[tex]\bf f(x)=x^{\cfrac{}{}\frac{1}{3}}(x+2)\\\\
-----------------------------\\\\
\cfrac{dy}{dx}=\cfrac{1}{3}x^{-\frac{2}{3}}\cdot (x+2)+x^{\frac{1}{3}}\cdot 1\implies
\cfrac{dy}{dx}=\cfrac{x+2}{3\sqrt[3]{x^2}}+\sqrt[3]{x}
\\\\\\
\cfrac{dy}{dx}=\cfrac{x+2+3x}{3\sqrt[3]{x^2}}\implies \cfrac{dy}{dx}=\cfrac{2(2x+1)}{3\sqrt[3]{x^2}}[/tex]
now, you get extremas when the derivative is 0, OR
when the derivative is "undefined", when is undefined, you get a "cusp", or an asymptote, so the graph goes to infinity
the derivative is undefined, when the denominator is 0
so, get the critical points from [tex]\bf \cfrac{dy}{dx}=\cfrac{2(2x+1)}{3\sqrt[3]{x^2}}\implies
\begin{cases}
0=\cfrac{2(2x+1)}{3\sqrt[3]{x^2}}
\\\\
or
\\\\
3\sqrt[3]{x^2}=0
\end{cases}[/tex]
so hmm get the critical points from those two cases, and then do a first derivative check in the regions left and right of the critical points
setting the derivative to 0, is easy to see the critical point, is just -1/2
so. you do a first derivative test left and right of that.. like hmmm -0.51 or so... or -3/4 to the left, or -1/4 to its right
a positive value for the derivative, means is increasing and a negative is decreasing, and the critical point, is well, the extrema :)
