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

1 Answer
Sep 17, 2017

Answer:

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]}