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Math Help - Find x-coordinate of stationary point

  1. #1
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    Find x-coordinate of stationary point

    Find the x-coordinate of the stationary point of the curve y=\frac{(x-1)^3}{x+1}, x>0
    What I did

    \frac{dy}{dx}=\frac{(x+1)(3)(x-1)^2-(x-1)^3}{(x+1)^2}

    Stationary Point, dy/dx=0

     \frac{(x+1)(3)(x-1)^2-(x-1)^3}{(x+1)^2}=0
    (x+1)(3)(x-1)^2-(x-1)^3=0
    3(x+1)(x-1)^2=(x-1)^3
    3(x+1)=(x-1)
    3x+3=x-1
    2x+4=0
    x=-2

    however question states x>0 and ans is 1!
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  2. #2
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    You've made a mistake when you divided the expression 3(x+1)(x-1)^2=(x-1)^3 with the term (x-1)^2.

    This is what you should do:
    (x+1)(3)(x-1)^2-(x-1)^3=0
    (x-1)^2(3(x+1)-(x-1))=0
    (x-1)^2(2x+4))=0
    2(x-1)^2(x+2))=0

    And now you see that the equality will hold if either x-1=0 or x+2=0, that is if x=1 or x=-2. Since you have a constraint x>0 the correct answer is x=1.
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  3. #3
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    Thank you! but why was that a mistake?
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  4. #4
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    Because you cannot be sure whether or not you carried out a division by zero! For an example, if you want to divide something with an expression (x-1) than you should point out that you exclude the possibility of x taking the value 1, thus you loose a possible solution, which exactly was the case here. And I suspect that wast the idea of the given problem, to see if one is aware of what I just said in the first two sentences of this post.
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  5. #5
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    Quote Originally Posted by Punch View Post
    Thank you! but why was that a mistake?
    As Mathoman said, you cannot divide by 0. What you could have done, rather than factor, was, at the point where you had
    3(x+ 1)(x- 1)^2= (x- 1)^3
    was think "If x- 1 is not 0, I can divide both sides by (x- 1)^2 to get 3(x+ 1)= x- 1, 3x+ 3= x- 1, 2x= -4, x= -2. But if x- 1= 0, then both sides would equal 0 so x= 1 is also a solution."

    If the condition "x> 0" had not been included, both x= -2 and x= 1 would have given stationary points. With that condition, only x= 1 does.
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