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Find inflection point?

  1. Jul 11, 2010 #1
    1. The problem statement, all variables and given/known data

    y=2√x-x


    3. The attempt at a solution

    First derivative:
    -2x-2-1
    Second derivative:
    4x-3

    4x-3=0
    No solutions?
     
  2. jcsd
  3. Jul 11, 2010 #2

    Mentallic

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    [tex]\sqrt{x}=x^{1/2}[/tex]

    So [tex]y=2x^{1/2}-x[/tex]

    Then [tex]\frac{dy}{dx}=x^{-1/2}-1[/tex]

    Which isn't what you got.
     
  4. Jul 11, 2010 #3

    Mentallic

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    But even so, if you correctly find the second derivative you'll see that it cannot equal 0, which means there is no inflection point. This means the graph of y=f(x) doesn't ever go from concave up to concave down or vice versa, it is always just one. Can you figure out whether it is always concave up or down without looking at the graph?
     
  5. Jul 11, 2010 #4
    So the first derivative is
    (x-1/2-1)
    I take a second derivative to find an inflection point.
    y''=-1/2x-3/2 Hence, x=0...

    What does it mean?...
     
  6. Jul 11, 2010 #5
    [tex]y'' = -\frac{1}{2}x^{-\frac{3}{2}} = -\frac{1}{2x^{3/2}}[/tex]

    x = 0 isn't in the domain because you would have 0 in the denominator, but that's not really important. This function y'' is never 0, it will never touch the x-axis, so y does not have an inflection point.
     
  7. Jul 11, 2010 #6
    Yes, I see I was wrong, I cannot have x=0 as a solution in this function.
     
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