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July 31, 2014

July 31, 2014

Posted by **Anonymous** on Wednesday, January 5, 2011 at 5:00pm.

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 -
**Reiny**, Wednesday, January 5, 2011 at 5:20pmWhere 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

- calculus -
**Anonymous**, Wednesday, January 5, 2011 at 5:30pmI 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 -
**Reiny**, Wednesday, January 5, 2011 at 5:54pmOk, 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

- calculus -
**Anonymous**, Wednesday, January 5, 2011 at 6:06pmOkay, so I was right apart from the x=0. Thank you

- calculus -
**Reiny**, Wednesday, January 5, 2011 at 6:08pmalso, change your answer for increasing and decreasing.

- calculus -
**Anonymous**, Wednesday, January 5, 2011 at 6:09pmNaturally, yes

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