Statement of Rolle's Theorem

Rolle's Theorem states:

Let $f$ be a function that satisfies the following conditions on a closed interval $[a, b]$:

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

Then there exists at least one point $c \in (a, b)$ such that $f'(c) = 0$.

Proof of Rolle's Theorem

1. Since $f$ is continuous on the closed interval $[a, b]$, by the Extreme Value Theorem, $f$ attains a maximum and a minimum value on $[a, b]$.

2. If $f$ is constant on $[a, b]$, then $f'(x) = 0$ for all $x \in (a, b)$, and the theorem is trivially satisfied since any $c \in (a, b)$ satisfies $f'(c) = 0$.

3. If $f$ is not constant on $[a, b]$, let $x = c$ be a point in $(a, b)$ where $f$ attains either a maximum or a minimum value.

4. Since $f(a) = f(b)$, this maximum or minimum value must occur at some point in $(a, b)$ (not at the endpoints).

5. By Fermat’s theorem, if $f$ has a local maximum or minimum at $c$ and $f$ is differentiable at $c$, then $f'(c) = 0$.

Thus, there exists a point $c \in (a, b)$ where $f'(c) = 0$, completing the proof.

Example of Rolle's Theorem

Consider the function $f(x) = x^2 - 4x + 3$ on the interval $[1, 3]$.

1. Continuity: The function $f(x) = x^2 - 4x + 3$ is a polynomial, so it is continuous on $[1, 3]$.
2. Differentiability: Since $f(x)$ is a polynomial, it is differentiable on $(1, 3)$.
3. Endpoints Equal: $f(1) = (1)^2 - 4(1) + 3 = 0$ and $f(3) = (3)^2 - 4(3) + 3 = 0$, so $f(1) = f(3)$.

Since all conditions of Rolle's Theorem are satisfied, there exists a point $c \in (1, 3)$ where $f'(c) = 0$.

Calculating $c$:

The derivative of $f(x)$ is: $f'(x) = 2x - 4$

Setting $f'(x) = 0$: $2x - 4 = 0 \implies x = 2$

Thus, $c = 2$, and indeed $f'(2) = 0$.

Geometric Interpretation

Rolle's Theorem implies that if a continuous and differentiable function has the same value at two endpoints, then there must be at least one point within the interval where the tangent to the curve is horizontal (i.e., the slope is zero).

In this example, the function $f(x) = x^2 - 4x + 3$ forms a parabola opening upward. The points $(1, 0)$ and $(3, 0)$ are on the x-axis, and the tangent to the curve at $x = 2$ is horizontal, indicating a local minimum.

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Would you like further details on Rolle's Theorem? Here are some related questions:

1. How does Rolle's Theorem relate to the Mean Value Theorem?
2. What happens if a function does not satisfy one of the conditions of Rolle's Theorem?
3. Can you give an example where a function does not meet Rolle's Theorem's differentiability condition?
4. How can Rolle's Theorem be applied to solve real-world problems?
5. How can Rolle's Theorem be extended to complex-valued functions?

Tip: Rolle’s Theorem only applies to functions that satisfy all three conditions. If any condition is not met, the theorem cannot be applied.