∫

∑

∂

∞

√

∏

Home/Calculus/Chapter 4/Mean Value Theorems

CALC-4.3

4-5 hours

Mean Value Theorems

Master the fundamental mean value theorems that connect function values to derivatives, enabling powerful applications in analysis and inequality proofs.

Learning Objectives

Understand and apply Fermat's theorem on local extrema

Prove and apply Rolle's theorem

Master Lagrange's mean value theorem and its applications

Apply Cauchy's mean value theorem

Understand Darboux's theorem on intermediate values of derivatives

Use mean value theorems to prove inequalities

Apply the derivative limit theorem

Solve problems involving multiple applications of mean value theorems

1. Fermat's Theorem

Definition 4.4: Local Extremum

A point $x_0$ is a local maximum of $f$ if there exists $\delta > 0$ such that $f(x) \leq f(x_0)$ for all $x \in (x_0-\delta, x_0+\delta)$ . Local minimum is defined similarly with $\geq$ .

Theorem 4.8: Fermat's Theorem

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

f'(x_0) = 0

Proof of Theorem 4.8:

Suppose $x_0$ is a local maximum. For $h > 0$ small: $\frac{f(x_0+h)-f(x_0)}{h} \leq 0$ .

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

Since $f'(x_0)$ exists, both one-sided limits equal $f'(x_0)$ , so $f'(x_0) = 0$ .

∎

Remark 4.1: Critical Points

Points where $f'(x) = 0$ are called critical points or stationary points.

Important Distinction:

Local extremum + differentiable ⟹ critical point
Critical point ⟹/ local extremum (e.g., $f(x) = x^3$ at $x = 0$ )

Example 4.8b: Fermat's Theorem Application

Problem: Find all critical points of $f(x) = x^4 - 4x^3 + 4x^2$ .

Solution:

f'(x) = 4x^3 - 12x^2 + 8x = 4x(x^2 - 3x + 2) = 4x(x-1)(x-2)

Setting $f'(x) = 0$ :

Critical points: $x = 0, 1, 2$

Analysis: $f(0) = 0$ (local min), $f(1) = 1$ (local max), $f(2) = 0$ (local min)

Example 4.8c: When Fermat's Theorem Doesn't Apply

Example: Consider $f(x) = |x|$ .

$f$ has a local minimum at $x = 0$ , but $f'(0)$ does not exist!

This shows that differentiability is essential in Fermat's theorem. Extrema can occur at points where the derivative doesn't exist.

Remark 4.1b: Finding Extrema: Complete Strategy

To find all extrema of $f$ on $[a, b]$ :

Find all critical points (where $f'(x) = 0$ )
Find all points where $f'$ doesn't exist
Evaluate $f$ at endpoints $a$ and $b$
Compare all values to find global max/min

Example 4.8d: Finding Global Extrema

Problem: Find the global max and min of $f(x) = x^3 - 3x^2 - 9x + 5$ on $[-2, 4]$ .

Solution:

Step 1: Find critical points.

f'(x) = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x-3)(x+1) = 0

Critical points: $x = -1, 3$ .

Step 2: Evaluate at critical points and endpoints.

$f(-2) = -8 - 12 + 18 + 5 = 3$
$f(-1) = -1 - 3 + 9 + 5 = 10$ (local max)
$f(3) = 27 - 27 - 27 + 5 = -22$ (local min)
$f(4) = 64 - 48 - 36 + 5 = -15$

Global max: 10 at $x = -1$ . Global min: -22 at $x = 3$ .

Remark 4.1c: Second Derivative Test Preview

At a critical point $x_0$ (where $f'(x_0) = 0$ ):

If $f''(x_0) > 0$ : $x_0$ is a local minimum
If $f''(x_0) < 0$ : $x_0$ is a local maximum
If $f''(x_0) = 0$ : test is inconclusive

2. Rolle's Theorem

Theorem 4.9: Rolle's Theorem

If $f \in C[a,b] \cap D(a,b)$ and $f(a) = f(b)$ , then there exists $\xi \in (a,b)$ such that:

f'(\xi) = 0

Proof of Theorem 4.9:

By the Extreme Value Theorem, $f$ attains max $M$ and min $m$ on $[a,b]$ .

Case 1: If $M = m$ , then $f$ is constant, so $f'(x) = 0$ for all $x \in (a,b)$ .

Case 2: If $M > m$ , since $f(a) = f(b)$ , at least one of $M, m$ is attained at some $\xi \in (a,b)$ . By Fermat's theorem, $f'(\xi) = 0$ .

∎

Remark 4.2: Geometric Interpretation of Rolle's Theorem

If a smooth curve starts and ends at the same height, there must be a point where the tangent is horizontal.

Conditions Check:

Continuity on [a,b]: No breaks in the curve
Differentiability on (a,b): No corners or cusps
f(a) = f(b): Same starting and ending height

Example 4.9b: Applying Rolle's Theorem

Problem: Show that $f(x) = x^3 - 3x + 1$ has at least one root in $(-1, 1)$ .

Solution:

Check: $f(-1) = -1 + 3 + 1 = 3$ , $f(1) = 1 - 3 + 1 = -1$ .

Since $f(-1) > 0 > f(1)$ and $f$ is continuous, by IVT, there exists $c \in (-1, 1)$ with $f(c) = 0$ .

Note: This uses the Intermediate Value Theorem, not Rolle's theorem directly!

Example 4.9c: Using Rolle's Theorem for Uniqueness

Problem: Show that $f(x) = x^3 + x - 1$ has exactly one real root.

Solution:

Existence: $f(0) = -1 < 0$ , $f(1) = 1 > 0$ . By IVT, at least one root exists.

Uniqueness: Suppose two roots $r_1 < r_2$ exist. Then $f(r_1) = f(r_2) = 0$ .

By Rolle's theorem, $\exists \xi \in (r_1, r_2)$ : $f'(\xi) = 0$ .

But $f'(x) = 3x^2 + 1 > 0$ for all $x$ . Contradiction!

Therefore, $f$ has exactly one real root.

Example 4.9d: Rolle's Theorem and Polynomial Roots

Problem: If $P(x)$ is a polynomial with $n$ distinct real roots, how many roots does $P'(x)$ have?

Solution:

Let the roots be $r_1 < r_2 < \cdots < r_n$ .

Between consecutive roots $r_i$ and $r_{i+1}$ , apply Rolle's theorem:

$P(r_i) = P(r_{i+1}) = 0$ , so $\exists \xi_i \in (r_i, r_{i+1})$ with $P'(\xi_i) = 0$ .

Conclusion: $P'(x)$ has at least $n - 1$ real roots.

Remark 4.2b: Generalization: Multiple Roots

If $f$ has $n+1$ points where $f$ takes the same value (i.e., $f(x_0) = f(x_1) = \cdots = f(x_n)$ ), then $f'$ has at least $n$ zeros, $f''$ has at least $n-1$ zeros, etc.

f^{(k)} \text{ has at least } n+1-k \text{ zeros for } k \leq n

Example 4.9e: Rolle's Theorem for Trigonometric Functions

Problem: Show that $f(x) = e^x \sin x$ has infinitely many critical points.

Solution:

$f(n\pi) = e^{n\pi} \sin(n\pi) = 0$ for all integers $n$ .

By Rolle's theorem, between consecutive zeros at $n\pi$ and $(n+1)\pi$ :

\exists \xi_n \in (n\pi, (n+1)\pi): f'(\xi_n) = 0

Since there are infinitely many such intervals, $f$ has infinitely many critical points.

Example 4.9f: Constructing Auxiliary Functions

Problem: If $f(0) = 0$ and $f(1) = 1$ , show $\exists a, b$ with $f'(a)f'(b) = 1$ .

Solution:

By MVT on $[0, 1]$ : $\exists c \in (0, 1)$ with $f'(c) = 1$ .

Consider $g(x) = f(x) - x$ .

$g(0) = 0$ and $g(1) = 0$ .

By Rolle's theorem: $\exists d \in (0, 1)$ with $g'(d) = 0$ , i.e., $f'(d) = 1$ .

Take $a = b = c$ or $a = b = d$ : $f'(a)f'(b) = 1 \cdot 1 = 1$ .

Remark 4.2c: Why Hypothesis Checking Matters

Counterexample when hypotheses fail:

Consider $f(x) = |x|$ on $[-1, 1]$ .

$f(-1) = f(1) = 1$ ✓
$f$ is continuous on $[-1, 1]$ ✓
$f$ is NOT differentiable at $x = 0$ ✗

There is no $\xi$ with $f'(\xi) = 0$ ! Rolle's theorem doesn't apply.

3. Lagrange's Mean Value Theorem

Theorem 4.10: Lagrange's Mean Value Theorem

If $f \in C[a,b] \cap D(a,b)$ , then there exists $\xi \in (a,b)$ such that:

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

Proof of Theorem 4.10:

Define $g(x) = f(x) - \frac{f(b)-f(a)}{b-a}(x-a)$ . Then $g \in C[a,b] \cap D(a,b)$ and $g(a) = f(a) = g(b)$ .

By Rolle's theorem, $\exists \xi \in (a,b)$ : $g'(\xi) = f'(\xi) - \frac{f(b)-f(a)}{b-a} = 0$ .

∎

Corollary 4.1: Constant Function Criterion

If $f \in C[a,b] \cap D(a,b)$ and $f'(x) = 0$ for all $x \in (a,b)$ , then $f$ is constant on $[a,b]$ .

Example 4.12: Proving Inequalities

Prove: For $x > -1, x \neq 0$ : $\frac{x}{1+x} < \ln(1+x) < x$

Apply MVT to $f(t) = \ln(1+t)$ on $[0, x]$ (assume $x > 0$ ):

\ln(1+x) - \ln 1 = \frac{1}{1+\xi} \cdot x \text{ for some } \xi \in (0, x)

Since $0 < \xi < x$ : $\frac{1}{1+x} < \frac{1}{1+\xi} < 1$ , giving $\frac{x}{1+x} < \ln(1+x) < x$ .

Remark 4.3: Geometric Interpretation of Lagrange's MVT

There exists a point where the tangent line is parallel to the secant line connecting $(a, f(a))$ and $(b, f(b))$ .

Secant line slope: $\frac{f(b) - f(a)}{b - a}$

Tangent line slope at ξ: $f'(\xi)$

MVT says these are equal for some $\xi \in (a, b)$ !

Corollary 4.2: Monotonicity Criterion

Let $f \in C[a,b] \cap D(a,b)$ . Then:

$f'(x) > 0$ for all $x \in (a,b)$ ⟹ $f$ is strictly increasing on $[a,b]$

$f'(x) < 0$ for all $x \in (a,b)$ ⟹ $f$ is strictly decreasing on $[a,b]$

Proof of Corollary 4.2 (Monotonicity):

For $x_1 < x_2$ in $[a,b]$ , by MVT:

f(x_2) - f(x_1) = f'(\xi)(x_2 - x_1)

Since $x_2 - x_1 > 0$ and $f'(\xi) > 0$ , we get $f(x_2) - f(x_1) > 0$ .

∎

Example 4.12b: Proving sin x < x for x > 0

Problem: Prove that $\sin x < x$ for all $x > 0$ .

Solution:

Apply MVT to $f(t) = \sin t$ on $[0, x]$ :

\sin x - \sin 0 = \cos \xi \cdot (x - 0) \text{ for some } \xi \in (0, x)

So $\sin x = x \cos \xi$ .

Since $0 < \xi < x$ and $\cos \xi \leq 1$ (with equality only at $\xi = 0$ ):

\sin x = x \cos \xi < x \cdot 1 = x

Example 4.12c: Lipschitz Condition

Problem: If $|f'(x)| \leq M$ for all $x$ , prove $|f(x) - f(y)| \leq M|x - y|$ .

Solution:

By MVT, for any $x < y$ :

f(y) - f(x) = f'(\xi)(y - x) \text{ for some } \xi \in (x, y)

Taking absolute values:

|f(y) - f(x)| = |f'(\xi)| \cdot |y - x| \leq M|y - x|

This is called the Lipschitz condition with Lipschitz constant $M$ .

Example 4.12d: e^x Inequality

Problem: Prove that $e^x \geq 1 + x$ for all $x \in \mathbb{R}$ .

Solution:

Case 1: $x \geq 0$ . Apply MVT to $f(t) = e^t$ on $[0, x]$ :

e^x - e^0 = e^\xi \cdot x \text{ for some } \xi \in (0, x)

Since $e^\xi > 1$ for $\xi > 0$ : $e^x - 1 > x$ , so $e^x > 1 + x$ .

Case 2: $x < 0$ . Apply MVT on $[x, 0]$ :

1 - e^x = e^\xi \cdot (-x) \text{ for some } \xi \in (x, 0)

Since $e^\xi < 1$ for $\xi < 0$ : $1 - e^x < -x$ , so $e^x > 1 + x$ .

Equality holds only when $x = 0$ .

Remark 4.3b: Applications of Lagrange's MVT

Lagrange's MVT is used to:

Prove inequalities involving functions
Establish monotonicity of functions
Prove functions are constant (if $f' = 0$ )
Derive Lipschitz bounds
Foundation for L'Hôpital's rule

Example 4.12e: Bounding sqrt(1+x)

Problem: Prove that for $x > 0$ : $\sqrt{1+x} < 1 + \frac{x}{2}$ .

Solution:

Apply MVT to $f(t) = \sqrt{1+t}$ on $[0, x]$ :

\sqrt{1+x} - 1 = \frac{1}{2\sqrt{1+\xi}} \cdot x \text{ for some } \xi \in (0, x)

Since $\xi > 0$ : $\sqrt{1+\xi} > 1$ , so $\frac{1}{2\sqrt{1+\xi}} < \frac{1}{2}$ .

\sqrt{1+x} - 1 < \frac{x}{2} \implies \sqrt{1+x} < 1 + \frac{x}{2}

Example 4.12f: Finite Difference Approximation

Connection to Numerical Methods:

MVT says that for some $\xi \in (a, b)$ :

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

This motivates the finite difference approximation:

f'(x) \approx \frac{f(x + h) - f(x)}{h}

The error in this approximation is related to $f''$ and $h$ by Taylor's theorem.

Corollary 4.3: Bounded Derivative Implies Continuous

If $f$ is differentiable on $(a, b)$ and $|f'(x)| \leq M$ for all $x$ , then $f$ is uniformly continuous on $(a, b)$ .

Example 4.12g: Proving Uniqueness of Fixed Points

Problem: If $|f'(x)| < 1$ for all $x$ , show $f$ has at most one fixed point.

Solution:

Suppose $f(p) = p$ and $f(q) = q$ with $p \neq q$ .

By MVT:

\frac{f(p) - f(q)}{p - q} = f'(\xi) \text{ for some } \xi

This gives $\frac{p - q}{p - q} = 1 = f'(\xi)$ .

But $|f'(\xi)| < 1$ . Contradiction!

4. Cauchy's Mean Value Theorem

Theorem 4.11: Cauchy's Mean Value Theorem

If $f, g \in C[a,b] \cap D(a,b)$ and $g'(x) \neq 0$ , then $\exists \xi \in (a,b)$ :

\frac{f'(\xi)}{g'(\xi)} = \frac{f(b) - f(a)}{g(b) - g(a)}

Proof of Theorem 4.11:

Note $g(b) \neq g(a)$ (by Rolle's theorem applied to $g$ ). Define:

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

Then $h(a) = f(a) = h(b)$ . By Rolle's theorem, $\exists \xi$ : $h'(\xi) = f'(\xi) - \frac{f(b)-f(a)}{g(b)-g(a)}g'(\xi) = 0$ .

∎

Remark 4.4: Why Cauchy's MVT Matters

Cauchy's MVT is crucial for proving L'Hôpital's Rule. It generalizes Lagrange's MVT to ratios of functions.

Relationship to Lagrange's MVT:

If we set $g(x) = x$ , then $g'(x) = 1$ and:

\frac{f'(\xi)}{g'(\xi)} = f'(\xi) = \frac{f(b) - f(a)}{b - a}

This is exactly Lagrange's MVT!

Example 4.11b: Cauchy MVT Application

Problem: Using Cauchy's MVT, show that for $0 < a < b$ :

\frac{b - a}{b} < \ln\frac{b}{a} < \frac{b - a}{a}

Solution:

Apply Cauchy's MVT to $f(x) = \ln x$ and $g(x) = x$ on $[a, b]$ :

\frac{1/\xi}{1} = \frac{\ln b - \ln a}{b - a} \text{ for some } \xi \in (a, b)

Since $a < \xi < b$ :

\frac{1}{b} < \frac{1}{\xi} < \frac{1}{a}

Therefore:

\frac{b-a}{b} < \ln b - \ln a < \frac{b-a}{a}

Example 4.11c: Connection to L'Hôpital

Preview: Cauchy's MVT is the foundation of L'Hôpital's Rule.

For the limit $\lim_{x \to a} \frac{f(x)}{g(x)}$ with $f(a) = g(a) = 0$ :

\frac{f(x)}{g(x)} = \frac{f(x) - f(a)}{g(x) - g(a)} = \frac{f'(\xi)}{g'(\xi)}

for some $\xi$ between $a$ and $x$ .

As $x \to a$ , $\xi \to a$ , suggesting:

\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{\xi \to a} \frac{f'(\xi)}{g'(\xi)}

Remark 4.4b: Parametric Interpretation

If we think of $(g(t), f(t))$ as a parametric curve, Cauchy's MVT says:

There's a point $t = \xi$ where the tangent to the curve is parallel to the chord from $t = a$ to $t = b$ .

\text{Slope of tangent} = \frac{f'(\xi)}{g'(\xi)} = \frac{f(b) - f(a)}{g(b) - g(a)} = \text{Slope of chord}

Example 4.11d: Comparing Rates of Change

Problem: Show that for $0 < a < b$ : $\frac{e^b - e^a}{b - a} > e^{(a+b)/2}$ .

Solution:

By MVT applied to $f(x) = e^x$ on $[a, b]$ :

\frac{e^b - e^a}{b - a} = e^\xi \text{ for some } \xi \in (a, b)

We need to show $\xi > \frac{a+b}{2}$ (i.e., $\xi$ is in the right half of $(a,b)$ ).

Since $e^x$ is convex ( $e'' > 0$ ), the average value of $e'$ is achieved to the right of the midpoint.

Example 4.11e: Ratio of Sine to Argument

Problem: Using Cauchy's MVT, show for $0 < x < \pi/2$ :

\frac{2}{\pi} < \frac{\sin x}{x} < 1

Solution:

Apply Cauchy's MVT to $f(t) = \sin t$ , $g(t) = t$ on $[0, x]$ :

\frac{\sin x - 0}{x - 0} = \frac{\cos \xi}{1} = \cos \xi \text{ for some } \xi \in (0, x)

Since $0 < \xi < x < \pi/2$ :

$\cos x < \cos \xi < \cos 0 = 1$
Taking $x \to \pi/2$ : $\cos(\pi/2) = 0$ , giving the bound $\frac{2}{\pi}$

Corollary 4.4: Ratio of Derivatives

If $f, g$ are differentiable with $g' \neq 0$ and $\frac{f'}{g'}$ is monotonic, then:

\min \frac{f'}{g'} \leq \frac{f(b) - f(a)}{g(b) - g(a)} \leq \max \frac{f'}{g'}

5. Darboux's Theorem

Theorem 4.12: Darboux's Theorem (Intermediate Value Property for Derivatives)

If $f$ is differentiable on $[a,b]$ , then $f'$ takes every value between $f'_+(a)$ and $f'_-(b)$ .

Significance: Even though $f'$ may not be continuous, it still has the intermediate value property. This means derivatives cannot have "jump discontinuities"—any discontinuity must be of the oscillating type.

Proof of Theorem 4.12 (Darboux's Theorem):

Let $f$ be differentiable on $[a, b]$ with $f'(a) < \gamma < f'(b)$ .

Define $g(x) = f(x) - \gamma x$ . Then:

$g'(a) = f'(a) - \gamma < 0$
$g'(b) = f'(b) - \gamma > 0$

Since $g'(a) < 0$ , $g$ is decreasing near $a$ , so $g$ attains its minimum at some $\xi \in (a, b)$ .

By Fermat's theorem: $g'(\xi) = 0$ , so $f'(\xi) = \gamma$ .

∎

Remark 4.5: Implications of Darboux's Theorem

Darboux's theorem places strong restrictions on what functions can be derivatives:

Cannot be a derivative:

Step functions, signum function, any function with jump discontinuities

Can be a derivative:

Continuous functions, functions with only removable or oscillating discontinuities

Example 4.12b: Function That Cannot Be a Derivative

Problem: Show that $g(x) = \text{sgn}(x)$ is not the derivative of any function.

Solution:

Suppose $g = f'$ for some $f$ .

$g(-1) = -1$ and $g(1) = 1$ .

By Darboux's theorem, $f'$ must take all values between $-1$ and $1$ .

In particular, $f'(\xi) = 0$ for some $\xi \in (-1, 1)$ .

But $g(x) = \pm 1$ for all $x \neq 0$ , never $0$ . Contradiction!

Example 4.12c: Derivative with Discontinuity

Example: The function $f(x) = x^2 \sin(1/x)$ (with $f(0) = 0$ ) has a discontinuous derivative:

For $x \neq 0$ :

f'(x) = 2x\sin\frac{1}{x} - \cos\frac{1}{x}

At $x = 0$ : $f'(0) = 0$ (by limit definition).

$f'$ is discontinuous at 0 (oscillates), but by Darboux's theorem, it still has the IVP!

Remark 4.5b: Baire Category and Derivatives

Advanced fact: If $f$ is differentiable, then $f'$ is a Baire class 1 function.

This means $f'$ is the pointwise limit of continuous functions. The set of discontinuities of $f'$ must be "small" in a precise topological sense.

Example 4.12d: Testing if a Function Can Be a Derivative

Problem: Can $g(x) = \begin{cases} 1 & x \geq 0 \\ -1 & x < 0 \end{cases}$ be expressed as $f'$ for some $f$ ?

Solution:

No! By Darboux's theorem, if $g = f'$ , then $g$ must have the IVP.

$g(-1) = -1$ and $g(1) = 1$ , but $g$ never takes the value $0$ .

This violates the IVP, so $g$ cannot be a derivative.

Remark 4.5c: Types of Discontinuities Allowed

Allowed:

• Removable discontinuities
• Oscillating discontinuities
• Essential discontinuities (some types)

Not allowed:

• Jump discontinuities
• Step-like behavior
• Monotonic jumps

6. Additional Applications

Theorem 4.13: Derivative Limit Theorem

If $f$ is continuous on $[a, b]$ , differentiable on $(a, b)$ , and $\lim_{x \to a^+} f'(x) = L$ , then $f'_+(a) = L$ .

Example 4.13: Using the Derivative Limit Theorem

Problem: Find $f'(0)$ for $f(x) = x^2 \sin(1/x)$ ( $f(0) = 0$ ).

Solution:

By definition:

f'(0) = \lim_{h \to 0} \frac{h^2\sin(1/h) - 0}{h} = \lim_{h \to 0} h\sin\frac{1}{h} = 0

by the squeeze theorem since $|h\sin(1/h)| \leq |h|$ .

Remark 4.6: Summary: Hierarchy of Mean Value Theorems

The mean value theorems form a hierarchy, each generalizing the previous:

Fermat's Theorem

Local extremum ⟹ derivative = 0

↓

Rolle's Theorem

f(a) = f(b) ⟹ f'(ξ) = 0

↓

Lagrange's MVT

f'(ξ) = [f(b) - f(a)]/(b - a)

↓

Cauchy's MVT

f'(ξ)/g'(ξ) = [f(b) - f(a)]/[g(b) - g(a)]

Example 4.13b: Combining Multiple MVT Applications

Problem: If $f \in C^2[a, b]$ and $f(a) = f(b) = 0$ , prove $\exists \xi$ : $|f(\xi)| \leq \frac{(b-a)^2}{8}\max|f''|$ .

Sketch of Solution:

By Rolle's theorem, $\exists c \in (a,b)$ : $f'(c) = 0$ .
Apply MVT to $f'$ on $[a, c]$ and $[c, b]$ .
Use bounds on $f''$ to bound $f'$ .
Apply MVT again to bound $f$ .

Example 4.13c: Proving arctan Inequality

Problem: Prove that for $x > 0$ : $\frac{x}{1+x^2} < \arctan x < x$ .

Solution:

Apply MVT to $f(t) = \arctan t$ on $[0, x]$ :

\arctan x - \arctan 0 = \frac{1}{1+\xi^2} \cdot x \text{ for some } \xi \in (0, x)

Since $0 < \xi < x$ :

\frac{1}{1+x^2} < \frac{1}{1+\xi^2} < 1

Multiplying by $x > 0$ :

\frac{x}{1+x^2} < \arctan x < x

Example 4.13d: Two Functions with Same Derivative

Problem: If $f'(x) = g'(x)$ for all $x \in (a, b)$ , what can we conclude?

Solution:

Let $h(x) = f(x) - g(x)$ . Then $h'(x) = 0$ for all $x \in (a, b)$ .

By the constant function criterion (Corollary of MVT):

h(x) = C \text{ (constant)}

Conclusion: $f(x) = g(x) + C$ for some constant $C$ .

Remark 4.7: Physical Interpretation of MVT

If a car travels from A to B, its instantaneous speed must equal its average speed at some moment.

Example: Drive 100 km in 2 hours (average 50 km/h).

By MVT, at some instant, the speedometer showed exactly 50 km/h!

Example 4.13e: Proving cos Inequality

Problem: Prove that $1 - \frac{x^2}{2} < \cos x < 1$ for $x \neq 0$ .

Solution:

The right inequality $\cos x < 1$ is immediate for $x \neq 0$ .

For the left: let $g(x) = \cos x - 1 + \frac{x^2}{2}$ .

$g(0) = 0$
$g'(x) = -\sin x + x$
We proved $\sin x < x$ for $x > 0$ , so $g'(x) > 0$ for $x > 0$

Therefore $g(x) > g(0) = 0$ for $x > 0$ , giving $\cos x > 1 - \frac{x^2}{2}$ .

Theorem 4.14: Extended Mean Value Theorem

If $f \in C^n[a,b]$ and $f^{(n+1)}$ exists on $(a,b)$ , with $f(x_0) = f(x_1) = \cdots = f(x_n) = 0$ for distinct points $x_i \in [a,b]$ , then:

\exists \xi \in (a,b): f^{(n)}(\xi) = 0

Example 4.13f: Mean Value Inequality

Problem: If $f''(x) \geq 0$ (f is convex), prove:

f\left(\frac{a+b}{2}\right) \leq \frac{f(a) + f(b)}{2}

Solution Sketch:

Let $c = \frac{a+b}{2}$ . Apply MVT on $[a, c]$ and $[c, b]$ :

f(c) - f(a) = f'(\xi_1)(c - a), \quad f(b) - f(c) = f'(\xi_2)(b - c)

with $\xi_1 < c < \xi_2$ .

Since $f'' \geq 0$ , $f'$ is increasing, so $f'(\xi_1) \leq f'(\xi_2)$ .

Adding and using $c - a = b - c$ gives the result.

Example 4.13g: Speed Trap Problem

Real-World Application:

A car enters a highway at 2:00 PM at marker A and exits at marker B (10 km away) at 2:06 PM. Speed limit is 100 km/h. Can police prove speeding?

Solution:

Average speed = $\frac{10\text{ km}}{6\text{ min}} = \frac{10}{0.1} = 100$ km/h.

By MVT, at some moment the instantaneous speed equaled 100 km/h exactly.

If the car was stationary at any point, it must have exceeded 100 km/h at some other moment!

Remark 4.8: Preview: Connection to Integration

The Mean Value Theorem for Integrals is a related result:

If $f$ is continuous on $[a, b]$ , then $\exists c \in (a, b)$ :

\int_a^b f(x)\,dx = f(c)(b - a)

This is the integral analog of Lagrange's MVT, connecting average values to function values.

Example 4.13h: Proving AM-GM Inequality

Problem: Prove that for $x > 0$ : $e^x \geq ex$ (with equality at $x = 1$ ).

Solution:

Let $g(x) = e^x - ex$ . Then $g'(x) = e^x - e$ .

$g'(x) = 0$ at $x = 1$ ; $g'(x) < 0$ for $x < 1$ ; $g'(x) > 0$ for $x > 1$ .

So $g$ has a global minimum at $x = 1$ : $g(1) = e - e = 0$ .

Therefore $g(x) \geq 0$ for all $x > 0$ , i.e., $e^x \geq ex$ .

Chapter Summary

This module covered the fundamental mean value theorems of differential calculus:

1.Fermat's Theorem identifies critical points as candidates for extrema.
2.Rolle's Theorem guarantees a horizontal tangent between equal function values.
3.Lagrange's MVT connects average and instantaneous rates of change.
4.Cauchy's MVT generalizes to ratios, enabling L'Hôpital's rule.
5.Darboux's Theorem shows derivatives have the intermediate value property.

Mean Value Theorems Practice

Questions

Correct

Accuracy

What are the conditions for Rolle's theorem?

Easy

Not attempted

f(x) = x^3 - 3x

[-2, 2]

, does Rolle's theorem apply?

Medium

Not attempted

Lagrange's MVT states that

\exists \xi \in (a,b)

such that:

Easy

Not attempted

f'(x) = 0

for all

x \in (a,b)

, what can we conclude?

Easy

Not attempted

Cauchy's MVT relates:

Medium

Not attempted

f

is differentiable on

[a,b]

, Darboux's theorem says:

Hard

Not attempted

Use Lagrange's MVT to prove that for

x > 0

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

Hard

Not attempted

f(x_0)

is a local maximum and

f'(x_0)

exists, then:

Easy

Not attempted

Use MVT: If

f'(x) > 0

for all

x

, then

f

is:

Easy

Not attempted

f(x) = x^5 - 5x

, how many real roots does

f'(x) = 0

have?

Medium

Not attempted

Prove

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

using:

Medium

Not attempted

For the circle

x^2 + y^2 = r^2

, MVT applied to

f(x) = \sqrt{r^2 - x^2}

[0, r]

gives:

Medium

Not attempted

f''(x) > 0

(a,b)

and

f(a) = f(b) = 0

, then for

x \in (a,b)

Hard

Not attempted

By Darboux's theorem, which function CANNOT be a derivative?

Hard

Not attempted

Frequently Asked Questions

What's the geometric meaning of Lagrange's MVT?

There exists a point where the tangent line is parallel to the secant line connecting (a, f(a)) and (b, f(b)).

How is Rolle's theorem related to Lagrange's MVT?

Rolle's theorem is a special case of Lagrange's MVT when f(a) = f(b), making the secant line horizontal.

Why is Darboux's theorem surprising?

It shows that derivatives have the intermediate value property even without being continuous. This is a strong constraint on what functions can be derivatives.

How do I apply MVT in proofs?

Identify the function and interval, verify hypotheses (continuity, differentiability), apply the theorem to get a point ξ with the desired property.

What's the difference between local and global extrema?

Local extrema are max/min in a neighborhood; global extrema are the largest/smallest values on the entire domain. Fermat's theorem applies to local extrema.

Can Rolle's theorem be applied if the endpoint condition fails?

No. If f(a) ≠ f(b), use Lagrange's MVT instead. Rolle's theorem specifically requires equal endpoint values.

Why can't step functions be derivatives?

By Darboux's theorem, derivatives must have the IVP. Step functions have jump discontinuities, violating this property.

How do I use MVT to prove inequalities?

Apply MVT to the relevant function on an appropriate interval. The existence of ξ with f'(ξ) = [f(b)-f(a)]/(b-a) often leads to bounds.

What's the importance of checking hypotheses?

MVT fails if hypotheses aren't met. Always verify: (1) continuity on [a,b], (2) differentiability on (a,b), (3) any additional conditions like f(a)=f(b) for Rolle's.

How does Cauchy's MVT generalize Lagrange's?

When g(x) = x, Cauchy's MVT reduces to Lagrange's. Cauchy's handles ratios of function changes, essential for L'Hôpital's rule.

Key Takeaways

Fermat's Theorem

Local extremum + differentiable ⟹ f'(x₀) = 0

Rolle's Theorem

f(a) = f(b) ⟹ ∃ξ: f'(ξ) = 0

Lagrange's MVT

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

Cauchy's MVT

\frac{f'(\xi)}{g'(\xi)} = \frac{f(b) - f(a)}{g(b) - g(a)}

Darboux's Theorem

Derivatives have the Intermediate Value Property

Monotonicity

f' > 0 ⟹ f increasing; f' < 0 ⟹ f decreasing

Study Tips

Proof Strategy

For inequality proofs: identify the function, choose the right interval, apply MVT, then use bounds on f' to conclude.

Hypothesis Check

Always verify: (1) continuity on closed interval, (2) differentiability on open interval. Theorems fail without these!

Uniqueness Arguments

To prove a function has exactly one root: show existence (IVT), then show uniqueness (Rolle's theorem by contradiction).

Auxiliary Functions

Create helper functions like g(x) = f(x) - cx or h(x) = f(x) - kx² to transform problems into forms where Rolle's theorem applies.

What's Next?

In the next module, you will learn about L'Hôpital's Rule for evaluating limits and Taylor's Theorem for polynomial approximations.

L'Hôpital's rule for 0/0 and ∞/∞ forms (uses Cauchy's MVT)
Taylor polynomials and remainder estimation
Maclaurin series for common functions
Applications to limit evaluation and approximation

Common Mistakes to Avoid

✗
Forgetting to check hypotheses
MVT fails if f is not continuous on [a,b] or not differentiable on (a,b).
✗
Confusing local and global extrema
Fermat's theorem only applies to local extrema in the interior.
✗
Thinking f'(x₀) = 0 implies extremum
The converse of Fermat's theorem is false! Consider f(x) = x³ at x = 0.
✗
Misapplying Rolle vs Lagrange
Use Rolle when f(a) = f(b); use Lagrange in general cases.