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

Question

2491 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-26T18:43:32

Find values in the domain of #f# at which #f'(x)# does not exist or #f'(x)=0# . See the explanation.

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

#"Dom"(f) = RR - {0}#

#f'(x) = 2/3x^(-1/3) - 1/3x^(-4/3)#

# =1/3 x^(-4/3) (2x-1)#

# = (2x-1)/(3x^(4/3))#

#f'(x)# does not exist at #x=0#, which is not in #"Dom"(f)#.

#f'(x) = 0# at #x=1/2# which is in #"Dom"(f)#.

The only critical number for #f# is #1/2#.