16.2 Tangent Lines, Derivatives, Monotonicity, and Extremum Analysis

16.2 Tangent Lines, Derivatives, Monotonicity, and Extremum Analysis

In Gaokao Mathematics, the derivative is the primary tool for analyzing function behavior. This section explores the dual nature of derivatives: their geometric representation as tangent slopes and their analytical power in determining monotonicity and local extrema.

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1. Tangent Lines and Normal Lines

1.1 Geometric Interpretation

For a curve $y = f(x)$, the derivative value $f'(x_0)$ represents the exact slope $k$ of the tangent line to the curve at point $P(x_0, f(x_0))$.

• Tangent Line Equation: $$y - f(x_0) = f'(x_0)(x - x_0)$$
• Normal Line Equation (the line perpendicular to the tangent at $P$, assuming $f'(x_0) \neq 0$): $$y - f(x_0) = -\frac{1}{f'(x_0)}(x - x_0)$$

1.2 Crucial Distinction in Gaokao Problems

Gaokao questions frequently distinguish between two phrasing types:

1. "Tangent AT point $P(x_0, y_0)$": Point $P$ lies on the curve ($y_0 = f(x_0)$), and the slope is directly $k = f'(x_0)$.
2. "Tangent PASSING THROUGH point $P(x_0, y_0)$": Point $P$ may NOT lie on the curve. In this case:
   • Assume the unknown point of tangency is $T(t, f(t))$.
   • Express the slope as $k = f'(t)$.
   • Set up the tangent line equation using $T$: $y - f(t) = f'(t)(x - t)$.
   • Substitute point $P(x_0, y_0)$ into the line equation to solve for $t$.

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2. Monotonicity Analysis via Derivatives

2.1 First Derivative Test for Monotonicity

Let $f(x)$ be continuous on $[a, b]$ and differentiable on $(a, b)$.

• Strictly Increasing: If $f'(x) > 0$ for all $x \in (a, b)$, then $f(x)$ is strictly increasing on $[a, b]$.
• Strictly Decreasing: If $f'(x) < 0$ for all $x \in (a, b)$, then $f(x)$ is strictly decreasing on $[a, b]$.
• Constant Function: If $f'(x) = 0$ for all $x \in (a, b)$, then $f(x)$ is constant on $[a, b]$.

Important Distinction: If $f(x)$ is strictly increasing on $(a,b)$, then $f'(x) \ge 0$ for all $x \in (a,b)$, provided $f'(x)$ is not identically zero on any subinterval. For example, $f(x) = x^3$ is strictly increasing on $\mathbb{R}$, even though $f'(0) = 0$.

2.2 Framework for Parametric Monotonicity Problems

When solving parametric monotonicity problems ($f(x)$ contains a parameter $a$):

1. Determine the domain $D$ of $f(x)$.
2. Compute $f'(x)$ and factorize or express it cleanly, e.g., $f'(x) = \frac{g(x)}{h(x)}$.
3. Identify roots of $f'(x) = 0$ (critical points).
4. Perform classification discussion (分类讨论) based on:
   • Sign of the leading coefficient of quadratic numerator/denominator.
   • Discriminant $\Delta$ of $f'(x) = 0$.
   • Location of roots relative to domain boundaries $D$.

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3. Local Extremum Analysis

3.1 Necessary Condition (Fermat's Theorem)

If $f(x)$ attains a local extremum (maximum or minimum) at $x_0$ and $f(x)$ is differentiable at $x_0$, then:

$$f'(x_0) = 0$$

Points where $f'(x_0) = 0$ are called stationary points or critical points. Note that $f'(x_0) = 0$ is a necessary, but not sufficient condition for a local extremum (e.g., $f(x) = x^3$ at $x=0$ has $f'(0)=0$, but no local extremum).

3.2 First Derivative Test for Local Extrema

Suppose $x_0$ is a critical point ($f'(x_0) = 0$) and $f(x)$ is differentiable in a neighborhood of $x_0$:

3.3 Second Derivative Test for Local Extrema

If $f'(x_0) = 0$ and $f''(x_0)$ exists:

• If $f''(x_0) < 0$, then $f(x)$ has a local maximum at $x_0$.
• If $f''(x_0) > 0$, then $f(x)$ has a local minimum at $x_0$.
• If $f''(x_0) = 0$, the test is inconclusive (must return to the first derivative test).

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4. Worked Gaokao Exam Problem

Problem Statement

Let $f(x) = \frac{1}{3}x^3 - \frac{1}{2}ax^2 + (a-1)x + 1$, where $a \in \mathbb{R}$.

1. Find the monotonic intervals of $f(x)$ in terms of parameter $a$.
2. If $f(x)$ has a local minimum equal to $\frac{5}{6}$, find the value of $a$.

Step-by-Step Solution

Step 1: Compute derivative and analyze roots.

$$f'(x) = x^2 - ax + (a-1)$$

Factorizing $f'(x) = 0$:

$$x^2 - ax + (a-1) = (x - 1)(x - (a-1)) = 0$$

The roots of $f'(x) = 0$ are $x_1 = 1$ and $x_2 = a - 1$.

Step 2: Classification discussion on root positions. Compare $x_1 = 1$ and $x_2 = a - 1$:

• $x_2 - x_1 = (a - 1) - 1 = a - 2$.

Case 1: When $a - 2 > 0 \implies a > 2$: Roots are $x_1 = 1 < x_2 = a - 1$.

• For $x \in (-\infty, 1)$ and $x \in (a-1, +\infty)$, $f'(x) > 0 \implies f(x)$ is strictly increasing.
• For $x \in (1, a-1)$, $f'(x) < 0 \implies f(x)$ is strictly decreasing.
• Local maximum at $x = 1$, local minimum at $x = a - 1$.

Case 2: When $a - 2 = 0 \implies a = 2$: Roots coincide: $x_1 = x_2 = 1$, so $f'(x) = (x - 1)^2 \ge 0$.

• $f'(x) > 0$ for all $x \neq 1$, so $f(x)$ is strictly increasing on $(-\infty, +\infty)$. No local extrema.

Case 3: When $a - 2 < 0 \implies a < 2$: Roots are $x_2 = a - 1 < x_1 = 1$.

• For $x \in (-\infty, a-1)$ and $x \in (1, +\infty)$, $f'(x) > 0 \implies f(x)$ is strictly increasing.
• For $x \in (a-1, 1)$, $f'(x) < 0 \implies f(x)$ is strictly decreasing.
• Local maximum at $x = a - 1$, local minimum at $x = 1$.

Step 3: Solve Part 2 (Local Minimum $= \frac{5}{6}$). From our classification:

• If $a > 2$, local minimum occurs at $x = a - 1$. $$f(a-1) = \frac{1}{3}(a-1)^3 - \frac{1}{2}a(a-1)^2 + (a-1)(a-1) + 1 = \frac{5}{6}$$ Evaluating at $a=3$: $x_2 = 2$. $f(2) = \frac{8}{3} - 6 + 4 + 1 = \frac{5}{3} \neq \frac{5}{6}$. Solving gives no valid $a > 2$.
• If $a < 2$, local minimum occurs at $x = 1$. $$f(1) = \frac{1}{3}(1)^3 - \frac{1}{2}a(1)^2 + (a-1)(1) + 1 = \frac{1}{3} - \frac{1}{2}a + a - 1 + 1 = \frac{1}{3} + \frac{1}{2}a$$ Set $\frac{1}{3} + \frac{1}{2}a = \frac{5}{6} \implies \frac{1}{2}a = \frac{5}{6} - \frac{2}{6} = \frac{3}{6} = \frac{1}{2} \implies a = 1$. Since $a = 1 < 2$, this solution is valid!

Final Answer: For $a > 2$, increasing on $(-\infty, 1)$ and $(a-1, +\infty)$, decreasing on $(1, a-1)$; for $a = 2$, increasing on $\mathbb{R}$; for $a < 2$, increasing on $(-\infty, a-1)$ and $(1, +\infty)$, decreasing on $(a-1, 1)$. The value for Part 2 is $a = 1$.