二元函数f(x,y)=4x-4y-x^2-y^2的驻点是
A: (2,2)
B: (-2,2)
C: (2,-2)
D: (-2,-2)
A: (2,2)
B: (-2,2)
C: (2,-2)
D: (-2,-2)
举一反三
- 二元函数f(x,y)=4x-4y-x^2-y^2的驻点是
- 在环形区域[img=136x26]18030733be53638.png[/img]上， 绘制函数图形[img=129x27]18030733c6c9cd6.png[/img] A: Plot3D[x^2+y^2,{x,-2,2},{y,-2,2},Exclusions→Function[{x,y},0.5<x^2+y^2<2]] B: Plot3D[x^2+y^2,{x,-2,2},{y,-2,2},RegionFunction→Function[{x,y},0.5<x^2+y^2<2]] C: Plot3D[x^2+y^2,{x,-2,2},{y,-2,2},RegionFunction→Function[{x,y},2>x^2+y^2>0.5]] D: Plot3D[x^2+y^2,{y,-2,2},{x,-2,2},Exclusions→Function[{x,y},0.5<x^2+y^2<2]]
- 求解方程组[img=218x63]1803072f0e0e849.png[/img]接近 (2,2) 的解 A: FindRoot[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,2},{y,2}] B: NSolve[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,2},{y,2}] C: FindRoot[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,y},{2,2}] D: FindRoots[{x^2+y^2=5Sqrt[x^2+y^2]-4x,y=x^2},{x,2},{y,2}]
- 求解方程组[img=218x63]1803072e5daced1.png[/img]接近 (2,2) 的解 A: NSolve[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,2},{y,2}] B: FindRoot[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,2},{y,2}] C: FindRoot[{x^2+y^2==5Sqrt[x^2+y^2]-4x,y==x^2},{x,y},{2,2}] D: FindRoots[{x^2+y^2=5Sqrt[x^2+y^2]-4x,y=x^2},{x,2},{y,2}]
- 若函数$f(x)$具有二阶导数，且$y=f({{x}^{2}})$，则$y'' =$( )。 A: $f'' ({{x}^{2}})$ B: $2f'’ ({{x}^{2}})$ C: $2f’ ({{x}^{2}})+4{{x}^{2}}f’' ({{x}^{2}})$ D: $4{{x}^{2}}f’ ({{x}^{2}})+2f'' ({{x}^{2}})$