Answer
Function $y$ has no endpoints. The critical points are $(0,0)$ and $(-1,-3)$ and $(1,-3)$.
At $x=0$, $y$ has the local maximum value of $0$ and at both $x=1$ and $x=-1$, $y$ has the local and absolute minimum value of $-3$. There is no absolute maximum value.
Work Step by Step
$$y=x^{2/3}(x^2-4)$$
- Domain: $R$. Therefore, there are no endpoints.
1) Find all the critical points of the function:
- Find $y'$: $$y'=(x^{2/3})'(x^2-4)+x^{2/3}(x^2-4)'$$ $$y'=\frac{2}{3}x^{-1/3}(x^2-4)+x^{2/3}(2x)$$ $$y'=\frac{2(x^2-4)}{3x^{1/3}}+2x^{5/3}$$ $$y'=\frac{2(x^2-4)+(3x^{1/3})(2x^{5/3})}{3x^{1/3}}$$ $$y'=\frac{2x^2-8+6x^2}{3x^{1/3}}$$ $$y'=\frac{8x^2-8}{3x^{1/3}}=\frac{8(x-1)(x+1)}{3x^{1/3}}$$
- We have $y'=0$ when $8(x-1)(x+1)=0$, which leads to $x=\pm1$.
- $y$ is undefined when $3x^{1/3}=0$, or $x=0$.
So the critical points of function $y$ are at $x=0$ and $x=\pm1$.
2) Evaluate $y$ at the critical points:
- For $x=-1$: $$y=(-1)^{2/3}\Big((-1)^2-4\Big)=1\Big(1-4\Big)=-3$$
- For $x=0$: $$y=(0)^{2/3}\Big((0)^2-4\Big)=0$$
- For $x=1$: $$y=(1)^{2/3}\Big((1)^2-4\Big)=1\Big(1-4\Big)=-3$$
So the critical points are $(0,0)$ and $(-1,-3)$ and $(1,-3)$.
Now take a look at the graph of $y$.
From the graph, we see that at $x=0$, $y$ has the local maximum value of $0$ and at both $x=1$ and $x=-1$, $y$ has the local and absolute minimum value of $-3$. There is no absolute maximum value.