calculus
posted by Anonymous .
state increasing, decreasing, max, min
for f(x)= x^(1/3) (x+4)
Here is my work:
f'(x)= 4/3x^(1/3) + 4/3x^(2/3)
4/3x^(1/3) + 4/3x^(2/3)=0
x=0 x= 1
Increasing: [0, ∞)
Decreasing: (∞, 0]
min @ x=0
max: none
Is this correct?

Where does the (4/3) come from in your answer?
the original was
f(x) = x^(1/3) (x+4) , do you have a typo?
using the product rule, I got
f '(x) = x^(1/3) + (x+4)(1/3)x^(2/3)
common factor of x^(2/3) ...
= x^(2/3) (x + x+4)
= x^(2/3) (2x+4)
or
(2x+4)/x^(2/3) = (2x+4)/(x^2)^(1/3)
the denominator is always positive
max/min ? when 2x+4 = 0
x = 2, the f(x) = 2(2)^1/3)
we can see that the curve crosses at 0 and 4
so x=2 would produce a minimum.
There is no maximum
looking at the f '(x) of (2x+4)/(x^(2/3))
we already saw that the bottom is always positive except x=0,
so all we need to do is look at the top
2x+4 > 0 for all x> 2
and
2x+4 < 0 for all x < 2 
I combined right off the bat, so
f(x)= x^(1/3) (x+4)
f(x)= x^(4/3) + 4x^(1/3)
f'(x)= (4/3)x^(1/3) + (4/3)x^(2/3)
(4/3)x^(1/3) + (4/3)x^(2/3)=0
x=0 x= 1
...I can't find a flaw in my calculations.. 
Ok, I see now,
so your derivative is ok
but your value of x=0 doesn't satisfy
(4/3)x^(1/3) + (4/3)x^(2/3)=0
let x = 0
LS = (4/3)(0) + (4/3)(1/0) which is undefined
(4/3)x^(1/3) + (4/3)x^(2/3)=0
divide both terms by 4/3
x^(1/3) + x^(2/3)=0
common factor of x^(2/3)
x^(2/3) ( x + 1) = 0
x = 1
ahhh, let's check mine, looks like I forgot the 1/3 when I factored.
f '(x) = x^(1/3) + (x+4)(1/3)x^(2/3)
common factor of x^(2/3) ...
= x^(2/3) (x + (1/3)(x+4))
= x^(2/3) (3x+x+4)/3
= x^(2/3)(4x+4)/3
which when I set it equal to zero gives me x=1 , the same as yours
the rest of my argument is valid
so x = 1 yields a minimum value of (1)^(1/3) (3)
= 3
increasing for x > 1
decreasing for x < 1 
Okay, so I was right apart from the x=0. Thank you

also, change your answer for increasing and decreasing.

Naturally, yes