Using the approximation, explain why the second derivative test works.

  1. 1. The problem statement, all variables and given/known data[/
    Using the approximation, explain why the second derivative test works

    approximation=f(x0+delta x, y0+delta y)

    delta x and delta y are small...


    2. Relevant equations

    f(x0+delta x,y0+delta y)

    3. The attempt at a solution

    ok so i know the first derivative of it is:

    fx(x0+y0)*delta x+fy(x0+y0)*delta y

    and second derivative is:

    fxx(x0,y0)*delta x^2+fxy(x0,y0)*delta x*delta y+fyy(x0,y0)*delta y^2

    it justs seems like since delta x and delta y are getting smaller,that everything will be going towards 0....so why does it work?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. SammyS

    SammyS 8,067
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    What is the problem as written?

    f appears to be a function of two variables. What is the second derivative test for a function of two variables?
     
  4. Using the approximation, explain why the second derivative works. Give three exam-
    ples for each scenario of the second derivative test.

    isnt that what the approximation is? the f(x+delta x,y +delta y)?

    its asking about finding local mins,local max and saddle points....now i can show those examples,but how can i explain how it works?its like a proof or something
     
  5. Ray Vickson

    Ray Vickson 6,019
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    You don't say what you are testing for. If you mean the second-order test for a maximum or minimum, you still need to be more specific: the necessary conditions and the (most common) sufficient conditions are a bit different. You need to tell us which ones you want.

    RGV
     
  6. SammyS

    SammyS 8,067
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    Mathematical language is very exacting. You need to say what you mean & mean what you say.

    f(x0+Δx, y0+Δy) is the (exact) value of the function, f, at the point (x0+Δx, y0+Δy).

    If the first derivatives of f are zero at the point, (x0, y0), then the following is an approximation to f at the point (x0+Δx, y0+Δy).

    f(x0+Δx, y0+Δy) ≈ f(x0, y0) + (1/2)[ fxx(x0, y0)(Δx)2 +2 fxy(x0, y0)(Δx)(Δy) + fyy(x0, y0)(Δy)2 ]
     
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