Rolle’s Theorem and The Mean Value Theorem

Rolle's Theorem is a special case of the Mean Value Theorem. It states that if a function f satisfies these three conditions:

1. f is continuous on the closed interval [a, b].
2. f is differentiable on the open interval (a, b).
3. f(a) = f(b) (the function has the same value at the endpoints).

Then there exists at least one number c in the open interval (a, b) such that f'(c) = 0. Geometrically, this means there's a point where the tangent line is horizontal.

The Mean Value Theorem is a more general theorem than Rolle's Theorem. It states that if a function f satisfies these two conditions:

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

Then there exists at least one number c in the open interval (a, b) such that:

f'(c) = [f(b) - f(a)] ÷ (b - a)

Geometrically, this means there's a point where the tangent line is parallel to the secant line connecting the endpoints (a, f(a)) and (b, f(b)). The right side of the equation is the slope of the secant line.

The main difference is the third condition in Rolle's Theorem (f(a) = f(b)). This condition is not required for the Mean Value Theorem. Because of this extra condition, Rolle's Theorem guarantees that the derivative at some point 'c' is zero (f'(c) = 0).

The Mean Value Theorem is more general, stating that the derivative at some point 'c' equals the *average rate of change* of the function over the interval [a, b] (which is the slope of the secant line). Rolle's Theorem is the special case where the average rate of change (and thus the derivative) is zero.

No, Rolle's Theorem is not the *same* as the Mean Value Theorem, but it is a specific instance or a special case of the MVT. The MVT is the more general theorem.

They become "equivalent" in the sense that Rolle's Theorem's conclusion is guaranteed *if* the Mean Value Theorem's conditions are met *and* the additional condition f(a) = f(b) is met. If f(a) = f(b), then the average rate of change [f(b) - f(a)] ÷ (b - a) becomes 0, and the Mean Value Theorem's conclusion f'(c) = [f(b) - f(a)] ÷ (b - a) simplifies to f'(c) = 0, which is exactly the conclusion of Rolle's Theorem.

The Mean Value Theorem can be proven by constructing an auxiliary function and applying Rolle's Theorem to it.

Let f be a function satisfying the MVT conditions. Define a new function g(x) = f(x) - [f(b) - f(a) ÷ b - a] * (x - a).

• This function g(x) is continuous on [a, b] and differentiable on (a, b) because f(x) is, and the second part is a simple linear function.
• Also, if you evaluate g(a) and g(b), you'll find that g(a) = f(a) and g(b) = f(b) - [f(b) - f(a) ÷ b - a] * (b - a) = f(b) - (f(b) - f(a)) = f(a). So, g(a) = g(b).

Since g(x) satisfies all conditions of Rolle's Theorem, there must exist a number c in (a, b) such that g'(c) = 0.

Now, find g'(x) = f'(x) - [f(b) - f(a) ÷ b - a]. Setting g'(c) = 0 gives f'(c) - [f(b) - f(a) ÷ b - a] = 0, which means f'(c) = [f(b) - f(a) ÷ b - a]. This proves the Mean Value Theorem.