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Loc Max, Min, increase & decrease!

  1. Jul 26, 2004 #1
    Hello, I have a question about these things

    What is the local Max, local Min, when is the function increasing and decreasing!

    the function is:

    G(x)= x - 4 sqrt[x]
     
  2. jcsd
  3. Jul 26, 2004 #2

    HallsofIvy

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    Why do you have questions about them? Is this homework? If so, it should be posted in the homework section and you should show what you have done so we will know what kind of hints will help.

    Here, I suggest that you write the function as G(x)= x- 4x2 (Is [x] just "parentheses" or "greatest integer function"? I'm assuming it's just parens.)

    The maximum value will occur where G'= 0 or G' does not exist (look carefully at x=0).
    The function is increasing where G'> 0 and decreasing where G'< 0.

    If this is really x- 4 sqrt("greatest integer less than or equal to x"), then it is not differentiable. I would recommend you graph it carefully.
     
  4. Jul 26, 2004 #3
    and a local minima when G ' ' > 0 and local maxima when G ' ' < 0, point of infelction when G ' ' = 0
     
  5. Jul 26, 2004 #4

    ShawnD

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    First find the domain of the function. You know the function is only real when x >= 0 (or at least in my reality :wink: )

    Secondly, find the derivative. I got this as the derivative:

    [tex]1 - 2x^{\frac{-1}{2}}[/tex]

    Just by looking at the equation you were given, you know G(x) is going to be negative at low x values. Since it starts decreasing, the local maximum is 0. To find the local minimum, set the derivative G'(x) equal to 0, solve for x, then fill that x value into your original formula G(x).

    To find when the function is increasing or decreasing, substitute x values into the derivative. Sub in an X value slightly less than where the derivative equals 0, then sub in an X value slightly more than where the derivative equals 0.

    I got these answers:
    Local Max G(x) = 0
    Local Min G(x) = -4
    Decreasing when 0 > x > 4
    Increasing when 4 > x > infinity
     
  6. Jul 26, 2004 #5

    Zurtex

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    Not always. For example G = x^6, at x = 0 there is a minimum yet G ' ' = 0 at x = 0.
     
  7. Jul 26, 2004 #6

    ShawnD

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    That's still an inflection point though.
     
  8. Jul 26, 2004 #7

    Zurtex

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    It is? :confused:

    I seem to have the wrong idea on what an inflection is then, could you please explain.
     
  9. Jul 26, 2004 #8

    ShawnD

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    The definition is somewhat subjective. Inflection is sometimes defined as simply when the second derivative is 0, but it can also be when concavity changes. For X^6, the second derivative is 0, but the concavity does not change.
    Depends on definition I guess.
     
  10. Aug 13, 2004 #9

    mathwonk

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    Interesting point. In real calculus I am used to defining an inflection point as one where curvature changes sign, but in complex calculus where that presumably makes less sense, I define inflection point as a point where the tangent line has order of contact higher than two, hence in this case where the second derivative vanishes also. I never realized before I am using different definitions in different settings.
     
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