What are the critical points of #f(x) = 6x^(2/3) + x^(5/3)#?

1 Answer
Oct 27, 2015

Answer:

Thee critical numbers for #f# are #-12/5# and #0#..

Explanation:

For #f(x) = 6x^(2/3) + x^(5/3)#, we have

#"Dom"(f)=(-oo,oo)# and

#f'(x) = 12/3x^(-1/3)+5/3x^(2/3)#

# = 1/3 x^(-1/3) (12+5x)#

# = (12+5x)/(3x^(1/3))#.

#f'(x)# is undefined at #x=0# and
#f'(x) = 0# at #x=-5/12#.

Both of these numbers are in #"Dom"(f)#, so both are critical numbers for #f#.

Alternative Method

Many students find it more clear to do the algebra differently.

#f'(x) = 12/3x^(-1/3)+5/3x^(2/3)#

# = 12/(3root3x)+(5root3x^2)/3#

# = 12/(3root3x)+(5root3x^2)/3 * root3x/root3x#.

# = 12/(3root3x)+(5x)/(3root3x)#

# = (12+5x)/(3root3x)#

Now find #0# and #-12/5# as above.