Using the Mean Value Theorem
AP Calculus BC Topic 5 9:33 English narration · English + 中文 subtitles burned in
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You drive two hundred kilometres in two hours.
你用两小时开了两百公里。
Your average speed was one hundred kilometres per hour.
你的平均速度是每小时一百公里。
Now here is a claim. At some moment during that drive, your speedometer read exactly one hundred.
现在有一个说法： 在这段行程中的某个时刻，你的速度表读数恰好是一百。
Not more, not less.
不多也不少。
That feels obvious, and it is true, and it has a name. The Mean Value Theorem. Watch the picture.
这听起来理所当然，而且它是对的，它还有个名字：中值定理。
Draw the straight line joining the start and the end.
看这张图。 画出连接起点和终点的直线。
Then slide a tangent along the curve.
然后让一条切线沿着曲线滑动。
Somewhere it becomes parallel to the straight line.
在某个地方，它会变得和那条直线平行。
This unit uses the derivative to describe the shape of a function.
本单元用导数来描述一个函数的形状。
Let's begin.
让我们开始吧。
Here is the theorem, and it comes with two hypotheses.
这就是定理，它带有两个前提条件。
Both must hold, and both must be named when you write it up.
两个都必须成立，而且写答案时两个都必须写出来。
First, the function is continuous on the closed interval, endpoints included.
第一，函数在闭区间上连续，包含端点。
Second, it is differentiable on the open interval, endpoints excluded.
第二，函数在开区间上可导，不含端点。
Notice the asymmetry.
注意这个不对称。
That is deliberate.
这是故意的。
The conclusion is then that there is at least one point strictly inside the interval, where the instantaneous rate equals the average rate over the whole interval.
结论是：至少存在一个严格位于区间内部的点， 在那里瞬时变化率等于整个区间上的平均变化率。
Like the Intermediate Value Theorem, this is an existence theorem.
和介值定理一样，这是一个存在性定理。
It promises a point exists.
它保证有这样一个点存在，但不告诉你在哪里。
It does not tell you where.
所以考题会要求你写出理由，而拿满分需要三个部分。
So exam questions ask you to justify, and full credit needs three parts.
第一，说明函数在闭区间上连续、 在开区间上可导。
State both hypotheses.
第二，算出平均变化率。
Compute the average rate of change.
第三，得出结论：根据中值定理， 在区间内部存在这样一个点。
Then conclude, by the Mean Value Theorem, that such a point exists inside the interval.
现在取 x 平方，区间从 1 到 3。
Now take x squared on the interval from one to three.
它是多项式，所以两个前提都成立。
It is a polynomial, so both hypotheses hold.
平均变化率是 9 减 1 再除以 2，等于 4。
The average rate is four.
令导数等于 4。
Set the derivative equal to four, and c is two, which does lie inside.
2c 等于 4，所以 c 等于 2， 而 2 确实位于区间内部。
Before we hunt for extremes, one theorem promises they are there at all.
在我们去寻找极值之前，有一个定理先保证它们确实存在。
The Extreme Value Theorem.
极值定理。
If a function is continuous on a closed interval, then it attains both an absolute maximum and an absolute minimum somewhere on that interval.
如果一个函数在一个闭区间上 连续，那么它一定在这个区间上取到绝对最大值和绝对最小值。
Look at the picture: the curve has a genuine highest point and a genuine lowest point, and neither one escapes.
看这张图：曲线有一个真正的最高点 和一个真正的最低点，两个都跑不掉。
Both hypotheses are doing real work here.
这里两个条件都在起实质作用。
Drop continuity, and a jump can leap straight over the top without ever reaching it.
去掉连续性，一个跳跃就能 直接越过最高处而永远达不到它。
Drop the closed interval, and the peak can run off toward an open end that the function never actually gets to.
去掉闭区间，最高点就可能跑向一个开口的端点，而函数永远 到不了那里。
And this is exactly what licenses the candidates test later: because the extremes are guaranteed to exist, all you have to do is find them.
这也正是后面候选点检验法成立的依据：因为极值一定存在，你要做的只是把它们找出来。
Next, extreme values, and a distinction the exam is strict about.
接下来是极值，以及一个考试要求非常严格的区分。
Local relative extrema — a local extremum is the highest or lowest value in a small neighbourhood.
局部极值是在一个小邻域内的最大值或最小值。
A global extremum is the highest or lowest over the whole interval.
整体极值，也叫绝对极值，是整个区间上的最大值或最小值。
A local peak may be nowhere near the global one.
一个局部的峰可能离整体的峰很远。
Where can they hide?
它们会藏在哪里？
At a critical point, meaning a point where the derivative is zero, or where it does not exist.
藏在临界点， 也就是定义域内导数为零、或者导数不存在的点。
A corner counts.
尖角算。
A vertical tangent counts.
竖直切线也算。
And on a closed interval, check the endpoints too, because a global extremum is happy to sit right at the edge.
而且在闭区间上，端点也必须检查，因为整体极值很乐意就坐在边缘上。
The first derivative controls the direction of the graph.
一阶导数决定了图像的走向。
Where the derivative is positive, the function is increasing.
导数为正的地方，函数递增。
Where the derivative is negative, the function is decreasing.
导数为负的地方，函数递减。
That is the whole rule, and the working method that goes with it is a sign chart.
规则就这么简单，而配套的做题方法是符号表。
Find every point where the derivative is zero or undefined.
找出所有导数为零或不存在的点。
Mark them on a number line.
把它们标在数轴上。
Then test one value in each interval between them, and record the sign.
然后在相邻两点之间各取一个值来测试，记下符号。
The chart is the argument, so write it down.
这张表就是你的论证过程，所以一定要写出来。
Now classify each critical point with the first derivative test.
现在用一阶导数检验法来判定每个临界点。
If the derivative changes from positive to negative there, the function rises then falls, so it is a local maximum.
如果导数在那里由正变负，函数先升后降， 那就是局部最大值。
If it changes from negative to positive, it falls then rises, so it is a local minimum.
如果由负变正，函数先降后升，那就是局部最小值。
And if the derivative does not change sign at all, it is neither.
如果导数根本不变号，那就两者都不是。
Work one.
做一道。
Take x cubed minus three x squared.
取 x 的三次方减 3x 平方。
The derivative factors as three x, times x minus two, which is zero at zero and at two.
导数分解成 3x 乘以 (x 减 2)，在 0 和 2 处为零。
The signs run positive, negative, positive.
符号依次是正、负、正。
So zero is a local maximum with value zero, and two is a local minimum with value minus four.
所以 0 是局部最大值，函数值为 0；2 是局部最小值，函数值为负 4。
For an absolute extremum on a closed interval there is a faster method, and it needs no sign chart at all.
要在闭区间上求绝对极值，有一个更快的方法，而且完全不需要符号表。
It is called the candidates test.
它叫做候选点检验法。
On a closed interval, an absolute extremum can only happen at a critical point or at an endpoint.
在闭区间上，绝对极值只可能出现在临界点或端点。
So, one. List every critical point inside the interval, and both endpoints.
所以：第一，列出区间内的每一个临界点，以及两个端点。
Two. Evaluate the function at each one. Three.
第二，在每一个点处求函数值。
The largest output is the absolute maximum, and the smallest is the absolute minimum.
第三，最大的输出就是绝对最大值，最小的就是绝对最小值。
Show the table of values, because the comparison itself is the argument.
要把这张数值表写出来，因为这个比较过程本身就是论证。
The second derivative describes how the graph bends.
二阶导数描述图像如何弯曲。
Where the second derivative is positive, the graph is concave up.
二阶导数为正的地方，图像凹向上。
It holds water, and the slope is increasing.
它能盛住水，斜率在增大。
Where the second derivative is negative, the graph is concave down.
二阶导数为负的地方，图像凹向下。
It spills water, and the slope is decreasing.
它会把水洒出去，斜率在减小。
And where concavity actually changes, you have a point of inflection.
而凹凸性真正发生改变的地方，就是拐点。
Be careful here.
这里要小心。
The second derivative being zero is not enough by itself.
二阶导数等于零本身并不够。
It must change sign there, exactly as with the first derivative test.
它必须在那里变号，就像一阶导数检验法一样。
Concavity gives a second way to classify a critical point, and it is often quicker.
凹凸性还给出了判定临界点的第二种方法，而且常常更快。
Suppose the first derivative is zero at a point.
假设一阶导数在某点为零。
If the second derivative is positive there, the graph is concave up, a valley, so it is a local minimum.
如果二阶导数在那里为正，图像凹向上，是一个谷，所以是局部最小值。
If it is negative, the graph is concave down, a hill, so it is a local maximum.
如果二阶导数在那里为负，图像凹向下，是一个山丘，所以是局部最大值。
But if the second derivative is zero, the test is inconclusive, and you must fall back on the first derivative test.
但如果二阶导数为零，检验无定论，你必须退回去用一阶导数检验法。
One useful shortcut.
有一个很有用的捷径。
If a continuous function has only one critical point on an interval, and it is a local extremum, it is also the absolute extremum there.
如果一个连续函数在某区间上只有一个临界点， 而且那个点是局部极值，那么它同时也是那里的绝对极值。
Here is the exam setup that catches people out.
这是最容易让人栽跟头的考试设置。
The question gives you the graph of the derivative, then asks about the original function.
题目给你的是导数的图像，然后问原函数的性质。
Do not read it as if it were the function.
不要把它当成函数本身来读。
Read it with two rules.
用两条规则来读。
The height of this graph tells you whether the function is increasing or decreasing.
这张图的高度告诉你函数是递增还是递减。 在横轴上方表示上升，在下方表示下降。
Above the axis means rising, below means falling, and a crossing with a sign change is an extremum.
而这张图带着变号穿过横轴的地方，函数有极值。 这张图的斜率告诉你凹凸性。
The slope of this graph tells you about concavity.
上升表示凹向上，下降表示凹向下。
Rising means concave up, falling means concave down, and its own peak is a point of inflection.
而这张图自己出现峰或谷的地方，函数有拐点。
The three graphs mirror each other, feature for feature.
这三张图一个特征对一个特征地互相对应。
Optimization is all of this machinery pointed at a real goal. Find the largest area, the smallest cost, the shortest time.
最优化就是把前面这一整套工具对准一个真实的目标：求最大面积、最小成本、最短时间。
The procedure has four steps.
这个流程有四步。
One. Write the quantity as a function of one single variable, using the constraint to eliminate the others. Two.
第一，用约束条件消去其他变量，把要优化的量写成只含一个变量的函数。
State the interval of allowed inputs.
第二，写出允许的输入区间。
Three. Find and test the critical points. Four.
第三，求出并检验临界点，用任一种导数检验法， 如果区间是闭的就用候选点检验法。
Answer the question that was actually asked, with units and interpretation.
第四，回答题目真正问的那个问题，带上单位和解释。
If the question asks for the area, do not stop at the value of x.
如果题目问的是面积，就不要只写到 x 的值为止。
Work the classic one.
来做那道最经典的题。
A farmer has one hundred metres of fence for a rectangular pen built against a wall, so only three sides need fencing.
一个农民有一百米栅栏，要沿着一堵墙围一个矩形的畜栏， 所以只需要围三条边。
Maximize the area.
求最大面积。
Let the two ends be x, and the far side be y.
设两条端边为 x，远端那条边为 y。
The constraint is two x plus y equals one hundred, so y is one hundred minus two x.
约束条件是 2x 加 y 等于 100，所以 y 等于 100 减 2x。
Now the area function is x times that, which expands to one hundred x minus two x squared.
于是面积函数是 x 乘以它，展开得到 100x 减 2x 平方。
Set the derivative to zero.
令导数为零。
One hundred minus four x equals zero, so x is twenty five.
100 减 4x 等于 0，所以 x 等于 25。
The second derivative is minus four, which is negative, so this really is a maximum.
二阶导数是负 4，是负的，所以这确实是最大值。
Then y is fifty, and the maximum area is one thousand two hundred and fifty square metres.
于是 y 等于 50，最大面积是 1250 平方米。
One last section.
最后一节。
All of this extends to implicitly defined relations too.
这一整套方法同样适用于隐式关系。
A critical point of an implicit relation is where the derivative is zero, giving a horizontal tangent, or where it is undefined, giving a vertical tangent.
隐式关系的临界点， 是导数为零的地方，给出水平切线；或者导数无定义的地方，给出竖直切线。
There is one extra complication.
这里多了一个复杂之处。
The first derivative is usually an expression in both x and y, so the second derivative involves the first derivative as well.
一阶导数通常是同时含 x 和 y 的表达式， 于是二阶导数还会含有 x、y 以及一阶导数本身。
After you differentiate a second time, substitute your first derivative back in wherever it appears.
所以第二次求导之后， 要在出现一阶导数的每一处把它代回去。
Only then read concavity from the sign.
只有这样，你才能根据符号来判断凹凸性。
Three marks students lose.
学生最常丢分的三个地方。
First, an extremum or an inflection point needs a sign change, not merely a zero.
第一，极值或拐点需要的是变号，而不只是等于零。
Second, for an absolute extremum on a closed interval, check the endpoints too.
第二，求闭区间上的绝对极值时，端点也要检查。
Third, justify every conclusion by citing the sign of the first or second derivative.
第三，每一个结论都要引用一阶或二阶导数的符号来说明理由。
The exam wants the reasoning, not just the answer.
考试要的是推理过程，不只是答案。