What are the critical points of #g(x)=x/3 + x^-2/3#?

Question

2008 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-07T21:36:22

#x_c=root(3)2# is a local minimum

We can find the critical points by equating the first derivative of #g(x)# to zero:

#g'(x) =1/3 -2/3x^(-3) = 0#

#1/3 =2/(3x^3)#

#x^3= 2#

#x_c=root(3)2#

We can easily see that #g'(x) < 0# for #x< x_c# and viceversa #g'(x) > 0# for #x>x_c# , so #x_c is a local minimum.

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