Answer
There are no domain endpoints. The critical points are $(0,0)$ and $(-4/5,1.0341)$.
At $x=0$, $y$ has the local minimum value of $0$ and at $x=-4/5$, $y$ has the local maximum value of $1.0341$. There are no absolute extreme values.
Work Step by Step
$$y=x^{2/3}(x+2)$$
- Domain: $R$. Therefore, there are no endpoints.
1) Find all the critical points of the function:
- Find $y'$: $$y'=(x^{2/3})'(x+2)+x^{2/3}(x+2)'$$ $$y'=\frac{2}{3}x^{-1/3}(x+2)+x^{2/3}$$ $$y'=\frac{2(x+2)}{3x^{1/3}}+x^{2/3}$$ $$y'=\frac{2(x+2)+3x^{1/3}x^{2/3}}{3x^{1/3}}$$ $$y'=\frac{2x+4+3x}{3x^{1/3}}$$ $$y'=\frac{5x+4}{3x^{1/3}}$$
- We have $y'=0$ when $5x+4=0$, or $x=-4/5$.
- $y$ is undefined when $3x^{1/3}=0$ or $x=0$.
So the critical points of function $y$ are at $x=0$ and $x=-4/5$.
2) Evaluate $y$ at the critical points:
- For $x=-4/5$: $$y=\Big(-\frac{4}{5}\Big)^{2/3}\Big(-\frac{4}{5}+2\Big)\approx0.8618\times1.2\approx1.0341$$
- For $x=0$: $$y=\Big(0\Big)^{2/3}\Big(0+2\Big)=0\times2=0$$
So the critical points are $(0,0)$ and $(-4/5,1.0341)$.
Now take a look at the graph of $y$.
From the graph, we see that at $x=0$, $y$ has the local minimum value of $0$ and at $x=-4/5$, $y$ has the local maximum value of $1.0341$. There are no absolute extreme values.
