# calculus

posted by .

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?

• calculus -

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

• calculus -

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..

• calculus -

Ok, I see now,

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

• calculus -

Okay, so I was right apart from the x=0. Thank you

• calculus -

• calculus -

Naturally, yes