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Homework Help: Find the X coordinate with slope of -2

  1. Dec 10, 2008 #1
    1. The problem statement, all variables and given/known data

    Function defined as:

    y = The integral from 0 to x2 of 1 / (1 - Sqrt(t) + t)

    There exists a point where the slope of the tangeant is = -2. Find the x coordinate at this point.

    2. Relevant equations

    Fundemental Theorem of Calculus

    3. The attempt at a solution

    If I use FTC and replace t with x2. Then I get two equations because sqrt(x2) is +/- x. I use those to create two seperate equations.

    y = 2x and y = 2x/1+2x

    These should be 2 equations for the tangeant line but I am unsure what to do next.

    This was on my exam yesterday and I tried some things, none of which seemed right. I know I lost mark just would like to know how many. :)
  2. jcsd
  3. Dec 10, 2008 #2


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    Assuming you mean: [tex]y(x)\equiv \int_0^{x^2} \frac{1}{1-\sqrt{t}+t}dt[/tex] and y'(x)=2x or 2x/(1+2x) , then you have the right idea. However, [tex]1-\sqrt{x^2}+x^2=1-|x|+x^2[/tex] not 1 or 1+2x!.....And y'(x) is the equation for the slope of the tangent line, not the tangent line itself!

    So, where does y'(x)=-2?
  4. Dec 10, 2008 #3
    I forgot one of the squares on the bottom, bad mistake.

    So my equation becomes:
    y' = 2x / 1 - |x| + x2

    -2 = 2x / 1 - |x| + x2 but we know x is positive because of the range of the original question so it becomes:

    -2 = 2x / 1 - x + x2

    Solve for x = 1? I still don't think that is quite right.
  5. Dec 10, 2008 #4


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    Where did this come from? You didn't say anything about a range or domain (which is what I think you really mean) before. The only reference to the variable was a "x2 in which x can be positive of negative.

    That isn't at all right because x= 1 does NOT satify that equation!

    But if you take x to be negative then you must have -2= 2x/(1+ x+ x2) so that -1- x- x2= x or x2+ 2x+ 1= 0. What negative solution does that have? Does it satisfy -2= 2x/(1- |x|+ x2)?
  6. Dec 10, 2008 #5


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    As Halls said, unless you are explicitly told that x must be positive, you must examine two cases: (1)x is positive and (2)x is negative.
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