How do you find the critical numbers of f(x)=x^(2/3)+x^(-1/3)?

Sep 17, 2017

This function has one critical value of $x$, at $x = \frac{1}{2}$, and gives a local minimum value there of $f \left(\frac{1}{2}\right) = \frac{3}{{2}^{\frac{2}{3}}} \approx 1.89$

Explanation:

The derivative of this function is $f ' \left(x\right) = \frac{2}{3} {x}^{- \frac{1}{3}} - \frac{1}{3} {x}^{- \frac{4}{3}}$. Getting a common denominator of $3 {x}^{\frac{4}{3}}$ allows us to write $f ' \left(x\right) = \frac{2 x - 1}{3 {x}^{\frac{4}{3}}}$. This is equal to zero when $2 x - 1 = 0$, which means $x = \frac{1}{2}$. It's also undefined at $x = 0$, though the original function is undefined there as well.

Hence, the value $x = \frac{1}{2}$ is the only critical value of $f$. The sign of $f '$ changes from negative to positive as $x$ increases through $\frac{1}{2}$, implying that there is a local (relative) minimum point at $x = \frac{1}{2}$.

The local minimum value is f(1/2)=1/2^(2/3)+2^(1/3)=(1+2)/(2^(2/3)) =3/(2^(2/3)) approx 1.89.

Here's the graph of this function:

graph{x^(2/3)+x^(-1/3) [-10, 10, -5, 5]}