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Help with derivative & points on curve

  1. Jun 27, 2007 #1
    Hey guys, really stuck on this problem, thanks for the help !

    Question:
    Find the values of x at which the slope of the tangent line to the curve defined by y=(x+1)1/3(x2-x-6) is:
    a) Undefined / vertical b) horizontal

    Answer / Attempted answer

    Firstly I took the derivative:

    Dy/dx = 1/3 (2x+1)^-2/3(2)(x2-x-6) + (2x-1)(2x+1) ^1/3
    Dy/dx = (2x+1) ^-2/3 [ 2/3(x-3)(x+2) + (2x-1)(2x+1) ^-1/2]


    Now here I am stuck, I don’t know how to get part a) and for part b) would I just set dy/dx to zero and solve for the x values?
     
    Last edited: Jun 27, 2007
  2. jcsd
  3. Jun 27, 2007 #2

    danago

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    Gold Member

    Well when is a fraction undefined?
     
  4. Jun 27, 2007 #3
    when the denom = 0?

    .. but there isnt one, or should i transfer the - exponents to the botton and create one?
     
  5. Jun 27, 2007 #4

    danago

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    Gold Member

    (x+1)1/3

    is that (x+1)1/3?

    Or is it the product of (x+1) and 1/3?
     
  6. Jun 27, 2007 #5

    HallsofIvy

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    Peculiarly stated! If the slope is "undefined", then the line itself is vertical. If the slope is 0, then the line itself is horizontal.
    I assume that the function given is y= (x+1)/(3(x^2- x-6)) is that correct?
    What is that extra "1" in the numerator? The other possible interpretation of what you wrote is y= (x+1)(1/3)(x^2- x- 6) but that seems unlikely. From what you write further, perhaps that "1/3(x2-x-6)" is [itex]^3\sqrt{x^2- x-6}= (x+1)(x^2-x-6)^{1/3}[/itex] but if so that really bad notation!

    Assuming [itex]y= (x+1) (x^2-x-6)^{1/3}[/itex] then [itex]dy/dx= (x^2-x-6)^{1/3}+ (1/3)(x+1)(x^2-x-6)^{-2/3}(2x-1)[/itex]

    That y' certainly does have a "denominator" because of that -2/3 power. For what value of x is that denominator = 0? And, for b, yes, you set the whole thing equal to 0 and solve for x.
     
  7. Jun 27, 2007 #6
    ok the original function was:

    [itex]y= (x+1)^{1/3} (x^2-x-6)[/itex]

    i then got
    dy/dx= (1/3)(2x+1)^{-2/3}(2)(x^2-x-6) + (2x-1)(2x+1)^{1/3}

    now here is my dy/dx , from here would i common factor out the (2x+1)^{-2/3} or just put it right to the bottom and get the demon. on one part and set that to 0 to see where it is undefined. (assuming this it would be definied at x=-1/2)
     
    Last edited: Jun 27, 2007
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