Course: 14.7 中间和极端价值理论

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中间和极端价值理论

While the idea of continuity may seem somewhat basic, when a function is continuous over a closed interval like $\begin{aligned} x \in [1, 4] \end{aligned}$ , you can actually draw some major conclusions. The conclusions may be obvious when you understand the statements and look at a graph, but they are powerful nonetheless.
::虽然连续性的概念看起来似乎有些基本,但当一个函数在像x[1,4]这样的封闭间隔内连续持续时,你实际上可以得出一些主要结论。当你理解这些声明并查看图表时,这些结论可能显而易见,但尽管如此,这些结论是强大的。

What can you conclude using the and the Extreme Value Theorem about a function that is continuous over the closed interval $\begin{aligned} x \in [1, 4] \end{aligned}$ ?
::您对关闭间隔 x {{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{}}) 连续的函数使用极端值理论和极端值理论得出什么结论?

The Intermediate and Extreme Value Theorems
::中间和极端价值理论

The Intermediate Value Theorem states that if a function is continuous on a closed interval 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内是连续的,而u是f(a)和f(b)之间的数值,那么就存在f(c)=u这样的 c[a,b]。

Simply stated, if a function is continuous between a low point and a high point, then it must be valued at each intermediate height in between the low and high points.
::简言之,如果一个函数在低点和高点之间是连续的,那么必须按低点和高点之间的每中间高度来估价。

The converse of an if then statement is a new statement with the hypothesis of the original statement switched with the conclusion of the original statement. In other words, the converse is when the if part of the statement and the then part of the statement are swapped. In general, the converse of a statement is not true.
::换句话说,如果当时的声明是一个新的声明,假设最初的声明与原先的声明的结束相转变。换句话说,相反的是,如果声明的一部分和当时的声明的一部分被转换,一般说来,声明的反对是不真实的。

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)之间每个u的f(c)=u,则函数是连续的。

This statement is false. In order to show the statement is false, all you need is one counterexample where every intermediate value is hit and the function is discontinuous. A counterexample to an if then statement is when the hypothesis (the if part of the sentence) is true, but the conclusion (the then part of the statement) is not true.
::此语句是虚假的。为了显示该语句是虚假的, 您需要的只是一个反例, 即每个中间值被击中, 而函数是不连续的。如果此语句是假设( 句子中的部分) 是真实的, 但结论( 句子中的那部分) 不真实。

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

The Extreme Value Theorem states that in every interval $\begin{aligned} [a, b] \end{aligned}$ where a function is continuous there is at least one maximum and one minimum. In other words, it must have at least two extreme values.
::极端值理论指出,在连续函数的每个间隔[a,b]中,至少有一个最大值和一个最低值。换句话说,它必须至少有两个极端值。

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。

This statement is false. In order 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.
::此语句是虚假的。为了显示该语句是虚假的, 您只需要一个反例即可。目标是在封闭的间隔[ a, b] 上找到一个函数, 该间隔至少有一个上限和最小值, 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。

Examples
::实例

Example 1
::例1

Earlier, you were asked to apply the Intermediate and Extreme Value Theorems to a function is continuous on the interval $\begin{aligned} x \in [1, 4] \end{aligned}$ . You can conclude by the Intermediate Value Theorem that there exists a $\begin{aligned} c \in [1, 4] \end{aligned}$ such that $\begin{aligned} f (c) = u \end{aligned}$ for every $\begin{aligned} u \end{aligned}$ between $\begin{aligned} f (1) \end{aligned}$ and $\begin{aligned} f (4) \end{aligned}$ . You can also conclude that on this interval the function has both a maximum and a minimum value by the Extreme Value Theorem.
::早些时候,您被要求对一个函数应用中间值和极端值理论,该函数在间距 x++++[1,4]。根据中间值理论,您可以得出结论,存在一个c+++[1,4],因此 f(c)=u f(1)和f(4)之间的每个u。您还可以得出结论,在此间距上,该函数的极限值和最小值值都与极端值理论值相同。

Example 2
::例2

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] 。

First note that the function is a cubic and is therefore continuous everywhere.
::首先要注意的是,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],这样f(c)=0,因为0介于-3至60之间。

Example 3
::例3

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

$\begin{aligned} x^{8} = 2^{x} \end{aligned}$
::x8=2x

First rewrite the equation: $\begin{aligned} x^{8} - 2^{x} = 0 \end{aligned}$
::第一次重写方程: x8- 2x=0

Then describe it as a continuous function: $\begin{aligned} f (x) = x^{8} - 2^{x} \end{aligned}$
::然后将其描述为一个连续函数: f(x)=x8-2x

This function is continuous because it is the difference of two continuous functions.
::这一职能是连续性的,因为它是两个连续职能的区别。

$\begin{aligned} f (0) = 0^{8} - 2^{0} = 0 - 1 = - 1 \end{aligned}$
::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 4
::例4

Show that there is at least one solution to the following equation.
::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

Write the equation as a continuous function: $\begin{aligned} f (x) = \sin x - x - 2 \end{aligned}$
::将方程式写成连续函数 : f( x) =sinx- x- 2

The function is continuous because it is the sum and difference of continuous functions.
::该功能是连续性的,因为它是连续功能的和差。

$\begin{aligned} f (0) = \sin 0 - 0 - 2 = - 2 \end{aligned}$
::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 5
::例5

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

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

The Intermediate Value Theorem states that 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 $\begin{aligned} c \end{aligned}$ in that interval 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)之间的一个函数是连续的,那么在这个间隔内就有一个c,即f(c)=u。

The converse of the Intermediate Value Theorem is false.
::中间价值理论的反比是虚假的。

The Extreme Value Theorem states that in every interval where a function is continuous, there is at least one maximum and one minimum.
::极端值理论指出,在函数连续的每一个间隔期间,至少有一个最大和一个最低。

The converse of the Extreme Value Theorem is also false.
::极端价值理论的反比也是虚假的。

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 (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

Section outline

General

中间和极端价值理论

The Intermediate and Extreme Value Theorems ::中间和极端价值理论

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Review ::回顾

Review (Answers) ::回顾(答复)