1.3 Mean Value Theorem, Rolle's Theorem, and Function Extrema

1.3 Mean Value Theorem, Rolle's Theorem, and Function Extrema

The Mean Value Theorem connects local derivative values to global interval changes. On the GRE Mathematics Subject Test, Rolle's Theorem and the MVT are tested through root counting, function bounding, and critical point classification.

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Rolle's Theorem and the Mean Value Theorem

Rolle's Theorem

Let $f: [a, b] \to \mathbb{R}$ satisfy three hypotheses:

1. $f$ is continuous on $[a, b]$.
2. $f$ is differentiable on $(a, b)$.
3. $f(a) = f(b)$.

Then there exists at least one $c \in (a, b)$ such that $f'(c) = 0$. Proof Blueprint: EVT guarantees extrema on $[a, b]$. If maximum equals minimum, $f$ is constant. If maximum or minimum occurs in $(a, b)$, Fermat's Theorem forces $f'(c) = 0$.

Mean Value Theorem (Lagrange MVT)

If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, there exists $c \in (a, b)$ with: $f'(c) = \frac{f(b) - f(a)}{b - a} \iff f(b) - f(a) = f'(c)(b - a).$

Core Corollaries

• Constancy: $f'(x) = 0$ on $(a, b) \implies f$ is constant.
• Monotonicity: $f'(x) > 0$ on $(a, b) \implies f$ is strictly increasing on $[a, b]$.
• Root Separation: Between any two distinct roots of a differentiable function $f$, there lies at least one root of $f'$.

Cauchy's Generalized MVT

If $f, g$ are continuous on $[a, b]$ and differentiable on $(a, b)$, then for some $c \in (a, b)$: $[f(b) - f(a)] g'(c) = [g(b) - g(a)] f'(c).$ When $g'(x) \neq 0$, this yields $\frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}$, which provides the theoretical foundation for L'Hôpital's Rule.

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Critical Points, Fermat's Theorem, and Extrema

A point $c \in \operatorname{Dom}(f)$ is a critical point if $f'(c) = 0$ or $f'(c)$ is undefined.

• Fermat's Theorem: If $f$ attains a local extremum at an interior point $c$ and $f'(c)$ exists, then $f'(c) = 0$.
• Converse Fails: $f(x) = x^3$ has $f'(0) = 0$, but $x = 0$ is an inflection point, not an extremum.

Closed Interval Method for Global Extrema

To find absolute extrema of continuous $f$ on $[a, b]$:

1. Locate all critical points $c_i \in (a, b)$.
2. Evaluate $f(c_i)$ at all critical points.
3. Evaluate $f(a)$ and $f(b)$ at boundary endpoints.
4. The maximum value is $\max {f(c_i), f(a), f(b)}$; the minimum is $\min {f(c_i), f(a), f(b)}$.

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Derivative Tests, Concavity, and Inflection Points

First Derivative Test

At critical point $c$:

• $f'$ flips from $+$ to $-$: local maximum.
• $f'$ flips from $-$ to $+$: local minimum.
• No sign flip: not an extremum.

Second Derivative Test

At stationary point $c$ ($f'(c) = 0$):

• $f''(c) > 0 \implies$ concave up $\implies$ local minimum.
• $f''(c) < 0 \implies$ concave down $\implies$ local maximum.
• $f''(c) = 0 \implies$ inconclusive.

Concavity and Inflection Points

• Concave Up: $f''(x) \ge 0$; graph lies above tangent lines.
• Concave Down: $f''(x) \le 0$; graph lies below tangent lines.
• Inflection Point: A point where the curve changes concavity. $f''(c) = 0$ is necessary (for $C^2$ functions) but not sufficient.

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Step-by-Step Worked Problems

Problem 1: Root Uniqueness via Rolle's Theorem

Prove $P(x) = x^7 + 4x^3 + 3x - 5 = 0$ has exactly one real solution.

Solution:

1. Existence: $P(0) = -5 < 0$ and $P(1) = 3 > 0$. By IVT, $P$ has at least one root in $(0, 1)$.
2. Uniqueness: $P'(x) = 7x^6 + 12x^2 + 3$. Since $x^6 \ge 0$ and $x^2 \ge 0$, $P'(x) \ge 3 > 0$ for all $x \in \mathbb{R}$.
3. If $P$ had two roots $x_1 < x_2$, Rolle's Theorem would guarantee $c \in (x_1, x_2)$ with $P'(c) = 0$, contradicting $P'(x) \ge 3$. Thus exactly one root exists.

Problem 2: Function Bounding via MVT

Show that $|\sin x - \sin y| \le |x - y|$ for all $x, y \in \mathbb{R}$.

Solution:

1. For $x = y$, the statement holds trivially. Assume $x < y$.
2. The function $f(t) = \sin t$ is continuous on $[x, y]$ and differentiable on $(x, y)$.
3. By MVT, $\frac{\sin y - \sin x}{y - x} = \cos c$ for some $c \in (x, y)$.
4. Taking absolute values gives $\frac{|\sin x - \sin y|}{|x - y|} = |\cos c| \le 1$.
5. Multiplying by $|x - y|$ yields $|\sin x - \sin y| \le |x - y|$.

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GRE Exam Traps & Pitfalls

Trap 1: Applying MVT Across Non-Differentiable Interior Points For $f(x) = |x|$ on $[-1, 2]$, the average slope is $\frac{2 - 1}{2 - (-1)} = \frac{1}{3}$. Yet $f'(x) \in {-1, +1}$ everywhere it exists. MVT fails because differentiability breaks down at $x = 0 \in (-1, 2)$.

Trap 2: Declaring Inflection Points Where $f''(c) = 0$ Without Sign Flip For $f(x) = x^4$, $f''(0) = 0$, but $f''(x) = 12x^2 \ge 0$ for all $x$. The graph remains concave up everywhere, so $(0, 0)$ is not an inflection point.

Trap 3: Forgetting Boundary Evaluations on Compact Sets On closed intervals, global extrema often occur at boundary endpoints rather than interior critical points. Always evaluate $f(a)$ and $f(b)$.