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Using the approximation, explain why the second derivative test works. 
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#1
Jul1811, 10:14 AM

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[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 


#2
Jul1811, 11:05 AM

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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? 


#3
Jul1811, 11:11 AM

P: 13

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 


#4
Jul1811, 11:14 AM

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


#5
Jul1811, 11:56 AM

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Mathematical language is very exacting. You need to say what you mean & mean what you say.
f(x_{0}+Δx, y_{0}+Δy) is the (exact) value of the function, f, at the point (x_{0}+Δx, y_{0}+Δy). If the first derivatives of f are zero at the point, (x_{0}, y_{0}), then the following is an approximation to f at the point (x_{0}+Δx, y_{0}+Δy). f(x_{0}+Δx, y_{0}+Δy) ≈ f(x_{0}, y_{0}) + (1/2)[ f_{xx}(x_{0}, y_{0})(Δx)^{2} +2 f_{xy}(x_{0}, y_{0})(Δx)(Δy) + f_{yy}(x_{0}, y_{0})(Δy)^{2} ] 


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