Finding the critical point and local extreme value? How would I find the critical point of the following function.
$f(x)=x^{2/3}+2x^{-1/3}$
This is what I did.
$\frac{2}{3}x^{-1/3}-\frac{2}{3}x{^\frac{-4}{3}}$
$\frac{2}{3\sqrt[3]{x}}-\frac{2}{3\sqrt[3]{x^4}}$
but how do I simplify it and get the critical point.
 A: Hint: 
$$
\frac{x^{-\frac{1}{3}}}{x^{-\frac{4}{3}}}=x^{\frac{4}{3}-\frac{1}{3}}=x\quad\forall x\neq 0.
$$
A: $$
\begin{align} 
f'(x) = \frac{2}{3}x^{-1/3}-\frac{2}{3}x^{-4/3} 
& = \frac 23\left(x^{-1/3} - x^{-4/3}\right) \\ \\
& = \frac 23\left(x^{3/3}x^{-4/3} - x^{-4/3}\right) \\ \\
& = \frac 23 x^{-4/3}\left(x^{3/3} - 1\right) \\ \\
& = \frac 23 x^{-4/3}(x - 1) \\ \\
& = \frac{2(x-1)}{3x^{4/3}} \\ \\ 
& = \frac{2(x-1)}{3\sqrt[\large 3]{x^4}} \\ 
\end{align}
$$
Now, we have critical points where 


*

*$f'(x)$ is undefined: at $x = 0$, as is $f(x)$

*$f'(x) = 0:$ This will occur if and only when the numerator is zero. In this case, $f'(x) = 0$ when $(x - 1) = 0 \implies x = 1$.



Let's look at the Wolgram Alpha graph below (the blue curve): We see that $f(x)\to \infty$ as $x \to 0$. That is, there exists a vertical asymptote at $x = 0$. We also see that at $x = 1$ we have a local minimum. You can use the derivative to tell you when $f(x)$ is increasing ($f'(x) > 0), when decreasing ($f'(x) \lt 0$), and you already know where it is neither increasing nor decreasing, given the points we've found. 

