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

Then there exists a number c in (a,b) such that f'(c) = 0.

Let f be a function that satisfies 2 conditions:

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

Then there exists a number c in (a,b) such that f'(c) = [f(b) - f(a)] / (b-a).

Mean Value Theorem is very important theorem because it is needed to graph functions using derivatives. That doesn't mean that Rolle's Theorem can be neglected. As you can see, Mean Value Theorem has same 2 conditions as Rolle's Theorem. Rolle's Theorem builds up Mean Value Theorem. Mean Value Theorem is the most important theorem in calculus, so you might want to memorize it.

Example 1.3.1:

Verify that it satisfies Rolle's Theorem.

f(x) = x² on [0,1]

Solution: You can see that f(x) is both continuous and differentiable (able to

take derivative).

Conditions: 1) f is continuous on [-1,1] since it's a polynomial.

2) f is differentiable on (-1,1) since f'(x) = 2x.

3) f(-1) = 1 = f(1)

Then there exist a number c in (-1,1) such that f'(c) = 0.

So, f'(c) = 2c = 0 implies c = 0.

Rolle's Theorem is verified since c is in (-1,1).

Example 1.3.2:

Verify that it satisfies Mean Value Theorem.

f(x) = x² on [0,1]

Solution: I have put same function so that you can compare the two.

I know that f(x) is both continuous and differentiable.

Conditions: 1) f is continuous on [-1,1] since it's a polynomial.

2) f is differentiable on (-1,1) since f'(x) = 2x.

Then there exists a number c in (-1,1) such that f'(c) = [f(1) - f(-1)]/[1 - (-1)].

So, f'(c) = (1 - 1)/(2) = 0.

Now, f'(c) = 2c = 0 implies c = 0. Verified since 0 is in (-1,1).

This c is same as when I verified Rolle's Theorem, so you can see that they are related.

Try these problems.

Problem 1.3.1:

Verify by both Rolle's Theorem and Mean Value Theorem.

Problem 1.3.2:

Verify by Mean Value Theorem.

f(x) = 1/x on [1,2]