高等数学基础归档 - 第26页共55页

问题

根据拉格朗日乘数法，若要求函数

u

=

f (x, y, z)

在

φ_{1} (x, y, z)

=

0

和

φ_{2} (x, y, z)

=

0

两个条件约束下的极值，如何构造拉格朗日函数

F (x, y, z)

选项

[A].

F (x, y, z)

=

λ

f (x, y, z)

+

λ_{1}

φ_{1} (x, y, z)

+

λ_{2}

φ_{2} (x, y, z)

[B].

F (x, y, z)

=

f (x, y, z)

-

λ_{1}

φ_{1} (x, y, z)

-

λ_{2}

φ_{2} (x, y, z)

[C].

F (x, y, z)

=

f (x, y, z)

+

λ_{1}

φ_{1} (x, y, z)

+

λ_{2}

φ_{2} (x, y, z)

[D].

F (x, y, z)

=

f (x, y, z)

+

\frac{1}{λ_{1}}

φ_{1} (x, y, z)

+

\frac{1}{λ_{2}}

φ_{2} (x, y, z)

答案

$F (x, y, z)$ $=$ $f (x, y, z)$ $+$ $λ_{1}$ $φ_{1} (x, y, z)$ $+$ $λ_{2}$ $φ_{2} (x, y, z)$

问题

根据拉格朗日乘数法，若要求函数

u

=

f (x, y, z)

在

φ (x, y, z)

=

0

条件约束下的极值，如何构造拉格朗日函数

F (x, y, z)

选项

[A].

F (x, y, z)

=

λ

f (x, y, z)

+

φ (x, y, z)

[B].

F (x, y, z)

=

f (x, y, z)

-

λ

φ (x, y, z)

[C].

F (x, y, z)

=

f (x, y, z)

+

φ (x, y, z)

[D].

F (x, y, z)

=

f (x, y, z)

+

λ

φ (x, y, z)

答案

$F (x, y, z)$ $=$ $f (x, y, z)$ $+$ $λ$ $φ (x, y, z)$

问题

根据拉格朗日乘数法，若要求函数

z

=

f (x, y)

在

φ (x, y)

=

0

条件约束下的极值，如何构造拉格朗日函数

F (x, y)

选项

[A].

F (x, y)

=

λ

f (x, y)

+

φ (x, y)

[B].

F (x, y)

=

f (x, y)

-

λ

φ (x, y)

[C].

F (x, y)

=

f (x, y)

+

λ

φ (x, y)

[D].

F (x, y)

=

f (x, y)

+

\frac{1}{λ}

φ (x, y)

答案

$F (x, y)$ $=$ $f (x, y)$ $+$ $λ$ $φ (x, y)$

极值存在的充分条件：判别公式中的 $A$ , $B$ , $C$ 都是多少？（B013）

问题

若已知函数

z

=

f (x, y)

在点

(x_{0}, y_{0})

的某邻域内有连续的二阶偏导数,且

f_{x}^{'} (x_{0}, y_{0})

=

0

f_{y}^{'} (x_{0}, y_{0})

=

0

; 则极值判别公式

A C

-

B^{2}

中的

A

B

和

C

各等于多少？

选项

[A].

{\begin{cases} A = f_{x}^{'} (x_{0}, y_{0}) \\ B = f_{x y}^{''} (x_{0}, y_{0}) \\ C = f_{y}^{'} (x_{0}, y_{0}) \end{cases}

[B].

{\begin{cases} A = f_{y x}^{''} (x_{0}, y_{0}) \\ B = f_{x y}^{''} (x_{0}, y_{0}) \\ C = f_{y y}^{''} (x_{0}, y_{0}) \end{cases}

[C].

{\begin{cases} A = f_{y y}^{''} (x_{0}, y_{0}) \\ B = f_{x y}^{''} (x_{0}, y_{0}) \\ C = f_{x x}^{''} (x_{0}, y_{0}) \end{cases}

[D].

{\begin{cases} A = f_{x x}^{''} (x_{0}, y_{0}) \\ B = f_{x y}^{''} (x_{0}, y_{0}) \\ C = f_{y y}^{''} (x_{0}, y_{0}) \end{cases}

答案

${\begin{cases} A = f_{x x}^{''} (x_{0}, y_{0}) \\ B = f_{x y}^{''} (x_{0}, y_{0}) \\ C = f_{y y}^{''} (x_{0}, y_{0}) \end{cases}$

问题

若已知函数

z

=

f (x, y)

在点

(x_{0}, y_{0})

的某邻域内有连续的二阶偏导数,且

f_{x}^{'} (x_{0}, y_{0})

=

0

f_{y}^{'} (x_{0}, y_{0})

=

0

;

A

=

f_{x x}^{''} (x_{0}, y_{0})

B

=

f_{x y}^{''} (x_{0}

y_{0})

C

=

f_{y y}^{''} (x_{0}, y_{0})

则以下哪个选项可以说明点 $(x_{0}, y_{0})$ 为函数 $z$ $=$ $f (x, y)$ 的一个极值大点或极小值点？

选项

[A].

A C

-

B^{2}

>

0

\Rightarrow

{\begin{cases} A > 0 \Rightarrow 极 小 值 点 \\ A < 0 \Rightarrow 极 大 值 点 \end{cases}

[B].

A C

-

B^{2}

<

0

\Rightarrow

{\begin{cases} A > 0 \Rightarrow 极 小 值 点 \\ A < 0 \Rightarrow 极 大 值 点 \end{cases}

[C].

A C

-

B^{2}

>

0

\Rightarrow

{\begin{cases} A > 1 \Rightarrow 极 小 值 点 \\ A < 1 \Rightarrow 极 大 值 点 \end{cases}

[D].

A C

-

B^{2}

>

0

\Rightarrow

{\begin{cases} A < 0 \Rightarrow 极 小 值 点 \\ A > 0 \Rightarrow 极 大 值 点 \end{cases}

答案

$A C$ $-$ $B^{2}$ $>$ $0$ $\Rightarrow$ ${\begin{cases} A > 0 \Rightarrow 极小值点 \\ A < 0 \Rightarrow 极大值点 \end{cases}$

问题

若已知函数

z

=

f (x, y)

在点

(x_{0}, y_{0})

的某邻域内有连续的二阶偏导数,且

f_{x}^{'} (x_{0}, y_{0})

=

0

f_{y}^{'} (x_{0}, y_{0})

=

0

;

A

=

f_{x x}^{''} (x_{0}, y_{0})

B

=

f_{x y}^{''} (x_{0}

y_{0})

C

=

f_{y y}^{''} (x_{0}, y_{0})

则以下哪个选项可以说明点 $(x_{0}, y_{0})$ 为函数 $z$ $=$ $f (x, y)$ 的一个极值点？

选项

[A].

A C

-

B^{2}

=

0

[B].

B C

-

A^{2}

<

0

[C].

A C

-

B^{2}

<

0

[D].

A B

-

C^{2}

>

0

[E].

A C

-

B^{2}

>

0

[F].

A B

-

C^{2}

=

0

答案

$A C$ $-$ $B^{2}$ $>$ $0$ $\Rightarrow$ 点 $(x_{0}, y_{0})$ 是极值点.

$A C$ $-$ $B^{2}$ $<$ $0$ $\Rightarrow$ 点 $(x_{0}, y_{0})$ 不是极值点.

$A C$ $-$ $B^{2}$ $=$ $0$ $\Rightarrow$ 不确定点 $(x_{0}, y_{0})$ 是否是极值点.

问题

设

z = f (x, y)

在点

(x_{0}, y_{0})

的一阶偏导数存在, 且

(x_{0}, y_{0})

是

z =

f (x, y)

的极值点, 则可以推出以下哪个选项所示的结论？

选项

[A].

{\frac{\partial z}{\partial x} |}_{(x, y)}

=

0

{\frac{\partial z}{\partial y} |}_{(x, y)}

=

0

[B].

{\frac{\partial z}{\partial x} |}_{(x_{0}, y_{0})}

\neq

{\frac{\partial z}{\partial y} |}_{(x_{0}, y_{0})}

[C].

{\frac{\partial z}{\partial x} |}_{(x_{0}, y_{0})}

=

1

{\frac{\partial z}{\partial y} |}_{(x_{0}, y_{0})}

=

1

[D].

{\frac{\partial z}{\partial x} |}_{(x_{0}, y_{0})}

=

0

{\frac{\partial z}{\partial y} |}_{(x_{0}, y_{0})}

=

0

答案

${\frac{\partial z}{\partial x} |}_{(x_{0}, y_{0})}$ $=$ $0$ , ${\frac{\partial z}{\partial y} |}_{(x_{0}, y_{0})}$ $=$ $0$

问题

以下哪个选项可以说明点

(x_{0}, y_{0})

是二元函数

z

=

f (x, y)

的极大值点（或极小值点）？

选项

[A]. 在点

(x_{0}, y_{0})

的邻域内有多个异于点

(x_{0}, y_{0})

的点

(x, y)

使得

f (x, y) < f (x_{0}, y_{0})

（或

f (x, y)

>

f (x_{0}, y_{0})

）成立

[B]. 在点

(x_{0}, y_{0})

的邻域内有两个异于点

(x_{0}, y_{0})

的点

(x, y)

使得

f (x, y) < f (x_{0}, y_{0})

（或

f (x, y)

>

f (x_{0}, y_{0})

）成立

[C]. 在点

(x_{0}, y_{0})

的邻域内任意一个异于点

(x_{0}, y_{0})

的点

(x, y)

, 都使得

f (x, y) < f (x_{0}, y_{0})

（或

f (x, y)

>

f (x_{0}, y_{0})

）成立

[D]. 在点

(x_{0}, y_{0})

的邻域内有一个异于点

(x_{0}, y_{0})

的点

(x, y)

使得

f (x, y) < f (x_{0}, y_{0})

（或

f (x, y)

>

f (x_{0}, y_{0})

）成立

答案

在点 $(x_{0}, y_{0})$ 的邻域内任意一个异于点 $(x_{0}, y_{0})$ 的点 $(x, y)$ , 都使得 $f (x, y) < f (x_{0}, y_{0})$ （或 $f (x, y)$ $>$ $f (x_{0}, y_{0})$ ）成立 $\Rightarrow$ $f (x_{0}, y_{0})$ 是函数 $z$ $=$ $f (x, y)$ 的一个极大值（或极小值）

三元隐函数的复合函数求导法则（B012）

问题

设由方程组

{\begin{cases} F (x, y, z) = 0 \\ G (x, y, z) = 0 \end{cases}

确定的隐函数为

y

=

y (x)

与

z

=

z (x)

, 则

\frac{d y}{d x}

和

\frac{d z}{d x}

可以通过解以下哪个线性方程组求出？

选项

[A].

{\begin{cases} F_{x} + F_{y}^{'} \frac{d y}{d x} + F_{x}^{'} \frac{d z}{d x} = 0 \\ G_{x} + G_{y}^{'} \frac{d y}{d x} + G_{z}^{'} \frac{d z}{d x} = 0 \end{cases}

[B].

{\begin{cases} F_{z}^{'} + F_{x}^{'} \frac{d y}{d x} + F_{y}^{'} \frac{d z}{d x} = 0 \\ G_{z}^{'} + G_{x}^{'} \frac{d y}{d x} + G_{y}^{'} \frac{d z}{d x} = 0 \end{cases}

[C].

{\begin{cases} F_{x}^{'} + F_{y}^{'} \frac{d z}{d x} + F_{x}^{'} \frac{d y}{d x} = 0 \\ G_{x}^{'} + G_{y}^{'} \frac{d z}{d x} + G_{z}^{'} \frac{d y}{d x} = 0 \end{cases}

[D].

{\begin{cases} F_{x}^{'} + F_{y}^{'} \frac{d y}{d x} + F_{z}^{'} \frac{d z}{d x} = 0 \\ G_{x}^{'} + G_{y}^{'} \frac{d y}{d x} + G_{z}^{'} \frac{d z}{d x} = 0 \end{cases}

答案

${\begin{cases} F_{x}^{'} + F_{y}^{'} \frac{d y}{d x} + F_{z}^{'} \frac{d z}{d x} = 0 \\ G_{x}^{'} + G_{y}^{'} \frac{d y}{d x} + G_{z}^{'} \frac{d z}{d x} = 0 \end{cases}$

三元复合函数求导法则（B012）

问题

已知函数

F (x, y, z)

=

0

, 若

F_{z}^{'}

\neq

0

, 则

\frac{\partial z}{\partial x}

=

?

\frac{\partial z}{\partial y}

=

?

选项

[A].

\frac{\partial z}{\partial x}

=

-

\frac{F_{x}^{'} (x, y, z)}{F_{z} (x, y, z)}

\frac{\partial z}{\partial y}

=

-

\frac{F_{y}^{'} (x, y, z)}{F_{z} (x, y, z)}

[B].

\frac{\partial z}{\partial x}

=

\frac{F_{x}^{'} (x, y, z)}{F_{z}^{'} (x, y, z)}

\frac{\partial z}{\partial y}

=

\frac{F_{y}^{'} (x, y, z)}{F_{z}^{'} (x, y, z)}

[C].

\frac{\partial z}{\partial x}

=

-

\frac{F_{x}^{'} (x, y, z)}{F_{z}^{'} (x, y, z)}

\frac{\partial z}{\partial y}

=

-

\frac{F_{y}^{'} (x, y, z)}{F_{z}^{'} (x, y, z)}

[D].

\frac{\partial z}{\partial x}

=

-

\frac{F_{z}^{'} (x, y, z)}{F_{x}^{'} (x, y, z)}

\frac{\partial z}{\partial y}

=

-

\frac{F_{z}^{'} (x, y, z)}{F_{y}^{'} (x, y, z)}

答案

$\frac{\partial z}{\partial x}$ $=$ $-$ $\frac{F_{x}^{'} (x, y, z)}{F_{z}^{'} (x, y, z)}$ , $\frac{\partial z}{\partial y}$ $=$ $-$ $\frac{F_{y}^{'} (x, y, z)}{F_{z}^{'} (x, y, z)}$

二元复合函数求导法则（B012）

问题

已知函数

F (x, y)

=

0

, 若

F_{y}^{'}

\neq

0

, 则

\frac{d y}{d x} =

?

选项

[A].

\frac{d y}{d x} =

\frac{F_{y}^{'} (x, y)}{F_{x}^{'} (x, y)}

[B].

\frac{d y}{d x} =

- \frac{F_{y}^{'} (x, y)}{F_{x}^{'} (x, y)}

[C].

\frac{d y}{d x} =

\frac{F_{x}^{'} (x, y)}{F_{y}^{'} (x, y)}

[D].

\frac{d y}{d x} =

- \frac{F_{x}^{'} (x, y)}{F_{y}^{'} (x, y)}

答案

$\frac{d y}{d x} =$ $- \frac{F_{x}^{'} (x, y)}{F_{y}^{'} (x, y)}$

二元三重复合函数求导法则（B012）

问题

设函数

z

=

f (x, u, v)

u

=

φ (x, y)

v

=

ψ (x, y)

, 则

\frac{\partial z}{\partial x}

=

?

\frac{\partial z}{\partial y}

=

?

选项

[A].

{\begin{cases} \frac{\partial z}{\partial x} = \frac{d f}{d x} + \frac{d f}{d u} \cdot \frac{\partial u}{\partial x} + \frac{d f}{d v} \cdot \frac{\partial v}{\partial x} \\ \frac{\partial z}{\partial y} = \frac{d f}{d u} \cdot \frac{\partial u}{\partial y} + \frac{d f}{d v} \cdot \frac{\partial v}{\partial y} \end{cases}

[B].

{\begin{cases} \frac{\partial z}{\partial x} = \frac{\partial f}{\partial x} \cdot \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x} \cdot \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial x} \\ \frac{\partial z}{\partial y} = \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial y} \cdot \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial y} \end{cases}

[C].

{\begin{cases} \frac{\partial z}{\partial x} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x} \cdot \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial x} \\ \frac{\partial z}{\partial y} = \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial y} \cdot \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial y} \end{cases}

[D].

{\begin{cases} \frac{\partial z}{\partial x} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial x} \\ \frac{\partial z}{\partial y} = \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial y} \end{cases}

答案

${\begin{cases} \frac{\partial z}{\partial x} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial x} \\ \frac{\partial z}{\partial y} = \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial y} \end{cases}$

二元二重复合函数求导法则（B012）

问题

设函数

z

=

f (u, v)

u

=

φ (x, y)

v

=

ψ (x, y)

, 则

\frac{\partial z}{\partial x}

=

?

\frac{\partial z}{\partial y}

=

?

选项

[A].

{\begin{cases} \frac{\partial z}{\partial x} = \frac{d z}{d u} \cdot \frac{d u}{d x} + \frac{d z}{d v} \cdot \frac{d v}{d x}, \\ \frac{\partial z}{\partial y} = \frac{d z}{d u} \cdot \frac{d u}{d y} + \frac{d z}{d v} \cdot \frac{d v}{d y} . \end{cases}

[B].

{\begin{cases} \frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x} \cdot \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial x}, \\ \frac{\partial z}{\partial y} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial y} \cdot \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial y} . \end{cases}

[C].

{\begin{cases} \frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \cdot \frac{d u}{d x} + \frac{\partial z}{\partial v} \cdot \frac{d v}{d x}, \\ \frac{\partial z}{\partial y} = \frac{\partial z}{\partial u} \cdot \frac{d u}{d y} + \frac{\partial z}{\partial v} \cdot \frac{d v}{d y} . \end{cases}

[D].

{\begin{cases} \frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial x}, \\ \frac{\partial z}{\partial y} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial y} . \end{cases}

答案

${\begin{cases} \frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial x}, \\ \frac{\partial z}{\partial y} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial z}{\partial v} \cdot \frac{\partial v}{\partial y} . \end{cases}$

一元二重复合函数求导法则（B012）

问题

设函数

z

=

f (u, v)

u

=

φ (x)

v

=

ψ (x)

, 则

\frac{d z}{d x}

=

?

选项

[A].

\frac{d z}{d x}

=

\frac{\partial z}{\partial u}

+

\frac{\partial z}{\partial v}

[B].

\frac{d z}{d x}

=

\frac{\partial z}{\partial u}

\cdot

\frac{d u}{d x}

\cdot

\frac{\partial z}{\partial v}

\cdot

\frac{d v}{d x}

[C].

\frac{d z}{d x}

=

\frac{\partial z}{\partial u}

\cdot

\frac{d u}{d x}

+

\frac{\partial z}{\partial v}

\cdot

\frac{d v}{d x}

[D].

\frac{d z}{d x}

=

\frac{\partial z}{\partial u}

\cdot

\frac{\partial u}{\partial x}

+

\frac{\partial z}{\partial v}

\cdot

\frac{\partial v}{\partial x}

答案

$\frac{d z}{d x}$ $=$ $\frac{\partial z}{\partial u}$ $\cdot$ $\frac{d u}{d x}$ $+$ $\frac{\partial z}{\partial v}$ $\cdot$ $\frac{d v}{d x}$

偏导数存在与可微之间的关系（B012）

问题

已知，若函数

z

=

f (x, y)

的偏导数存在，则这两个偏导数分别记为

\frac{\partial z}{\partial x}

和

\frac{\partial z}{\partial y}

, 则，以下选项中，正确的是哪些？（多选）

选项

[A]. 偏导数存在且连续

\Rightarrow

不一定可微

[B]. 偏导数存在

\Rightarrow

不一定可微

[C]. 偏导数存在

\Rightarrow

一定可微

[D]. 可微

\Rightarrow

偏导数一定存在

[E]. 偏导数存在且连续

\Rightarrow

一定可微

[F]. 可微

\Rightarrow

偏导数不一定存在

答案

函数 $z$ $=$ $f (x, y)$ 在点 $(x, y)$ 处可微 $\Rightarrow$ 偏导数 $\frac{\partial z}{\partial x}$ 和 $\frac{\partial z}{\partial y}$ 必存在.

偏导数 $\frac{\partial z}{\partial x}$ 和 $\frac{\partial z}{\partial y}$ 存在 $\Rightarrow$ 函数 $z$ $=$ $f (x, y)$ 在点 $(x, y)$ 处不一定可微.

偏导数 $\frac{\partial z}{\partial x}$ 和 $\frac{\partial z}{\partial y}$ 在点 $(x, y)$ 的某邻域内存在且连续 $\Rightarrow$ 函数 $z$ $=$ $f (x, y)$ 在点 $(x, y)$ 处可微.