Question

Find the local​ extrema, the open intervals on which the function is​ increasing, and the open intervals on which the function is decreasing.
f(x)=(x^(1/3))(x+2)
 

Answer 1

$\bf f(x)=x^{\cfrac{}{}\frac{1}{3}}(x+2)\\\\ -----------------------------\\\\ \cfrac{dy}{dx}=\cfrac{1}{3}x^{-\frac{2}{3}}\cdot (x+2)+x^{\frac{1}{3}}\cdot 1\implies \cfrac{dy}{dx}=\cfrac{x+2}{3\sqrt[3]{x^2}}+\sqrt[3]{x} \\\\\\ \cfrac{dy}{dx}=\cfrac{x+2+3x}{3\sqrt[3]{x^2}}\implies \cfrac{dy}{dx}=\cfrac{2(2x+1)}{3\sqrt[3]{x^2}}$

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   $\bf \cfrac{dy}{dx}=\cfrac{2(2x+1)}{3\sqrt[3]{x^2}}\implies \begin{cases} 0=\cfrac{2(2x+1)}{3\sqrt[3]{x^2}} \\\\ or \\\\ 3\sqrt[3]{x^2}=0 \end{cases}$

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 :)