Analysing Functions: The Mean Value Theorem
AP Calculus AB Topic 5 10:27 English narration · English + 中文 subtitles burned in
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Imagine a road trip.
想象一次公路旅行。
Over two hours, you cover one hundred and twenty kilometres — an average speed of sixty.
两小时里，你行驶了一百二十公里——平均速度是六十。
Here is a surprising promise: at some instant during that trip, your speedometer read exactly sixty. Not more, not less.
这里有一个令人惊讶的保证：在旅途中的某一瞬间，你的速度表恰好显示六十，不多也不少。
The average rate must be matched, somewhere, by the instantaneous rate.
平均变化率一定会在某处被瞬时变化率所匹配。
This is the Mean Value Theorem — the bridge between average and instantaneous change.
这就是中值定理——平均变化与瞬时变化之间的桥梁。
Today we use derivatives to analyse shape — where a function rises and falls, its peaks and valleys, its bends, and its very best values.
今天我们用导数来分析函数的形状——它在哪里上升、在哪里下降，它的峰与谷， 它的弯曲，以及它的最优值。
Let's begin.
让我们开始吧。
Watch the idea geometrically.
从几何上看这个想法。
Draw the straight line joining the two endpoints — that is the average rate of change.
画出连接两个端点的直线——那就是平均变化率。
The Mean Value Theorem promises that somewhere inside, there is a tangent line exactly parallel to it.
中值定理保证：在内部某处，有一条切线恰好与它平行。
At that point, the instantaneous rate equals the average rate.
在那一点，瞬时变化率等于平均变化率。
It works whenever the function is continuous on the closed interval, and differentiable inside.
只要函数在闭区间上连续、在内部可导，它就成立。
To use the theorem on the exam, you must justify.
要在考试中使用这个定理，你必须给出理由。
State that the function is continuous on the closed interval, and differentiable on the open one.
说明函数在闭区间上连续，在开区间上可导。
Compute the average rate — the change in output over the change in input.
计算平均变化率——输出的变化除以输入的变化。
Then conclude: by the Mean Value Theorem, there is a point c where the derivative equals that value.
然后得出结论：由中值定理， 存在一点 c，使得导数等于那个值。
For x squared on the interval one to three, the average rate is four, and setting the derivative equal to four gives c equals two — safely inside.
对区间一到三上的 x 平方，平均变化率是四， 令导数等于四得到 c 等于二——安全地落在内部。
Here is the picture behind the algebra.
这就是那些代数背后的图像。
Draw the secant line joining the two endpoints — its slope is the AVERAGE rate of change over the interval.
画出连接两个端点的割线—— 它的斜率就是整个区间上的平均变化率。
The theorem promises that somewhere strictly inside, at least one tangent line is exactly PARALLEL to that secant.
定理保证在区间内部至少存在一条切线，与那条割线恰好平行。
Parallel means equal slope, and the slope of the tangent is the instantaneous rate.
平行意味着斜率相等，而切线的斜率就是瞬时变化率。
So the geometric statement and the algebraic one are the same sentence: somewhere inside, the instantaneous rate equals the average rate.
所以几何的表述和代数的表述是同一句话： 在内部某处，瞬时变化率等于平均变化率。
Verify the Mean Value Theorem for f of x equals x squared on the closed interval one to three.
对 f x 等于 x 平方，在闭区间一到三上验证中值定理。
Step one, the hypotheses: it is a polynomial, so continuous everywhere and differentiable everywhere, which covers both conditions.
第一步，检验前提：它是多项式，所以处处连续、处处可导，两个条件都满足了。
Step two, the average rate is f of three minus f of one over three minus one — nine minus one over two, which is four.
第二步，平均变化率是 f 三减 f 一，除以三减一—— 九减一再除以二，等于四。
Step three, set f prime of c equal to that: two c equals four, so c equals two.
第三步，令 f 撇 c 等于这个值：二 c 等于四，所以 c 等于二。
And finish the sentence properly: two lies strictly inside the open interval one to three, which is the point the theorem guaranteed.
然后要把话说完整：二严格位于开区间一到三之内， 这正是定理所保证的那个点。
Checking that c is genuinely inside is part of the answer, not an afterthought.
检验 c 确实在区间内部是答案的一部分，而不是可有可无的补充。
Now, finding the important points.
现在来找那些重要的点。
The Extreme Value Theorem promises that a continuous function on a closed interval reaches both a highest and a lowest value.
极值定理保证：闭区间上的连续函数一定同时取到最高值和最低值。
Where can those extremes hide?
这些极值会藏在哪里呢？
At a critical point — an interior point where the derivative is zero, or where it does not exist.
藏在临界点——那是导数为零、或导数不存在的内部点。
All local relative extrema — every local peak or valley — occur at critical points.
所有局部相对极值——每一个局部的峰或谷——都出现在临界点。
But be careful: not every critical point is a peak or a valley.
但要小心： 并非每个临界点都是峰或谷。
Critical points are only the candidates — you must test each one.
临界点只是候选——你必须逐个检验。
The first derivative tells you the shape.
一阶导数告诉你形状。
Where the derivative is positive, the function is increasing — climbing uphill.
导数为正的地方，函数在递增——向上爬坡。
Where the derivative is negative, the function is decreasing — going downhill.
导数为负的地方，函数在递减——向下走。
So to find where a function rises or falls, find the critical points, then test the sign of the derivative between them.
所以要找函数在哪里上升或下降， 先找临界点，再检验它们之间导数的符号。
And always justify with the sign of the derivative — a stated reason, not just an interval.
并且永远用导数的符号来说明理由—— 要给出明确的理由，而不只是一个区间。
To classify a critical point, watch how the derivative changes sign.
要判定一个临界点，观察导数如何改变符号。
If it turns from positive to negative, the function stops climbing and starts falling — a local maximum.
如果它从正变负，函数就停止上升、开始下降—— 这是局部极大值。
If it turns from negative to positive, a local minimum.
如果它从负变正，就是局部极小值。
If it does not change sign at all, the point is neither.
如果它根本不变号，那这个点两者都不是。
For x cubed minus three x squared, the derivative is zero at zero and at two: a maximum at zero, and a minimum at two.
对 x 立方 减 三 x 平方，导数在零和二处为零：零处是极大值，二处是极小值。
The second derivative describes bending.
二阶导数描述弯曲。
Where it is positive, the curve is concave up — shaped like a cup.
它为正的地方，曲线上凹——形状像杯子。
Where it is negative, concave down — like a frown.
它为负的地方，下凹——像皱眉。
A point of inflection is where the concavity changes, where the second derivative changes sign.
拐点是凹凸性改变的地方，也就是二阶导数变号的地方。
This also gives a quick test for a critical point: if the second derivative is positive there, it is a minimum; if negative, a maximum; if zero, the test is inconclusive.
这还给出一个判定临界点的快捷方法： 如果二阶导数在那里为正，就是极小值；为负，就是极大值；为零，这个检验就失效了。
Read the bending off the picture.
从图上把弯曲方向读出来。
Where the curve holds water like a cup, it is concave up, and f double-prime is positive there.
曲线像杯子一样能盛住水的地方是上凹， 那里的 f 二阶导为正。
Where it spills, it is concave down and f double-prime is negative.
倒扣着会把水洒出去的地方是下凹，f 二阶导为负。
The point where it FLIPS is the point of inflection.
弯曲方向翻转的那一点就是拐点。
Now the trap, and it costs marks every year: an inflection point is where f double-prime CHANGES SIGN, not merely where f double-prime equals zero.
现在说那个陷阱，它每年都要丢分： 拐点是 f 二阶导变号的地方，而不仅仅是 f 二阶导等于零的地方。
A function can have f double-prime equal to zero and carry straight on bending the same way.
一个函数可以在某点 f 二阶导等于零，却仍旧朝同一个方向继续弯下去。
Report the x-coordinate, and justify with the sign change.
要报出 x 坐标，并用变号来给出理由。
There is a faster way to classify a critical point where f prime of c is zero — just look at the bending there.
要给 f 撇 c 等于零的临界点分类，还有一条更快的路——只看该处的弯曲方向。
If f double-prime of c is positive, concave up means a local MINIMUM.
如果 f 二阶导 c 为正，上凹就意味着极小值。
If it is negative, concave down means a local MAXIMUM.
如果为负，下凹就意味着极大值。
And if f double-prime of c is zero, the test is inconclusive, so you fall back on the first derivative test and its sign change.
而如果 f 二阶导 c 等于零，这个判别法失效， 于是退回到一阶导数判别法，用它的变号来判断。
One extra result worth carrying: if a continuous function has only ONE critical point on an interval and that point is a local extremum, then it is also the ABSOLUTE extremum there — which saves you building a candidates table.
还有一条结论值得随身带着：如果一个连续函数在某区间上只有一个临界点， 而且那个点是极值点，那么它同时也是该区间上的最值—— 这样你就不必再列候选点表了。
For the absolute — the global — highest and lowest on a closed interval, there is a simple test.
对于闭区间上的绝对——也就是全局——最高值和最低值，有一个简单的检验。
The candidates are the critical points, plus the two endpoints.
候选点是临界点，加上两个端点。
List them all, evaluate the function at each, and compare.
把它们全列出来，在每个点求函数值，然后比较。
The largest output is the absolute maximum; the smallest is the absolute minimum.
最大的输出是绝对最大值；最小的是绝对最小值。
Show the table — the comparison itself is your argument.
列出表格——这个比较本身就是你的论证。
And never forget the endpoints.
而且永远不要忘记端点。
A favourite exam trick: give you the graph of the derivative, and ask about the function itself.
一个考试最爱的花招：给你导数的图象，却问原函数本身。
Read it this way.
这样读它。
Where the derivative graph is above the axis, the function is increasing.
导数图象在横轴上方的地方，函数在递增。
Where the derivative crosses zero with a sign change, the function has a peak or a valley.
导数带着变号穿过零点的地方，函数有峰或谷。
And where the derivative is itself increasing, the function is concave up.
而导数本身在递增的地方，函数上凹。
You describe the function using the height and the slope of the derivative graph.
你用导数图象的高度和斜率，来描述原函数。
The three graphs are the same information told three ways, and the exam loves handing you one and asking about another.
这三张图是同一份信息的三种讲法，而考试最喜欢给你其中一张、问你另一张。
Learn them as equivalences.
要把它们当作等价关系来记。
f increasing means f prime above the axis.
f 递增，就是 f 撇在横轴上方。
f has a local maximum where f prime crosses from plus to minus.
f 在 f 撇由正变负穿过横轴处取得极大值。
f concave up means f prime is INCREASING, which means f double-prime is above the axis.
f 上凹，就是 f 撇在递增，也就是 f 二阶导在横轴上方。
And f has an inflection point exactly where f prime has a local extremum, which is where f double-prime crosses zero.
而 f 的拐点恰好在 f 撇取得极值的地方，也就是 f 二阶导穿过零点的地方。
Notice the pattern: a question about f's SHAPE becomes a question about f prime's HEIGHT, and a question about f's BENDING becomes a question about f prime's SLOPE.
注意这个规律：问 f 的形状，就变成问 f 撇的高度； 问 f 的弯曲，就变成问 f 撇的斜率。
All of this powers optimization — finding the best possible value.
这一切都为最优化提供了动力——寻找最好的可能取值。
Say we want the box of largest volume from a fixed sheet of card.
比如，我们想用一张固定的卡纸， 做出体积最大的盒子。
We write the volume as a function of one variable, then use the derivative to find its maximum.
我们把体积写成一个变量的函数，再用导数找它的最大值。
The same critical-point machinery, aimed at a real goal: the most volume, the least cost, the greatest area.
同样的临界点机制，瞄准一个真实的目标：最大的体积、最小的成本、最大的面积。
Let's optimise.
我们来优化。
A farmer has one hundred metres of fence for a rectangular pen against a wall — so only three sides need fencing.
一位农夫有一百米栅栏，要沿一堵墙围一个矩形畜栏——所以只需要围三面。
What dimensions give the largest area?
什么尺寸能给出最大面积？
Let the two ends be x, and the side along the wall be y.
设两端为 x，沿墙的一边为 y。
The fence constraint says two x plus y equals one hundred.
栅栏约束是 二 x 加 y 等于一百。
Substitute to get the area as one variable: x times one hundred minus two x.
代入把面积写成一个变量：x 乘以 一百减二 x。
Differentiate, set it to zero, and x is twenty-five.
求导，令它为零，得到 x 是二十五。
So y is fifty, and the largest area is one thousand two hundred and fifty square metres.
于是 y 是五十，最大面积是一千二百五十平方米。
Optimization is the machinery you already have, aimed at a real goal, and a dependable procedure keeps it from going wrong.
最优化就是你已经掌握的那套工具，用来对准一个实际目标， 而一套可靠的步骤能让它不出错。
One: write the quantity you are optimizing as a function of ONE variable, using the constraint equation to eliminate the others.
第一：把你要优化的量写成一个变量的函数，用约束方程把其他变量消掉。
Two: state the interval of allowed inputs, which is where the physical situation limits you.
第二：写出自变量的允许区间，也就是实际情形对你的限制所在。
Three: find and TEST the critical points — first derivative test, second derivative test, or the candidates test if the interval is closed.
第三：求出临界点并加以检验——一阶导数判别法、二阶导数判别法， 或者当区间是闭区间时用候选点法。
Four, and this is where marks are dropped: answer the question that was ASKED, with units and an interpretation in context.
第四，也正是丢分的地方：回答题目实际问的那个问题， 带上单位和在具体情境中的解释。
The maximum AREA, the minimum COST — not a bare number.
是最大的面积、最小的成本——而不是一个光秃秃的数字。
Everything here extends to implicitly defined relations, with two adjustments.
这里所有的内容都可以推广到隐函数关系，只需要两处调整。
A critical point is where d y d x equals zero, giving a horizontal tangent, OR where d y d x is undefined, giving a VERTICAL tangent — that second case does not arise for ordinary functions, so it is easy to forget.
临界点是 d y d x 等于零的地方，对应水平切线； 或者是 d y d x 无定义的地方，对应竖直切线—— 后面这种情形在普通函数里不会出现，所以很容易被忘掉。
And because d y d x is usually an expression in both x and y, the second derivative involves x, y and d y d x together.
而由于 d y d x 通常是同时含有 x 和 y 的表达式， 二阶导数就会同时牵涉 x、y 和 d y d x。
So when you differentiate again, substitute the first derivative expression back in before you try to reason about the sign.
所以再次求导之后，要先把一阶导数的表达式代回去， 再去讨论它的符号。
Then concavity follows from that sign exactly as before.
之后凹凸性就完全照旧由这个符号决定。
Before you go, three marks to keep.
结束之前，三个要守住的分。
First, the sign of the first derivative decides increasing or decreasing — always cite it.
第一，一阶导数的符号决定递增还是递减——一定要引用它。
Second, classify a critical point by the sign change of the first derivative, or the sign of the second.
第二，用一阶导数的变号，或二阶导数的符号，来判定临界点。
Third, for an absolute extremum on a closed interval, always check the endpoints too, not just the critical points.
第三，对闭区间上的绝对极值， 一定也要检查端点，而不只是临界点。
Justify every step — the exam rewards the reasoning.
每一步都给出理由——考试奖励的是推理。
This topic is yours.
这个专题就是你的了。