F(X) = (2/3)(X-2)^(-1/3)
F'(X)= (-2/9)(X-2)^(-4/3)
F''(X) = (8/27)(X-2)^(-5/3) Is F''(X) correct???
Just substitute X=1, then
$\displaystyle F(1) = \frac{2}{3}*(1-2)^{-1/3} = \frac{2}{3}*(-1)^{-1/3}$
We know that $\displaystyle a^{-1} = \frac{1}{a}$
Therefor
$\displaystyle F(1) = \frac{2}{3}*(-1)^{-1/3} = \frac{2}{3}*\frac{1}{\displaystyle(-1)^{1/3}}$
Furthermore we know that $\displaystyle x^{1/2} = \sqrt{x}$ and $\displaystyle x^{1/3} = \sqrt[3]{x}$
Thus
$\displaystyle F(1) = \frac{2}{3}*\frac{1}{\displaystyle(-1)^{\displaystyle1/3}} = \frac{2}{3}*\frac{1}{\sqrt[3]{-1}} = \frac{2}{3}*(-1) = - 2/3 < 0 $
Same here
$\displaystyle F'(1) = (-2/9) (1-2)^{-4/3} = -2/9*(-1)^{\displaystyle-4/3} = -2/9 * \frac{1}{(-1)^{\displaystyle4/3}}$
$\displaystyle = -2/9 * \frac{1}{\displaystyle((-1)^4)^{\displaystyle1/3}} = -2/9*\frac{1}{1^{\displaystyle1/3}} = -2/9 * \frac{1}{\sqrt[3]{1}} = -2/9*1 = -2/9 <0$