Section 1: Fermat's Theorem

Theorem 1.1: Fermat's Theorem

If $f$ has a local extremum at $x_0$ and $f'(x_0)$ exists, then:

f'(x_0) = 0

Proof of Fermat's Theorem:

WLOG, assume $x_0$ is a local maximum. Then for $h$ small enough:

$f(x_0 + h) \leq f(x_0)$

For $h > 0$ : $\frac{f(x_0 + h) - f(x_0)}{h} \leq 0$

For $h < 0$ : $\frac{f(x_0 + h) - f(x_0)}{h} \geq 0$

Taking limits: $f'(x_0) \leq 0$ and $f'(x_0) \geq 0$ , so $f'(x_0) = 0$ . ∎

∎

Section 2: Rolle's Theorem

Theorem 2.1: Rolle's Theorem

If $f$ is continuous on $[a, b]$ , differentiable on $(a, b)$ , and $f(a) = f(b)$ , then:

\exists c \in (a, b): f'(c) = 0

Proof of Rolle's Theorem:

By EVT, $f$ attains its maximum $M$ and minimum $m$ on $[a, b]$ .

If $M = m$ , then $f$ is constant, so $f'(x) = 0$ for all $x$ .

Otherwise, at least one extremum occurs at an interior point $c \in (a, b)$ .

By Fermat's theorem, $f'(c) = 0$ . ∎

∎

Example 2.1: Rolle's Theorem Application

Show: $x^3 - 3x + 1 = 0$ has at most one root in $[0, 1]$ .

Proof: Suppose it has two roots $a, b$ in $[0, 1]$ with $a < b$ .

By Rolle's theorem, $\exists c \in (a, b)$ with $f'(c) = 3c^2 - 3 = 0$ .

But $3c^2 - 3 = 0$ implies $c = \pm 1$ , neither in $(0, 1)$ . Contradiction! ∎

Section 3: Lagrange's Mean Value Theorem

Theorem 3.1: Lagrange's Mean Value Theorem

If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ , then:

\exists c \in (a, b): f'(c) = \frac{f(b) - f(a)}{b - a}

Proof of Lagrange's MVT:

Define $g(x) = f(x) - \frac{f(b) - f(a)}{b - a}(x - a)$ .

Then $g(a) = f(a)$ and $g(b) = f(b) - (f(b) - f(a)) = f(a)$ .

By Rolle's theorem, $\exists c \in (a, b)$ with $g'(c) = 0$ .

Since $g'(c) = f'(c) - \frac{f(b) - f(a)}{b - a} = 0$ , we have $f'(c) = \frac{f(b) - f(a)}{b - a}$ . ∎

∎

Corollary 3.1: Constant Function

If $f'(x) = 0$ for all $x \in (a, b)$ , then $f$ is constant on $(a, b)$ .

Example 3.1: MVT Application

Prove: $|\sin x - \sin y| \leq |x - y|$

Proof: By MVT, $\sin x - \sin y = \cos \xi (x - y)$ for some $\xi$ .

Since $|\cos \xi| \leq 1$ , we have $|\sin x - \sin y| \leq |x - y|$ . ∎

Section 4: Cauchy's Mean Value Theorem

Theorem 4.1: Cauchy's Mean Value Theorem

If $f$ and $g$ are continuous on $[a, b]$ and differentiable on $(a, b)$ , and $g'(x) \neq 0$ for all $x \in (a, b)$ , then:

\exists c \in (a, b): \frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}

Proof of Cauchy's MVT:

Define $h(x) = f(x) - \frac{f(b) - f(a)}{g(b) - g(a)}[g(x) - g(a)]$ .

Then $h(a) = f(a)$ and $h(b) = f(b) - (f(b) - f(a)) = f(a)$ .

By Rolle's theorem, $\exists c \in (a, b)$ with $h'(c) = 0$ .

Since $h'(c) = f'(c) - \frac{f(b) - f(a)}{g(b) - g(a)}g'(c) = 0$ , the result follows. ∎

∎

Section 5: Applications of MVT

Theorem 5.1: Constant Function Theorem

If $f'(x) = 0$ for all $x$ in an interval, then $f$ is constant on that interval.

Proof:

For any $a < b$ in the interval, by MVT: $f(b) - f(a) = f'(c)(b-a) = 0$ .

Therefore $f(b) = f(a)$ , so $f$ is constant.

∎

Section 6: Monotonicity via Derivatives

Theorem 6.1: Monotonicity Test

If $f'(x) > 0$ on an interval, then $f$ is strictly increasing.

If $f'(x) < 0$ on an interval, then $f$ is strictly decreasing.

Proof:

For $a < b$ , by MVT: $f(b) - f(a) = f'(c)(b-a)$ for some $c \in (a,b)$ .

If $f'(c) > 0$ , then $f(b) > f(a)$ .

∎

Section 7: Darboux's Theorem

Theorem 7.1: Darboux's Theorem (Intermediate Value Property for Derivatives)

If $f$ is differentiable on $[a,b]$ , then $f'$ has the intermediate value property.

That is, if $f'(a) < k < f'(b)$ , then $\exists c \in (a,b)$ with $f'(c) = k$ .

Practice Quiz: Mean Value Theorems

10

Questions

0

Correct

0%

Accuracy

1

What are the conditions for Rolle's theorem?

Easy

Not attempted

2

Lagrange's MVT states that

\exists \xi \in (a,b)

such that:

Easy

Not attempted

3

If

f'(x) = 0

for all

x \in (a,b)

, what can we conclude?

Easy

Not attempted

4

If

f(x_0)

is a local maximum and

f'(x_0)

exists, then:

Easy

Not attempted

5

Use MVT: If

f'(x) > 0

for all

x

, then

f

is:

Easy

Not attempted

6

Cauchy's MVT relates:

Medium

Not attempted

7

Prove

|\sin x - \sin y| \leq |x - y|

using:

Medium

Not attempted

8

If

f(x) = x^3 - 3x

on

[-2, 2]

, does Rolle's theorem apply?

Medium

Not attempted

9

Use Lagrange's MVT to prove that for

x > 0

:

\frac{x}{1+x} < \ln(1+x) < x

.

Hard

Not attempted

10

If

f''(x) > 0

on

(a,b)

and

f(a) = f(b) = 0

, then for

x \in (a,b)

:

Hard

Not attempted

Frequently Asked Questions

What are the conditions for Rolle's theorem?

Rolle's theorem requires: (1) f continuous on [a,b], (2) f differentiable on (a,b), (3) f(a) = f(b). Under these conditions, there exists c ∈ (a,b) with f'(c) = 0.

How is Lagrange's MVT different from Rolle's theorem?

Rolle's theorem is a special case of Lagrange's MVT when f(a) = f(b). Lagrange's MVT states that there exists c ∈ (a,b) such that f'(c) = [f(b) - f(a)]/(b - a), which is the slope of the secant line.

What is Cauchy's mean value theorem?

Cauchy's MVT generalizes Lagrange's MVT to two functions. If f and g are continuous on [a,b] and differentiable on (a,b) with g'(x) ≠ 0, then there exists c ∈ (a,b) such that [f'(c)]/[g'(c)] = [f(b) - f(a)]/[g(b) - g(a)].

What is Fermat's theorem?

Fermat's theorem states that if f has a local extremum at x₀ and f'(x₀) exists, then f'(x₀) = 0. This is the basis for finding critical points.

Can we use MVT to prove inequalities?

Yes! MVT is often used to prove inequalities. For example, to prove |sin x - sin y| ≤ |x - y|, apply MVT to f(t) = sin t and use |cos ξ| ≤ 1.