Analytical Applications of Differentiation

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

If $f$ is continuous on $[a,b]$ and differentiable 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, the instantaneous rate equals the average rate. Geometrically, some tangent line is parallel to the line joining the endpoints.

The Mean Value Theorem: some tangent is parallel to the secant over the interval

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. For $f(x)=x^2$ on $[1,3]$ the average rate is $\dfrac{9-1}{2}=4$; setting $f'(c)=2c=4$ gives $c=2$, which lies in $(1,3)$ – the guaranteed point.

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

At a maximum or minimum the derivative is zero

A critical point 临界点 is an interior point 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. So critical points are the candidates; you must test each.

The first derivative 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 as your justification.

Worked example. For $f(x)=x^3-3x^2$, $f'(x)=3x(x-2)$ is zero at $x=0,2$. Signs give $+,-,+$, so $x=0$ is a local maximum ($f=0$) and $x=2$ a local minimum ($f=-4$).

Where $f' > 0$ the graph rises and where $f' < 0$ it falls; a local max sits where $f'$ turns $+$ to $-$, a local min where it turns $-$ to $+$.

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

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.

The second derivative 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 where concavity changes, i.e. where $f''$ changes sign (not merely where $f''=0$). Report its $x$-coordinate and justify with the sign change of $f''$.

Concavity comes from the sign of the second derivative; it flips 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 – fall back on the first derivative test.

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.

Key features of $f$, $f'$, and $f''$ mirror each other. To sketch or read graphs:

• $f$ increasing $\Leftrightarrow$ $f'$ above the axis; $f$ has a local max $\Leftrightarrow$ $f'$ crosses from $+$ to $-$.
• $f$ concave up $\Leftrightarrow$ $f'$ increasing $\Leftrightarrow$ $f''$ above the axis; $f$ has an inflection point $\Leftrightarrow$ $f'$ has a local extremum $\Leftrightarrow$ $f''$ crosses zero.

This is the skill of reading one graph to describe another. A very common exam setup gives the graph of $f'$ and asks about $f$: where is $f$ increasing (where $f'>0$), where are $f$'s extrema (where $f'$ crosses zero, with a sign change), where is $f$ concave up (where $f'$ is increasing). Answer questions about $f$ using the height and slope of the $f'$ graph.

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 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. With $100\ \text{m}$ of fence for a rectangular pen against a wall (only three sides fenced), let the ends be $x$ and the far side $y=100-2x$. The area $A(x)=x(100-2x)=100x-2x^2$ has $A'(x)=100-4x=0$ at $x=25$; since $A''=-4<0$ this is the maximum, giving $y=50$ and $A=1250\ \text{m}^2$.

All of this extends to implicitly defined relations. A critical point of an implicit relation is where $\dfrac{dy}{dx}=0$ (horizontal tangent) or is undefined (vertical tangent). 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 ($f''>0$ 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.