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Using the approximation, explain why the second derivative test works. |
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| Jul18-11, 10:14 AM | #1 |
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Using the approximation, explain why the second derivative test works.
[b]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 |
| Jul18-11, 11:05 AM | #2 |
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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? |
| Jul18-11, 11:11 AM | #3 |
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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 |
| Jul18-11, 11:14 AM | #4 |
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Recognitions:
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Using the approximation, explain why the second derivative test works.RGV |
| Jul18-11, 11:56 AM | #5 |
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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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