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

1 Answer
Sep 17, 2017

This function has one critical value of x, at x=1/2, and gives a local minimum value there of f(1/2)=3/(2^(2/3)) approx 1.89

Explanation:

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

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