微分的分析性应用

The Mean Value Theorem 中值定理 (MVT) links the average rate of change 平均变化率 to an instantaneous one:

If $f$ is continuous 连续 on $[a,b]$ and differentiable 可导 on $(a,b)$, then there is at least one point $c$ in $(a,b)$ where

$$f'(c) = \frac{f(b)-f(a)}{b-a}.$$

In words: somewhere inside the interval 区间, the instantaneous rate equals the average rate. Geometrically, some tangent line 切线 is parallel 平行 to the line joining the endpoints 端点.

Exam skill. Like the IVT, the MVT is an existence theorem, and questions ask you to justify. Full credit needs: (1) state $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$; (2) compute the average rate $\frac{f(b)-f(a)}{b-a}$; (3) conclude “by the MVT there is a $c$ in $(a,b)$ with $f'(c)$ equal to that value.” Both hypotheses must be named.

Worked example. Verify the MVT for $f(x)=x^2$ on $[1,3]$. It is a polynomial, so continuous and differentiable everywhere. The average rate is $\dfrac{f(3)-f(1)}{3-1}=\dfrac{9-1}{2}=4$. Setting $f'(c)=2c=4$ gives $c=2$, which does lie in $(1,3)$ – the guaranteed point.

Why both hypotheses matter. Take $f(x)=|x|$ on $[-1,1]$. It is continuous, and the average rate is $\dfrac{f(1)-f(-1)}{1-(-1)}=\dfrac{1-1}{2}=0$. But $f'(x)$ is $-1$ or $1$ everywhere it exists, never $0$. The MVT does not apply, because $f$ is not differentiable at $x=0$, which is inside $(-1,1)$.

The Extreme Value Theorem 极值定理 (EVT) guarantees extremes exist: a function continuous on a closed interval 闭区间 $[a,b]$ attains both an absolute maximum 最大值 and an absolute minimum 最小值 on it.

A critical point 临界点 is an interior point 内点 of the function's domain 定义域 where $f'(x)=0$ or $f'(x)$ does not exist. All local (relative) extrema 局部极值 occur at critical points – but not every critical point is an extremum 极值. So critical points are the candidates 候选点; you must test each.

The first derivative 一阶导数 tells you where $f$ rises or falls:

• $f'(x) > 0$ on an interval $\Rightarrow$ $f$ is increasing 递增 there;
• $f'(x) < 0$ $\Rightarrow$ $f$ is decreasing 递减.

On the exam, “find the intervals where $f$ is increasing” means: find the critical points, then test the sign of $f'$ between them, and justify with the sign of $f'$ (a stated reason, not just an interval).

To classify a critical point $x=c$ as a local max, local min, or neither, check how $f'$ changes sign there:

• $f'$ changes $+$ to $-$ at $c$ $\Rightarrow$ local maximum 极大值;
• $f'$ changes $-$ to $+$ at $c$ $\Rightarrow$ local minimum 极小值;
• $f'$ does not change sign $\Rightarrow$ neither.

Always state the sign change 符号变化 as your justification 理由.

Worked example. Classify the critical points of $f(x)=x^3-3x^2$. Here $f'(x)=3x^2-6x=3x(x-2)$, zero at $x=0$ and $x=2$. Testing signs: $f'>0$ for $x<0$, $f'<0$ on $(0,2)$, and $f'>0$ for $x>2$. So $f'$ turns $+\to-$ at $x=0$ (a local maximum, $f(0)=0$) and $-\to+$ at $x=2$ (a local minimum, $f(2)=-4$).

Common mistakes. Justifying with “$f'(0)=0$” alone: a zero of $f'$ only makes $x=0$ a candidate; the reason is the sign change of $f'$. Testing the sign of $f$ instead of $f'$. Giving the $x$-value when the question asks for the maximum value ($f(0)=0$), or the reverse.

On a closed interval, absolute (global) extrema occur only at critical points or endpoints. The candidates test 候选点判别法:

1. List all critical points in $[a,b]$ and the two endpoints.
2. Evaluate $f$ at each candidate.
3. The largest output is the absolute maximum; the smallest is the absolute minimum.

Show the table of values – the comparison is the argument.

Worked example. Find the absolute maximum and minimum of $f(x)=x^3-3x^2$ on $[1,4]$.

$f'(x)=3x(x-2)$ is zero at $x=0$ and $x=2$. Only $x=2$ lies in $[1,4]$, so the candidates are $x=1$, $x=2$ and $x=4$:

The absolute maximum is $16$, at $x=4$; the absolute minimum is $-4$, at $x=2$. The maximum is at an endpoint where $f'(4)=24\neq 0$: this is why the endpoints must be on the list.

The second derivative 二阶导数 describes bending:

• $f'' > 0$ $\Rightarrow$ $f$ is concave up 上凹 (curving like a cup; $f'$ is increasing);
• $f'' < 0$ $\Rightarrow$ $f$ is concave down 下凹 ($f'$ is decreasing).

A point of inflection 拐点 is a continuous point where concavity changes. Where the second derivative exists on neighbouring intervals, use its sign change; a zero alone is insufficient. Report its $x$-coordinate and justify with the sign change of $f''$.

Worked counterexample: a zero is only a candidate

For $h(x)=x^4$, $h''(x)=12x^2$ is positive on both sides of 0, so there is no inflection. Yet $h(x)\ge h(0)=0$, so 0 is a minimum even though $h''(0)=0$. For $k(x)=x^3$, $k''(x)=6x$ changes from negative to positive, giving an inflection at 0. Check neighbouring concavity; the second derivative need not exist at the point itself. The function must be continuous at an inflection point.

An alternative way to classify a critical point $c$ where $f'(c)=0$:

• $f''(c) > 0$ $\Rightarrow$ concave up $\Rightarrow$ local minimum;
• $f''(c) < 0$ $\Rightarrow$ concave down $\Rightarrow$ local maximum;
• $f''(c) = 0$ $\Rightarrow$ the test is inconclusive 无法判定 – fall back on the first derivative test 一阶导数判别法.

Special case: if a continuous function has only one critical point on an interval and it is a local extremum, that point is also the absolute extremum there.

Worked example: connect the same three graphs

For $f(x)=x^3-3x^2$, the derivative is $f'(x)=3x(x-2)$ and the second derivative is $f''(x)=6x-6$.

On $(-1,0)$, $(0,2)$ and $(2,3)$, the signs of $f'$ are positive, negative and positive. Hence $f$ increases, decreases and increases. The sign changes give a local maximum at $(0,0)$ and minimum at $(2,-4)$. On $(-1,0)$ the function is increasing but concave down: $f'>0$ while $f'$ decreases. Concavity changes at 1 because $f''$ changes from negative to positive; the inflection is $(1,-2)$, where $f'(1)=-3$.

Adding 5 to $f$ leaves both derivatives unchanged. The shape and turning-point x coordinates stay the same; the vertical position needs a given function value.

Original 2018 AB Q 3(c–d): read the displayed derivative

The given $g$ is $f'$. For $f$ to be increasing and concave up, $g$ must be positive and increasing: $(0,1)$ and $(4,6)$. At 4, $g$ changes from decreasing to increasing, so $f$ has an inflection. Since $g$ is positive on both sides of 4, $f$ has no extremum there. The official scheme awards both the interval answer and its reason.

Original 2025 AB Q 4B: the notation changes

Here $g(x)=\int_6^x f(t)\,dt$, so $g'=f$. The inflections of $g$ occur at $x=-3,3,6$: the displayed $f$ changes decreasing/increasing behaviour there. At 0, $f$ increases on both sides, so $g$ remains concave up. At 6 the change of concavity is valid even though $f'$ is undefined at its corner. The official scheme requires evidence tied to the given graph, not simply naming a change in concavity.

Common mistakes when the graph shown is $f'$. Reading it as if it were $f$. On a graph of $f'$: $f$ increases where the graph lies above the axis (not where it rises); $f$ has a local extremum where the graph crosses the axis; and $f$ has an inflection point where the graph of $f'$ turns (changes from increasing to decreasing, or the reverse).

Optimization 最优化 uses the derivative to find the largest or smallest value of a quantity on an interval. It is the candidates/derivative-test machinery applied to a real goal.

A dependable procedure:

1. Write the quantity to optimize as a function of one variable (use a constraint 约束条件 equation to eliminate extras).
2. State the interval of allowed inputs.
3. Find critical points ($f'=0$ or undefined) and test them (first- or second-derivative test, or candidates test if the interval is closed).
4. Answer the question asked, with units and interpretation in context – the maximum area, the minimum cost, etc.

Worked example. A farmer has $100\ \text{m}$ of fence for a rectangular pen against a wall, so only three sides need fencing. Maximize the area. Let the two ends be $x$ and the side parallel to the wall be $y$; the constraint is $2x+y=100$, so $y=100-2x$. The area is

Common mistakes. Trying to optimize a function of two variables without using the constraint to remove one; skipping the justification that $x=25$ gives a maximum (here $A''<0$); and stopping at $x=25$ when the question asks for the largest area.

All of this extends to implicitly defined 隐式定义 relations. A critical point of an implicit relation is where $\dfrac{dy}{dx}=0$ (horizontal tangent 水平切线) or is undefined 无定义 (vertical tangent 竖直切线). Because $\dfrac{dy}{dx}$ is usually a relation in $x$ and $y$, and the second derivative involves $x$, $y$, and $\dfrac{dy}{dx}$, substitute your first-derivative expression back in when finding $\dfrac{d^2y}{dx^2}$, then reason about concavity from its sign.

• $f'>0$ means increasing, $f'<0$ decreasing; candidates for extrema are where $f'=0$ or is undefined.
• Classify a critical point with the first-derivative sign change or the second-derivative test 二阶导数判别法 ($f''>0$ minimum, $f''<0$ maximum).
• $f''>0$ is concave up, $f''<0$ concave down; a point of inflection is where concavity changes ($f''$ changes sign).
• For an absolute extremum on a closed interval, also check the endpoints.
• Justify every conclusion by citing the sign of $f'$ or $f''$ — the exam demands the reasoning, not just the answer.