Course: 15.8 中间和极端价值理论

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中间和极端价值理论
Introduction
::导言

While the idea of continuity may seem somewhat basic, when a function is continuous over a closed interval like $\begin{aligned} x \in [0, 20] \end{aligned}$ , you can actually draw some major conclusions. For instance, April takes a run through the Central Park for 20 minutes, and then returns to where she started. Suppose the path of April's runs is modeled by a continuous function. What can be concluded regarding the and Extreme Value Theorem about April's run?
::虽然连续性的概念看起来似乎有些基本,但当一个函数在x[0,20]这样的封闭间隔内连续运行时,你实际上可以得出一些主要结论。例如,4月在中央公园运行20分钟,然后返回到她开始的位置。假设四月运行路径是由一个连续函数模拟的。关于四月运行的“极端价值”和“极端价值”理论可以得出什么结论?

Intermediate Value Theorem
::中间值定理

Intermediate Value Theorem
::中间值定理

If a function $\begin{aligned} f (x) \end{aligned}$ is continuous on a closed interval $\begin{aligned} [a, b], \end{aligned}$ and $\begin{aligned} u \end{aligned}$ is a value between $\begin{aligned} f (a) \end{aligned}$ and $\begin{aligned} f (b) \end{aligned}$ , then there exists a $\begin{aligned} c \in [a, b] \end{aligned}$ such that $\begin{aligned} f (c) = u \end{aligned}$ .

Simply stated, if a function is continuous between a low point and a high point, then there must be a $\begin{aligned} y \end{aligned}$ -value in between the low and high points that has a corresponding $\begin{aligned} x \end{aligned}$ -value that is between $\begin{aligned} x \end{aligned}$ -values of the low and high points.
::简言之,如果一个函数在低点和高点之间具有连续性,那么低点和高点之间必须有Y值,其相应的X值在低点和高点的X值之间。

The following video provides examples of the Intermediate Value Theorem:
::以下录像提供了中间价值理论的例子:

Play, Learn, and Explore the Intermediate Value Theorem:
::游戏、学习和探索中间价值理论:

Extreme Value Theorem
::极端价值定理

Extreme Value Theorem
::极端价值定理

If a function $\begin{aligned} f (x) \end{aligned}$ is continuous on a closed interval $\begin{aligned} [a, b] \end{aligned}$ , then there is at least one maximum and one minimum value. That is, there exist numbers $\begin{aligned} c \end{aligned}$ and $\begin{aligned} d \end{aligned}$ in $\begin{aligned} [a, b], \end{aligned}$ such that $\begin{aligned} f (d) \leq f (x) \leq f (c) for all x \in [a, b] \end{aligned}$ .
::如果函数f(x)在封闭间隔[a,b]上是连续的,则至少有一个最大值和一个最低值。也就是说,在[a,b]中存在数字c和d,因此,f(d)-f(x)-f(c)对所有 x__[a,b]都存在数字f(d)-f(x)-f(c)。

Examples
::实例

Example 1
::例1

Show that the converse of the Intermediate Value Theorem is false.
::显示中间值定理的反比为假。

Solution:
::解决方案 :

The converse of the Intermediate Value Theorem is: If there exists a value $\begin{aligned} c \in [a, b] \end{aligned}$ such that $\begin{aligned} f (c) = u \end{aligned}$ for every $\begin{aligned} u \end{aligned}$ between $\begin{aligned} f (a) \end{aligned}$ and $\begin{aligned} f (b) \end{aligned}$ , then the function is continuous.
::中间值理论的反义词是:如果存在数值c[a,b],那么f(a)和f(b)之间的每个u,f(c)=u,则该函数为连续函数。

To show the statement is false, all you need is one counterexample where every intermediate value is hit, and the function is discontinuous.
::要显示该语句是虚假的, 您只需要一个对应示例, 每个中间值被击中, 而函数是不连续的。

This function is discontinuous on the interval $\begin{aligned} [0, 10] \end{aligned}$ , but every intermediate value between the first height at $\begin{aligned} (0, 0) \end{aligned}$ and the height of the last point $\begin{aligned} (10, 5) \end{aligned}$ is hit.
::3⁄4 ̄ ̧漯B

Example 2
::例2

Show that the converse of the Extreme Value Theorem is false.
::显示极端值定理的反比为假。

Solution:
::解决方案 :

The converse of the Extreme Value Theorem is : If there is at least one maximum and one minimum in the closed interval $\begin{aligned} [a, b] \end{aligned}$ , then the function is continuous on $\begin{aligned} [a, b] \end{aligned}$ .
::极端值理论的反义词是:如果在封闭间隔[a,b]中至少有一个最大值和一个最低值,则该函数在[a,b]上为连续函数。

To show the statement is false, all you need is one counterexample. The goal is to find a function on a closed interval $\begin{aligned} [a, b] \end{aligned}$ that has at least one maximum and one minimum and is also discontinuous.
::要显示该语句是虚假的, 您只需要一个反例即可。目标是在关闭间隔[ a, b] 上找到一个函数, 该函数至少有一个最大值和最小值, 并且也是不连续的。

On the interval $\begin{aligned} [0, 10] \end{aligned}$ , the function attains a maximum at $\begin{aligned} (5, 5) \end{aligned}$ and a minimum at $\begin{aligned} (0, 0) \end{aligned}$ , but is still discontinuous.
::在间隔[0,10]时,函数达到最高值(5,5)和最低值(0,0),但仍不连续。

Example 3
::例3

Use the Intermediate Value Theorem to show that the function $\begin{aligned} f (x) = (x + 1)^{3} - 4 \end{aligned}$ has a zero on the interval $\begin{aligned} [0, 3] \end{aligned}$ .
::使用中间值定理来显示函数 f(x) =(x+1)3 - 4 的间隔为零 [0] 。

Solution:
::解决方案 :

First note that the function is a cubic, and therefore is continuous everywhere.
::第一点指出,该函数是立方体,因此在任何地方都是连续的。

$\begin{aligned} f (0) = (0 + 1)^{3} - 4 = 1^{3} - 4 = - 3 \end{aligned}$

::f( 0) = ( 0+1) 3- 4= 13- 4 3

$\begin{aligned} f (3) = (3 + 1)^{3} - 4 = 4^{3} - 4 = 60 \end{aligned}$

::f(3)=( 3+1) 3- 4= 43-4=60

By the Intermediate Value Theorem, there must exist a $\begin{aligned} c \in [0, 3] \end{aligned}$ such that $\begin{aligned} f (c) = 0, \end{aligned}$ since 0 is between -3 and 60.
::根据中间值理论,必须有一个c[0],即f(c)=0,因为0介于-3至60之间。

Example 4
::例4

Recall the question from the Introduction: April takes a run through the Central Park for 20 minutes and then returns to where she started. What can be concluded regarding the Intermediate Value Theorem and Extreme Value Theorem about April's run?
::回顾导言中的问题:4月经过中央公园跑步20分钟,然后回到她起步的地方。关于四月跑步的中间价值理论和极端价值理论可以得出什么结论?

Solution:
::解决方案 :

April's run can be represented by a continuous function on the interval $\begin{aligned} x \in [0, 20] \end{aligned}$ . By the Intermediate Value Theorem, it can be concluded that at some point during the 20-minute run, the path was either higher or lower than April's starting and ending position. At that height, there is some corresponding time value between 0 and 20 minutes. By the Extreme Value Theorem, it can be concluded that there was at least one maximum height and one minimum height along the path during the 20-minute run.
::April 运行可以通过间距 x{{{{{{{0,20} 上的连续函数表示。根据中间值理论,可以得出结论,在20分钟运行的某个时刻,路径要么高于要么低于 April 的起始和结束位置。在那个高度,一定的时间值在0到20分钟之间。根据极端值理论,可以得出结论,在20分钟运行的路径上至少有一个最高高度和一个最低高度。

Example 5
::例5

Use the Intermediate Value Theorem to show that the equation below has at least one real solution.
::使用中间值定理来显示以下方程式至少有一个真实的解决方案。

$\begin{aligned} x^{8} = 2^{x} \end{aligned}$

::x8=2x

Solution :
::解决方案 :

First, write the equation as a continuous function.
::首先,将方程式写成连续函数。

$\begin{aligned} f (x) = x^{8} - 2^{x} \end{aligned}$

:xx) =x8-2x

T his function is continuous, because the difference of two continuous functions is continuous.
::这一职能是连续的,因为两个连续职能的区别是连续的。

Then a pply the Intermediate Value Theorem.
::然后适用中间值定理。

$\begin{aligned} f (0) = 0^{8} - 2^{0} = 0 - 1 = - 1 \end{aligned}$

::f(0)=08-20=0-11

$\begin{aligned} f (2) = 2^{8} - 2^{2} = 256 - 4 = 252 \end{aligned}$

::f(2)=28-22=256-4=252

By the Intermediate Value Theorem, there must exist a $\begin{aligned} c \end{aligned}$ such that $\begin{aligned} f (c) = 0 \end{aligned}$ because $\begin{aligned} - 1 < 0 < 252 \end{aligned}$ . The number $\begin{aligned} c \end{aligned}$ is one solution to the initial equation.
::根据中间值理论,必须存在一种c,即f(c)=0,因为-1<0<252。数字c是初始方程的一种解决办法。

Example 6
::例6

Show that there is at least one solution to the equation below.
::显示对以下方程式至少有一个解决方案。

$\begin{aligned} \sin x = x + 2 \end{aligned}$

::exix=x+2

Solution :
::解决方案 :

First, w rite the equation as a continuous function.
::首先,将方程式写成连续函数。

$\begin{aligned} f (x) = \sin x - x - 2 \end{aligned}$

:xx) =sinx-x-2

This function is continuous because the difference of two continuous functions is continuous.
::这一职能是连续的,因为两个连续职能的差别是连续的。

Then apply the Intermediate Value Theorem.
::然后适用中间值定理。

$\begin{aligned} f (0) = \sin 0 - 0 - 2 = - 2 \end{aligned}$

::f(0) =sin_0-0-22

$\begin{aligned} f (- π) = \sin (- π) + π - 2 = 0 + π - 2 > 0 \end{aligned}$

:)=sin() 2=02>0

By the Intermediate Value Theorem, there must exist a $\begin{aligned} c \end{aligned}$ such that $\begin{aligned} f (c) = 0 \end{aligned}$ because $\begin{aligned} - 2 < 0 < π - 2 \end{aligned}$ . The number $\begin{aligned} c \end{aligned}$ is one solution to the initial equation.
::根据中间值理论,必须存在一种c,即f(c)=0,因为-2<0.%2。数字c是初始方程的一种解决办法。

Example 7
::例7

When are you not allowed to use the Intermediate Value Theorem?
::何时不准使用中间值理论?

Solution:
::解决方案 :

The Intermediate Value Theorem should not be applied when the function is not continuous over the interval.
::当函数间隔期间不连续时,不应适用中间值定理。

Summary
::摘要

The Intermediate Value Theorem states: If a function is continuous on a closed interval $\begin{aligned} [a, b] \end{aligned}$ and $\begin{aligned} u \end{aligned}$ is a value between $\begin{aligned} f (a) \end{aligned}$ and $\begin{aligned} f (b) \end{aligned}$ , then there exists a $\begin{aligned} c \in [a, b] \end{aligned}$ such that $\begin{aligned} f (c) = u \end{aligned}$ .
::中间值理论指出:如果一个函数在封闭间隔[a、b]和u之间的一个数值为f(a)和f(b)之间,则存在f(c)=u的 c[a、b]。

The Extreme Value Theorem states: If a function is continuous on a closed interval $\begin{aligned} [a, b] \end{aligned}$ , then there is at least one maximum and one minimum value.

::极端值定理表示:如果一个函数在封闭间隔[a,b] 连续,则至少有一个最大值和一个最低值。

Review
::回顾

Use the Intermediate Value Theorem to show that each equation has at least one real solution.
::使用中间值定理来显示每个方程式至少有一个真实的解决方案。

1. $\begin{aligned} \cos x = - x \end{aligned}$
::1. COsxxxxx

2. $\begin{aligned} \ln (x) = e^{- x} + 1 \end{aligned}$
::2. In(x)=e-x+1

3. $\begin{aligned} 2 x^{3} - 5 x^{2} = 10 x - 5 \end{aligned}$
::3. 2x3-5x2=10x-5

4. $\begin{aligned} x^{3} + 1 = x \end{aligned}$
::4. x3+1=x

5. $\begin{aligned} x^{2} = \cos x \end{aligned}$
::5. x2=cosx

6. $\begin{aligned} x^{5} = 2 x^{3} + 2 \end{aligned}$
::6. x5=2x3+2

7. $\begin{aligned} 3 x^{2} + 4 x - 11 = 0 \end{aligned}$
::7. 3x2+4x-11=0

8. $\begin{aligned} 5 x^{4} = 6 x^{2} + 1 \end{aligned}$
::8. 5x4=6x2+1

9. $\begin{aligned} 7 x^{3} - 18 x^{2} - 4 x + 1 = 0 \end{aligned}$
::9. 7x3-18x2-4x+1=0

10. Show that $\begin{aligned} f (x) = \frac{2 x - 3}{2 x - 5} \end{aligned}$ has a real root on the interval $\begin{aligned} [1, 2] \end{aligned}$ .
::10. 显示 f(x) = 2x-32x- 5 在间隔[ 1, 2] 上有一个真正的根。

11. Show that $\begin{aligned} f (x) = \frac{3 x + 1}{2 x + 4} \end{aligned}$ has a real root on the interval $\begin{aligned} [- 1, 0] \end{aligned}$ .
::11. 显示 f(x) =3x+12x+4 在间隔[-1,0] 上有一个真正的根。

12. True or false: A function has a maximum and a minimum in the closed interval $\begin{aligned} [a, b] \end{aligned}$ ; therefore, the function is continuous.
::12. 真实或虚假:一个函数在封闭间隔[a,b]内有最大和最小值;因此,该函数是连续的。

13. True or false: A function is continuous over the interval $\begin{aligned} [a, b] \end{aligned}$ ; therefore, the function has a maximum and a minimum in the closed interval.
::13. 真实的或虚假的:一个函数在间隔[a,b] 期间是连续的;因此,该函数在封闭间隔内有上限和最小值。

14. True or false: If a function is continuous over the interval $\begin{aligned} [a, b] \end{aligned}$ , then it is possible for the function to have more than one relative maximum in the interval $\begin{aligned} [a, b] \end{aligned}$ .
::14. 真实的或虚假的:如果一个函数在[a,b]间隔内连续存在,则该函数在[a,b]间隔内可能具有一个以上的相对最大值。

15. What do the Intermediate Value and Extreme Value Theorems have to do with continuity?
::15. 中间价值和极端价值理论与连续性有什么关系?

Review (Answer s)
::回顾(答复)

Please see the Appendix.
::请参看附录。

Section outline

General

中间和极端价值理论

Introduction ::导言

Intermediate Value Theorem ::中间值定理

Intermediate Value Theorem ::中间值定理

Extreme Value Theorem ::极端价值定理

Extreme Value Theorem ::极端价值定理

Examples ::实例

Summary ::摘要

Review ::回顾

Review (Answer s) ::回顾(答复)

Introduction
::导言

Intermediate Value Theorem
::中间值定理

Intermediate Value Theorem
::中间值定理

Extreme Value Theorem
::极端价值定理

Extreme Value Theorem
::极端价值定理

Examples
::实例

Summary
::摘要

Review
::回顾

Review (Answer s)
::回顾(答复)