...: 3x(x+2) = 0 --> x = 0, -2 3y(y-2) = 0 --> y = 0, 2 four critical points : (0, 0) saddle point (0, 2) local minimum (-2, 0) local maximum (-2, 2) saddle point
f(x,y) = (x+y)e^{1-x^2-y^2} 對 x 偏微分, 就是把 y 看成是常數, 所以 f_x(x,y) = e^{1-x^2-y^2} + (x+y)e^{1-x^2-y^2}(-2x) = [1-2x(x+y)]e^{1-x^2-y^2} 對 y 偏微, 是把 x 看成是常數, 因此 f_y(x,y) = e^{1-x^2-y^2} + (x+y)e^{1-x^2-y^2}(-2y) = [1-2y(x+y)]e^{1-x^2-y^2} 臨界點, 在函數處處可微的情況如本例, 就是 f_x(x,y)=0=f_y(x,y) 而由於 e...
1. First find its critical points where f_x=0 and f_y=0. i.e 2xy-2y=0 and x^2-2x-3+2y...1<0 so (3,0,f(3,0)) is a saddle point , not local extremal point ; D(-1,0)-16<...
1.Find critical points or f(x)=x^(-1/3)+ x^(2/3)x≠0 for f(x)=∞f(1/2...23.f(x)=(x-2)/(x+1)f(-1,Left)=∞f(-1,right)=-∞ Critical point : x=-14.f(x)=x*Ln(x)>0f'(x)=1+Ln...
參考看看吧!! 圖片參考：http://i580.photobucket.com/albums/ss244/linch_1/picture-398.jpg http://i580.photobucket.com/albums/ss244/linch_1/picture-398.jpg
x^2-y^2+4x-8y-11 為什麼答案是 no critical point 有saddle point (-2,-4,1)? 令 f(x,y) = x^2-y^2+4x...0, 依雙變數極值之第2階導數測驗, f(x,y) 在 critical point (-2,-4) 有一 saddle point , (-2,-4,f(-2,-4)) = (-2,-4...
...x4- 8x2+ 3y2- 6yfx(x,y)=4x3-16xfy(x,y)=6y-6f之 critical points 滿足4x3-16x=0和6y-6=0解得(x,y)=(0,1),(-2...y3- 3xy + 5 fx(x,y)=3x2-3yfy(x,y)=3y2-3xf之 critical points 滿足3x2-3y=0和3y2-3x=0解得(x,y)=(0,0),(1...
...1)=3(x+2)-9. Letting f '(x)=0 then (x+2)=3 so that x =-2√3 ( critical points ). f(x)=x+3x-15x-9 ; f '(x)=3x+6x-15=3(x+2x-5)=3(x+1)-18. Letting...
第一題,首先注意到 domain of f ={x|x 大於,等於0}.所以 x = 0 為 critical point . (詳解如下): 圖片參考：http://img225.imageshack.us/img225/5380/24981831gl5.jpg...
