9}x^{-4/3}-\frac{20}{9}x^{-1/3}=-\frac{10}{9}x^{-4/3}(1+2x) \end{eqnarray}
ADDENDUM: Note that since $f^\prime(x)$ is undefined at $x=0$ and $f^{\prime\prime}(x)=0$ at $x=-\frac{1}{2}$ you should check the concavity on the intervals $\left(-\infty,-\frac{1}{2}\right)$, $\left(-\frac{1}{2},0\right)$ and $(0,\infty)$
Here is the graph with the visually indiscernible inflection at $\left(-\frac{1}{2}, 3\sqrt[3]{2}\right)$.