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Math Help - Why can I not find the top point of this quadratic equation?

  1. #1
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    Why can I not find the top point of this quadratic equation?

    Given the equation:

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

    I want to find the x-value of the top point of this equation, which as a graph I've found to be x=\frac{1}{3}

    Why can I not find the top point of this quadratic equation?-graphtop.gif

    However when I try to find this value, I don't end up with \frac{1}{3}

    First I differentiate the equation:

    f'(x)=\frac{d}{dx}1+x-\frac{x^2}{1}-\frac{x^3}{1}=1-2\cdot x-3\cdot x^2

    Now I put the differentiated equation equal to 0 and isolate x:

    1-2\cdot x-3\cdot x^2=0
    x=-1

    Why am I not getting x=\frac{1}{3}? According to my graph, that is the correct value?
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  2. #2
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    Re: Why can I not find the top point of this quadratic equation?

    Hi!

    Quote Originally Posted by MathIsOhSoHard View Post
    Given the equation:

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

    I want to find the x-value of the top point of this equation, which as a graph I've found to be x=\frac{1}{3}

    Click image for larger version. 

Name:	graphtop.gif 
Views:	3 
Size:	5.6 KB 
ID:	23730

    However when I try to find this value, I don't end up with \frac{1}{3}

    First I differentiate the equation:

    f'(x)=\frac{d}{dx}1+x-\frac{x^2}{1}-\frac{x^3}{1}=1-2\cdot x-3\cdot x^2

    Now I put the differentiated equation equal to 0 and isolate x:

    1-2\cdot x-3\cdot x^2=0
    x=-1
    How did you do that?

    Quote Originally Posted by MathIsOhSoHard View Post
    Why am I not getting x=\frac{1}{3}? According to my graph, that is the correct value?
    You can find out by substituting 1/3 for x

    =1-2\cdot (\frac{1}{3})-3\cdot (\frac{1}{3})^2

    =1- (\frac{2}{3})-3\cdot \frac{1}{9}

    =1- (\frac{2}{3})-\cdot \frac{3}{9}

    =1- (\frac{2}{3})-\cdot \frac{1}{3} = 0

    So x=1/3 is a solution of the quadratic equation.

    By the way are you sure x=1/3 is the one and only solution?
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  3. #3
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    Re: Why can I not find the top point of this quadratic equation?

    Quote Originally Posted by Quant View Post
    By the way are you sure x=1/3 is the one and only solution?
    No, there are probably more solutions to it. However I really want to know how to get to 1/3.
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  4. #4
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    Re: Why can I not find the top point of this quadratic equation?

    Your quadratic factorises (1+x)(1-3x)=0 So x=1/3 or -1
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  5. #5
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    Re: Why can I not find the top point of this quadratic equation?

    There are many ways to get this solutions. One of them:

    You need to solve

     1-2x-3x^2 = 0

    We devide by -3

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

    We add a zero

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

    Some rearranging

     -\frac{1}{3}+\frac{2}{3}x+(\frac{2}{6})^2+x^2 = +(\frac{2}{6})^2


     -\frac{1}{3}+\frac{2}{3}x+(\frac{2}{6})^2+x^2 = \frac{4}{36}

     \frac{2}{3}x+(\frac{2}{6})^2+x^2 = \frac{4}{36} +\frac{1}{3}

     \frac{2}{3}x+(\frac{2}{6})^2+x^2 = \frac{4}{36} +\frac{1}{3}

     \frac{2}{3}x+(\frac{1}{3})^2+x^2 = \frac{4}{36} +\frac{1}{3}

    As you know it is  (a+b)^2 = a^2+2ab+b^2 (do you know this theoreme? Respectively do you see how we find a and b?)

    thus

     \frac{2}{3}x+(\frac{2}{6})^2+x^2

    is the same as

     (x+\frac{1}{3})^2

    Using this in  \frac{2}{3}x+(\frac{1}{3})^2+x^2 = \frac{4}{36} +\frac{1}{3}

    leads us to

     (x+\frac{1}{3})^2 = \frac{4}{36} +\frac{1}{3}

    and so on...

     (x+\frac{1}{3})^2 = \frac{1}{9} +\frac{1}{3}

     (x+\frac{1}{3})^2 = \frac{1}{9} +\frac{3}{9}

     (x+\frac{1}{3})^2 = \frac{4}{9}


     x+\frac{1}{3} = \sqrt{\frac{4}{9}}

     x+\frac{1}{3} = \pm \frac{2}{3}

     x = -\frac{1}{3} \pm \frac{2}{3}

    so -1 and 1/3 are the solutions you are looking for
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