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Calculus and Beyond Homework Help
Find the D value of critical point and the type of critical point
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[QUOTE="coolusername, post: 4586766, member: 489841"] [h2]Homework Statement [/h2] The function f(x,y) = [e^(-y^2)]cos(4x) has a critical point (0,0) [h2]Homework Equations[/h2] Find the D value at the critical point. What type of critical point is it? (max, min, saddle or none) [h2]The Attempt at a Solution[/h2] I know that to find the D value I must compute the partial derivatives (fox, fey, fxy and fyx) fx = [-4e^(-y^2)]sin(4x) fy = [-2ye^(-y^2)]cos(4x) fxx = [-16e^(-y^2)](cos(4x) fyy = [4(y^2)e^(-y^2)]cos(4x) fxy = [8ye^(-y^2)]sin(4x) fxy = [8ye^(-y^2)]sin(4x) D = (fxx)(fyy) - (fxy)^2 = [-16e^(-y^2)](cos(4x)[4(y^2)e^(-y^2)]cos(4x) - {[8ye^(-y^2)]sin(4x)}^2 evaluated at the critical point (0,0) = (-16)(0) - (0)(0) = 0 This means that the 2nd derivative test gave no info. But somehow it's not the right answer. Did I find the correct value of D? Is my approach correct? Thanks [/QUOTE]
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Calculus and Beyond Homework Help
Find the D value of critical point and the type of critical point
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