Mean Value Theorem and Rolle's Theorem: 4.2

4.2: Mean Value Theorem

Introduction

This section explores two fundamental results in differential calculus: Rolle's Theorem and the Mean Value Theorem (MVT). Both theorems provide important connections between the behavior of a function and its derivative on a closed interval. Understanding these theorems is essential for analyzing the properties of differentiable functions and for proving further results in calculus.

Rolle's Theorem

Statement of Rolle's Theorem

• Rolle's Theorem states: Let f be continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, with $f(a) = f(b)$. Then there exists at least one point $c$ in $(a, b)$ such that $f'(c) = 0$.

Key Points:

• Continuity on $[a, b]$ is required.

• Differentiability on $(a, b)$ is required.

• The function values at the endpoints must be equal: $f(a) = f(b)$.

• There exists at least one $c$ in $(a, b)$ where the tangent is horizontal: $f'(c) = 0$.

Example: Consider a function representing the height of a thrown ball at time $t$; if the ball starts and ends at the same height, Rolle's Theorem guarantees that at some point, its vertical velocity (derivative) is zero.

Example 1: Application of Rolle's Theorem

• Given $f(x) = x(x-1)^2$ on $[0, 1]$:

  • Check continuity and differentiability: $f$ is a polynomial, so it is continuous and differentiable everywhere.

  • Check $f(0) = f(1) = 0$.

  • Find $c$ such that $f'(c) = 0$:

  • $f'(x) = 1 \cdot (x-1)^2 + x \cdot 2(x-1) = (x-1)^2 + 2x(x-1)$

  • Simplify: $f'(x) = (x-1)^2 + 2x(x-1) = (x-1)[(x-1) + 2x] = (x-1)(3x-1)$

  • Set $f'(c) = 0$: $(c-1)(3c-1) = 0 \implies c = 1$ (endpoint, not valid), $c = \frac{1}{3}$ (valid, since $\frac{1}{3} \in (0,1)$)

• Conclusion: Rolle's Theorem applies, and $c = \frac{1}{3}$ is the guaranteed point.

• Given $f(x) = 3x + x^{5/3}$ on $[-1, 1]$:

  • Check continuity: $f$ is defined for all $x$ in $[-1, 1]$.

  • Check differentiability: $f'(x) = 3 + \frac{5}{3}x^{2/3}$, which is not defined at $x = 0$.

  • Conclusion: Not differentiable on $(-1, 1)$, so Rolle's Theorem does not apply.

Mean Value Theorem (MVT)

Statement of the Mean Value Theorem

• Mean Value Theorem: Let f be continuous on $[a, b]$ and differentiable on $(a, b)$. Then there exists at least one point $c$ in $(a, b)$ such that:

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

• This means that at some point, the instantaneous rate of change (the derivative) equals the average rate of change over $[a, b]$.

Example: If you drive 100 miles in 2 hours, your average speed is 50 mph. The MVT guarantees that at some instant, your instantaneous speed was exactly 50 mph.

Example 2: Application of the Mean Value Theorem

• Given $f(x) = 7 - x^2$ on $[-1, 2]$:

  • Check if MVT applies: $f$ is a polynomial, so it is continuous and differentiable everywhere.

  • Compute $f'(c) = \frac{f(2) - f(-1)}{2 - (-1)} = \frac{(7 - 4) - (7 - 1)}{3} = \frac{3 - 6}{3} = -1$

  • Set $f'(x) = -2x = -1 \implies x = \frac{1}{2}$

  • Conclusion: $c = \frac{1}{2}$ is the guaranteed point.

• Given $f(x) = x + \frac{1}{x}$ on $[1, 3]$:

  • Check if MVT applies: $f$ is continuous and differentiable on $[1, 3]$ (since $x \neq 0$ and $0 \notin [1, 3]$).

  • Compute $f'(c) = \frac{f(3) - f(1)}{3 - 1} = \frac{(3 + \frac{1}{3}) - (1 + 1)}{2} = \frac{\frac{10}{3} - 2}{2} = \frac{\frac{4}{3}}{2} = \frac{2}{3}$

  • Set $f'(x) = 1 - \frac{1}{x^2} = \frac{2}{3}$

  • Solve: $1 - \frac{1}{x^2} = \frac{2}{3} \implies \frac{1}{x^2} = 1 - \frac{2}{3} = \frac{1}{3} \implies x^2 = 3 \implies x = \sqrt{3}$

  • Conclusion: $c = \sqrt{3}$ is the guaranteed point.

Summary Table: Conditions for Rolle's Theorem and MVT

Theorem	Continuity on $[a, b]$	Differentiability on $(a, b)$	Additional Condition	Conclusion
Rolle's Theorem	Required	Required	$f(a) = f(b)$	$\exists c \in (a, b): f'(c) = 0$
Mean Value Theorem	Required	Required	None	$\exists c \in (a, b): f'(c) = \frac{f(b) - f(a)}{b - a}$

Key Takeaways

• Both theorems require continuity on $[a, b]$ and differentiability on $(a, b)$.

• Rolle's Theorem is a special case of the Mean Value Theorem where $f(a) = f(b)$.

• These theorems are foundational for understanding the behavior of differentiable functions and for proving further results in calculus.