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Extrema of Quadratic functions

  1. Oct 31, 2013 #1

    Qube

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    1. The problem statement, all variables and given/known data

    Does every quadratic function have a relative extrema?

    2. Relevant equations

    Quadratic function: ax^2 + bx + c. Aka a polynomial.

    Polynomials are continuous through all real numbers.

    3. The attempt at a solution

    It seems as if all quadratic functions would have a relative extrema since the basic shape of a quadratic function is U-shaped and it's graphically obvious that the second derivative changes sign; all quadratic functions have a vertex whose x coordinate is given by -b/2a and this vertex is also the location of a horizontal tangent line and always represents either the max or min of the quadratic function (depending on orientation). And taking the derivative of the general form of a quadratic function yields 2ax + b where a and b are constants and it would appear that one can easily make the derivative both positive and negative given that the domain of quadratic functions is all real numbers.

    Are there any exceptions? (I'm guessing no).
     
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  3. Oct 31, 2013 #2

    mfb

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    Right

    No. Just make sure that "quadratic functions" implies a!=0 in your formula (otherwise it is not a quadratic function).
     
  4. Oct 31, 2013 #3

    vela

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    The second derivative is constant. You mean the first derivative changes sign.

    By the way, extrema is the plural of extremum. You should say that quadratic functions have a relative extremum.
     
  5. Oct 31, 2013 #4

    Qube

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    Right, and looking at the first derivative, it can only be positive since it's the product of three squared terms. So there can be no sign change regardless.
     
  6. Oct 31, 2013 #5

    mfb

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    I guess this belongs to your other thread.
     
  7. Oct 31, 2013 #6

    Mark44

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    ???
    "it" = what?
    If y = ax2 + bx + c, then y' = 2ax + b

    Where are you getting the three squared terms? mfb seems to know, but I don't recall seeing that other thread.

    If "it" refers to y', the derivative will always change sign in a quadratic function.

    If "it" refers to y'', that's 2a, so I still don't see where the three squared terms business comes in.
     
  8. Oct 31, 2013 #7

    Qube

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    Whoa LOL yes this belongs in the other thread. Confused. My apologies.
     
  9. Nov 1, 2013 #8

    mfb

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