calculus

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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 -

    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

  • 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,
    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 -

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

  • calculus -

    also, change your answer for increasing and decreasing.

  • calculus -

    Naturally, yes

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