How do you find the relative extrema for #f(x) =2x- 3x^(2/3) +2# on the interval [-1,3]?

1 Answer
May 25, 2015

Find and test the critical numbers for #f#.

#f(x) =2x- 3x^(2/3) +2#

Critical numbers for #f# are values for #x# that are:

  1. In the domain of #f#, and

  2. at which #f'(x) = 0# or #f'(x)# does not exist.

(This is a wonderful example of why you cannot ignore the second kind of critical numbers.)

For this function, we have:

#f'(x) = 2-2x^(-1/3) = 2(1-x^(-1/3)) = 2((root(3)x -1)/root(3)x)#

#f'(x) = 0# #color(white)"sssssssss"# #f'(x)# does not exist

#x=1# #color(white)"ssssssssssssss"# # x =0#

Using the first derivative we find that #f# is

increasing on #(-oo, 0)# (Yes, I know the domain has been restricted, but it is not necessary.)

decreasing on #(0,1) so

#f(0) = 2# is a relative maximum.

#f# is increasing on #(1,oo), so

#f(1) = 1# is a relative minimum.