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Finding the Inflection points of x^2-4√x

  1. Apr 8, 2016 #1
    1. The problem statement, all variables and given/known data
    find the inflection points of x^2-4√x

    2. Relevant equations


    3. The attempt at a solution
    Okay, I started with finding the derivatives;
    f'(x)=2x-2/√x
    f''(x)=2+1/√x^3
    and made the second derivative =0
    (2+1/√x^3=0)(√x^3)
    2√x^3+1=0
    (√x^3=-1/2)^2
    x^3=1/4
    x=cube root(1/4)
    x=0.63
    But when I enter '0.63' into the second derivative i get '4' (2+1/(√0.63^3)=4) Should i not get 0? Does this mean there are no inflection points?
     
  2. jcsd
  3. Apr 8, 2016 #2

    andrewkirk

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    It's unclear what you are doing here:
    Note that for the given expression x^2-4√x to be a function, we need to choose a branch of the square root. The convention is to choose the positive branch unless explicitly stated otherwise. Then when we get to (√x)^3=-1/2 we see that there is no solution, because in the positive branch (√x)^3>0. So there are no inflection points.

    The solution you found is for the negative branch of the square root, which does have an inflection point.
     
  4. Apr 9, 2016 #3

    ehild

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    It was a real function, the domain is x>0. That means, the second derivative can not be zero. When you take the square of the equation, you introduce false root.
     
  5. Apr 9, 2016 #4

    Ray Vickson

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    You have ##f(x) = f_1(x) + f_2(x)##, where ##f_1(x) = x^2## and ##f_2(x) = -2 \sqrt{x}##. We have ##f_1''(x)>0## and ##f_2''(x)>0## on ##\mathbb{R}_+ = \{ x > 0 \}##, so there are no inflection points. Basically, ##f## is the sum of two strictly convex functions on ##\mathbb{R}_+##, so is itself strictly convex on that set.
     
  6. Apr 10, 2016 #5
    great, thanks
     
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