Course: 4.4 第一次衍生物试验

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第一次衍生试验
If you look at any function curve, you can determine visually whether the function is increasing, decreasing, or remaining constant over an interval. You can also see approximately were the function attains high points and low points. There is a way to determine mathematically, using the derivative, what your visual observations provide. How does knowing the slope of a tangent line help you to precisely calculate what your visual observations can only estimate?
::如果查看任何函数曲线,您可以直观地确定函数是增加、减少,还是保持一个间隔不变。您也可以大致地看到函数达到高点和低点。有办法用数学来确定,使用衍生物,你的视觉观察能提供什么。知道正切线的斜坡如何帮助您精确地计算视觉观察只能估计到什么?

The First Derivative Test
::第一次衍生试验

Consider the graphs of the two functions below.
::考虑以下两个函数的图表。

Both functions are continuous over the intervals shown. In the first, $\begin{aligned} f (x) = x^{3} \end{aligned}$ , the function values are always increasing as $\begin{aligned} x \end{aligned}$ increases. In the second, the piece-wise function values increase over the interval [-5, -2], stay the same over the interval [-2, 2], then resume increasing for [2, 5].
::两个函数都是在所显示的间隔内连续的。在第一个间隔内, f(x) =x3, 函数值总是随着 x 的增加而增加。在第二个间隔内, 字符串函数值在[ 5 - 2] 间段内增加, 在[ 2 - 2] 间段内保持不变, 然后在 [ 2 5] 间段内恢复增加。

The difference between these two cases of increasing function values motivates the need for the following distinctions:
::这两种功能值不断提高的情况之间的差异促使有必要作出以下区分:

A function $\begin{aligned} f \end{aligned}$ is said to be increasing on $\begin{aligned} [a, b] \end{aligned}$ contained in the domain of $\begin{aligned} f \end{aligned}$ if $\begin{aligned} f (x_{1}) \leq f (x_{2}) \end{aligned}$ whenever $\begin{aligned} x_{1} \leq x_{2} \end{aligned}$ for all $\begin{aligned} x_{1}, x_{2} \in [a, b] . \end{aligned}$
::如果 f (x1), f (x2) 则 f 域所含的 f 函数 f 的 [a, b] 上据说正在增加,只要所有 x1, x2 的 x1x2 = [a, b] 。

If $\begin{aligned} f (x_{1}) < f (x_{2}) \end{aligned}$ whenever $\begin{aligned} x_{1} < x_{2} \end{aligned}$ for all $\begin{aligned} x_{1}, x_{2} \in [a, b], \end{aligned}$ then we say that $\begin{aligned} f \end{aligned}$ is strictly increasing on $\begin{aligned} [a, b] . \end{aligned}$
::如果 f( x1) < f( x2) 时所有 x1, x2\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

In similar manner:
::以类似方式:

A function $\begin{aligned} f \end{aligned}$ is said to be decreasing on $\begin{aligned} [a, b] \end{aligned}$ contained in the domain of $\begin{aligned} f \end{aligned}$ if $\begin{aligned} f (x_{1}) \geq f (x_{2}) \end{aligned}$ whenever $\begin{aligned} x_{1} \geq x_{2} \end{aligned}$ for all $\begin{aligned} x_{1}, x_{2} \in [a, b] \end{aligned}$ .
::如果 f (x1), 函数 f (x2) 在 f 域所含的 [a, b] 上, 当所有 x1, x2 的 x1x2 时, 函数 f 就会递减。 [a, b]

If $\begin{aligned} f (x_{1}) > f (x_{2}) \end{aligned}$ whenever $\begin{aligned} x_{1} > x_{2} \end{aligned}$ for all $\begin{aligned} x_{1}, x_{2} \in [a, b] \end{aligned}$ . Then we say that $\begin{aligned} f \end{aligned}$ is strictly decreasing on $\begin{aligned} [a, b] . \end{aligned}$
::如果 f( x1) > f( x2) , 则所有 x1, x2\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

Note that the symbols $\begin{aligned} ϵ \end{aligned}$ and $\begin{aligned} \in \end{aligned}$ are equivalent and denote that a particular element is contained within a particular set.
::请注意符号和符号等同,表示特定元素包含在特定集中。

Using the above terminology, we say the function $\begin{aligned} f (x) = x^{3} \end{aligned}$ is strictly increasing over the given interval, and the piece-wise function is increasing over the interval.
::使用上述术语,我们说,函数f(x)=x3在给定间隔内严格递增,而字符串函数在间隔内递增。

Now look at the of the two functions graphed above.
::现在看看上面图表显示的两个函数。

We can now state a theorem that relates the derivative of a function to the increasing/decreasing properties of the function.
::我们现在可以说明一个理论,将函数的衍生物与函数的不断增长/减少的特性联系起来。

If $\begin{aligned} f \end{aligned}$ is continuous on interval $\begin{aligned} [a, b] \end{aligned}$ , and differentiable on $\begin{aligned} (a, b) \end{aligned}$ then:
::如果 f 是按[a,b]间隔连续的,且在(a,b)间隔上可加以区别,则:

If $\begin{aligned} f^{'} (x) > 0 \end{aligned}$ for every $\begin{aligned} x \in (a, b) \end{aligned}$ , then $\begin{aligned} f \end{aligned}$ is increasing in $\begin{aligned} [a, b] \end{aligned}$ .
::如果 f_(x) > 0 代表每 x% (a, b), f 则在 [a, b] 中增加。

If $\begin{aligned} f^{'} (x) < 0 \end{aligned}$ for every $\begin{aligned} x \in (a, b) \end{aligned}$ , then $\begin{aligned} f \end{aligned}$ is decreasing in $\begin{aligned} [a, b] \end{aligned}$ .
::如果 f} (x) < 0 代表每 x(a,b), f 则在 [a,b] 中呈下降趋势。

Take the function $\begin{aligned} f (x) = x^{3} - 3 x^{2} - 6 x + 8 \end{aligned}$ . To find the intervals on which $\begin{aligned} f \end{aligned}$ is increasing and the intervals on which $\begin{aligned} f \end{aligned}$ is decreasing, first note that the function $\begin{aligned} f (x) \end{aligned}$ is continuous everywhere. The derivative of the function is $\begin{aligned} f^{'} (x) = 3 x^{2} - 6 x - 6 = 3 (x^{2} - 2 x - 2) \end{aligned}$ , which is a parabola with two $\begin{aligned} x \end{aligned}$ -intercepts (critical numbers of $\begin{aligned} f \end{aligned}$ ) at $\begin{aligned} x = 1 \pm \sqrt{3} \end{aligned}$ . Evaluation of $\begin{aligned} f^{'} (x) \end{aligned}$ in the three intervals that are defined by the two roots provides the following information:
::使用 f( x) =x3 - 3x2 - 6x+8. 函数 f( x) =x3 - 3x2 - 6x+8. 查找 f 增加的间隔和 f 减少的间隔,首先注意, f( x) 函数的衍生物是 f* (x) = 3x2 - 6x - 6= 3( x2 - 2x-2) 。该函数的衍生物是具有x= 1 =1 3 三个间隔内两个 X 界面( 临界数 f) 的抛光线。

$\begin{aligned} f^{'} (x) = {\begin{cases} > 0, x < 1 - \sqrt{3} f (x) is increasing \\ = 0, x = 1 - \sqrt{3} \\ < 0, (1 - \sqrt{3}) < x < (1 + \sqrt{3}) f (x) is decreasing \\ = 0, x = 1 + \sqrt{3} \\ > 0, x > 1 + \sqrt{3} f (x) is increasing \end{cases} \end{aligned}$

::f*(xx) =0,x <1 -3f(x) 正在递增=0,x=1 -3 <0,(1-3), <x <(1+3),f(x) 正在递减=0,x=1+3>0,x>1+3f(x) 正在递增

Changes in the derivative of a function from positive to negative, or negative to positive may indicate the presence of a local extremum.
::函数从正到负或从负到正的衍生物的变化可能表明存在局部末端。

The First Derivative Test describes where these extrema are and what type they are. The First Derivative Test is as follows:
::第一次衍生试验描述了这些外形的位置和种类。第一次衍生试验如下:

Suppose that $\begin{aligned} f \end{aligned}$ is a continuous function and that $\begin{aligned} x = c \end{aligned}$ is a critical value of $\begin{aligned} f \end{aligned}$ , then:
::假设 f 是连续函数, x=c 是 f 的临界值, 那么 :

If $\begin{aligned} f^{'} \end{aligned}$ changes from positive to negative at $\begin{aligned} x = c \end{aligned}$ , then $\begin{aligned} f \end{aligned}$ has a local maximum at $\begin{aligned} x = c \end{aligned}$ .
::如果f`字在x=c时从正数改为负数,那么f则在x=c时有当地最大值。

If $\begin{aligned} f^{'} \end{aligned}$ changes from negative to positive at $\begin{aligned} x = c \end{aligned}$ , then $\begin{aligned} f \end{aligned}$ has a local minimum at $\begin{aligned} x = c \end{aligned}$ .
::如果f==x=c时从负改为正,那么f=x=c时的当地最低值为x=c。

If $\begin{aligned} f^{'} \end{aligned}$ does not change sign at $\begin{aligned} x = c \end{aligned}$ , then $\begin{aligned} f \end{aligned}$ has neither a local maximum nor minimum at $\begin{aligned} x = c \end{aligned}$ .
::如果 f` 不更改 x=c 上的符号, 那么 f 在 x=c 上既没有本地最大值, 也没有本地最小值。

We can observe the consequences of this theorem by observing the tangent lines of the following graph in each of the intervals $\begin{aligned} (0, a) \end{aligned}$ , $\begin{aligned} (a, b) \end{aligned}$ , $\begin{aligned} (b, + \infty) \end{aligned}$ .
::我们可以通过在间隔(0,a)、(a,b)、(b)和(b)之间观察下图的相切线来观察这个理论的后果。

Note first that we have a relative maximum at $\begin{aligned} x = a \end{aligned}$ and a relative minimum at $\begin{aligned} x = b \end{aligned}$ . The slopes of the tangent lines change from positive for $\begin{aligned} x \in (0, a) \end{aligned}$ to negative for $\begin{aligned} x \in (a, b) \end{aligned}$ and then back to positive for $\begin{aligned} x \in (b, + \infty) \end{aligned}$ .
::首先请注意,我们在 x=a 时的相对最大值和在 x=b 时的相对最低值。正切线的斜度从x(0,a) 时的正值变为 x(a,b) 时的负值,然后返回到x(b,) 时的正值。

Examples
::实例

Example 1
::例1

Earlier, you were asked how the slope of the tangent line relates to whether or not a function is increasing or decreasing . Positive slope means an increasing function; negative slope means a decreasing function. Often the slope transition from positive (negative) to negative (positive) indicates the location of an extremum.
::早些时候,有人问到,正斜线的斜坡与函数是否在增加或减少有何关系。正斜坡意味着功能在增加;负斜坡意味着功能在减少。从正(负)向负(正)向负(正)的斜坡往往表示矩形的位置。

Example 2
::例2

Consider the function graph below. Determine whether the function where the function is strictly increasing or decreasing.
::考虑下方的函数图。确定该函数是严格增加还是减少的函数。

The function indicated here is strictly increasing on $\begin{aligned} (0, a) \end{aligned}$ and $\begin{aligned} (b, c), \end{aligned}$ and strictly decreasing on $\begin{aligned} (a, b) \end{aligned}$ and $\begin{aligned} (c, d) \end{aligned}$ .
::这里所指的功能正在严格增加(0,a)和(b,c),并严格减少(a,b)和(c,d)。

Example 3
::例3

Let's consider the function $\begin{aligned} f (x) = x^{2} + 6 x - 9 \end{aligned}$ and observe the graph around $\begin{aligned} x = - 3 \end{aligned}$ . What happens to the first derivative near this value?
::让我们考虑 F( x) =x2+6x- 9 的函数, 观察 x\\\ 3 周围的图形。接近此值的第一个衍生物会怎样 ?

With $\begin{aligned} f (x) = x^{2} + 6 x - 9 \end{aligned}$ , the derivative is $\begin{aligned} f^{'} (x) = 2 x + 6 \end{aligned}$ . Notice that the following condition applies to the slopes of tangent lines:
::F(x) =x2+6x-9, 衍生物为 f_(x) =2x+6. 注意以下条件适用于正切线的斜坡:

$\begin{aligned} f^{'} (x) = {\begin{cases} < 0, x < - 3 \\ = 0, x = 0 \\ > 0, x > - 3 \end{cases} \end{aligned}$

::f_(x)0, x_3=0, x=0>0, x_3

We observe that the slopes of the tangent lines to the graph change from negative to positive at $\begin{aligned} x = - 3 \end{aligned}$ . The first derivative test states that $\begin{aligned} f (x) \end{aligned}$ has a local minimum at $\begin{aligned} x = - 3 \end{aligned}$ . The function graph verifies this.
::我们注意到,图的正切线斜坡从x3的负向向正向变化。第一项衍生物试验指出,f(x)在 x3 的当地最低值为 f(x) 。函数图证实了这一点。

Review
::回顾

For #1-2, identify the intervals where the function is increasing, decreasing, or is constant. (Units on the axes indicate single units).
::对于 # 1-2, 确定函数正在增加、减少或恒定的间隔。 (轴上的单位表示单单位) 。

Give the sign of the following quantities for the graph in problem 2.

$\begin{aligned} f^{'} (- 3) \end{aligned}$
::f_( - 3)

$\begin{aligned} f^{'} (1) \end{aligned}$
::f*(1)

$\begin{aligned} f^{'} (3) \end{aligned}$
::f§(3)

$\begin{aligned} f^{'} (4) \end{aligned}$
::f§(4) =======================================

::给问题2. f`(- 3) f (1) f (3) f (4) 中的图表显示以下数量的符号。

For #4-6 , determine the intervals in which the function is increasing and those in which it is decreasing, then sketch the graph of each function.
::对于#4-6, 确定函数增加的间隔和减少的间隔, 然后绘制每个函数的图表。

$\begin{aligned} f (x) = x^{2} - \frac{1}{x} \end{aligned}$
:xx) =x2 - 1x

$\begin{aligned} f (x) = (x^{2} - 1)^{5} \end{aligned}$
:xx)=(x2-1)5

$\begin{aligned} f (x) = (x^{2} - 1)^{4} \end{aligned}$
:xx)=(x2-1)4

For #7-10 :
::7-10:

Use the First Derivative Test to find the intervals where the function increases and/or decreases
::使用第一个衍生工具测试查找函数增加和/或减少的间隔

Identify all max, min, or relative max and min values
::标记所有最大值、最小值或相对最大值和最小值

Sketch the graph
::绘制图形

$\begin{aligned} f (x) = - x^{2} - 4 x - 1 \end{aligned}$
:xx) x2 - 4x - 1

$\begin{aligned} f (x) = x^{3} + 3 x^{2} - 9 x + 1 \end{aligned}$
:x) =x3+3x2-9x+1

$\begin{aligned} f (x) = x^{\frac{2}{3}} (x - 5) \end{aligned}$
:xx)=x23(x-5)

$\begin{aligned} f (x) = 2 x \sqrt{x^{2} + 1} \end{aligned}$
:xx)=2x22+1

Use the first derivative test to classify the critical numbers $\begin{aligned} x = 0 \end{aligned}$ and $\begin{aligned} x = \frac{3 π}{2} \end{aligned}$ of $\begin{aligned} f (x) = \cos^{2} (x) \end{aligned}$ as local maxima or minima.
::使用第一个衍生物测试将 f( x) = cos2 = (x) 的关键数字 x=0 和 x= 32 分类为本地最大值或最小值。

Find the critical numbers of $\begin{aligned} f (x) = x^{5} - 20 x - 2 \end{aligned}$ and classify them as local maxima, minima or neither.
::查找 f( x) =x5-20x-2 的临界值,并将其分类为本地最大值、 minima 或两者兼而有之。

Find the local extrema of $\begin{aligned} f (x) = x + \sin (x) \end{aligned}$ on the interval $\begin{aligned} (- 2 π, 2 π) \end{aligned}$ .
::查找时间间隔(- 2 2) f(x) =x+sin(x) 的局部矩形。

Find the global and local extrema of $\begin{aligned} f (x) = \frac{3}{4} x^{4} + 4 x^{3} - 6 x^{2} - 48 x - 50 \end{aligned}$ .
::查找 f( x) = 34x4+4x3 - 6x2 - 48x- 50 的全球和本地的 extrema。

Find the global and local extrema of $\begin{aligned} f (x) = \frac{x^{2} - x - 6}{x^{2} + x - 6} \end{aligned}$ .
::查找 f( x) =x2- x-6x2+x-6 的全球和本地的 extrema。

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

第一次衍生试验

The First Derivative Test ::第一次衍生试验

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Review ::回顾

Review (Answers) ::回顾(答复)

The First Derivative Test
::第一次衍生试验

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Review
::回顾

Review (Answers)
::回顾(答复)