Mean Value Theroems

In this chapter, we will study fundamental theorems that describe the behavior of differentiable functions over an interval. Starting with Rolle’s Theorem, which guarantees a stationary point when a function has equal values at two endpoints, we then generalize it to the Mean Value Theorem (MVT), which states that at some point in an interval, the instantaneous rate of change of a function matches its average rate of change. Further extending this idea, we arrive at Cauchy’s Mean Value Theorem (CMVT), which applies to two functions simultaneously and leads to deeper results.

These theorems are essential in proving many important results in calculus and analysis. The Mean Value Theorem, in particular, plays a crucial role in understanding the behavior of functions, leading to results such as Taylor’s Theorem and L’Hôpital’s Rule. By studying these theorems, we will develop a clearer and more rigorous way of reasoning about differentiable functions.

Rolle’s Theorem

Let \( f \) be a function defined on a closed interval \([a, b]\) that satisfies the following conditions:

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

Then, there exists some \( c \in (a, b) \) such that

\[ f'(c) = 0. \]

This theorem guarantees the existence of at least one point in the open interval where the derivative of \( f \) vanishes, meaning the function has a horizontal tangent at some point.

Rolle’s Theorem states that if a function satisfies the given three conditions—continuity on \([a, b]\), differentiability on \((a, b)\), and equal function values at the endpoints—then we can conclude that there exists at least one point \( c \in (a, b) \) where \( f'(c) = 0 \). That is, the function must have a horizontal tangent at some point in the interval.

Consider the function

\[ f(x) = x^3 - x + 2. \]

We will check whether Rolle’s Theorem applies on the interval \([-1,1]\).

• Since \( f(x) \) is a polynomial, it is continuous on \([-1,1]\) and differentiable on \((-1,1)\).

• Evaluating at the endpoints,

  \[ f(-1) = (-1)^3 - (-1) + 2 = -1 + 1 + 2 = 2, \]
  \[ f(1) = (1)^3 - (1) + 2 = 1 - 1 + 2 = 2. \]

  Since \( f(-1) = f(1) \), the third hypothesis of Rolle’s Theorem is also satisfied.

Since all conditions of the theorem hold, we can conclude that there exists some \( c \in (-1,1) \) such that

\[ f'(c) = 0. \]

Computing the derivative,

\[ f'(x) = 3x^2 - 1. \]

Setting \( f'(c) = 0 \),

\[ 3c^2 - 1 = 0 \implies 3c^2 = 1 \implies c^2 = \frac{1}{3}. \]

Thus,

\[ c = \pm \frac{1}{\sqrt{3}}. \]

Since \( c \in (-1,1) \), we conclude that Rolle’s Theorem guarantees at least one such point in the interval where the derivative is zero. \(\blacksquare\)

However, the theorem does not tell us what happens if any of the hypotheses are not satisfied. If a function is not continuous on \([a, b]\), not differentiable on \((a, b)\), or does not satisfy \( f(a) = f(b) \), we cannot conclude anything. It may still happen that there exists some \( c \) such that \( f'(c) = 0 \), but this is not guaranteed.

For example, consider the function

\[ f(x) = |x| \]

on \([-1,1]\). It is continuous everywhere and satisfies \( f(-1) = f(1) \), but it is not differentiable at \( x = 0 \). In this case, we see that there is no \( c \) in \( (-1,1) \) where \( f'(c) = 0 \), because the derivative does not even exist at \( x = 0 \).

Proof of Rolle’s Theorem:

Since \( f \) is continuous on the closed interval \([a,b]\), by the Extreme Value Theorem, \( f \) must attain both a maximum and a minimum value at some points within the interval \([a,b]\).

Consider two possible cases for the location of the maximum and minimum:

Case 1: Suppose \( f \) is constant on \([a,b] \).
Then clearly \( f'(x) = 0 \) for every \( x \in (a,b) \). In this trivial case, the theorem is obviously true.

Case 2: Suppose \( f \) is not constant on \([a,b] \).
Then \( f \) attains either a maximum or a minimum (or both) at some point \( c \in (a,b) \). Without loss of generality, assume that the maximum is attained at the interior point \( c \). Since \( c \) is an interior maximum, it follows from basic properties of differentiable functions that: [ f'(c) = 0. ]

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

This completes the proof.

Example

Consider the function defined by

\[ f(x) = \log(\sin x), \quad x \in \left[\frac{\pi}{4}, \frac{3\pi}{4}\right]. \]

We examine whether Rolle’s theorem is applicable to this function:

• The function \( f(x) = \log(\sin x) \) is defined and continuous on the interval \(\left[\frac{\pi}{4}, \frac{3\pi}{4}\right]\), because \(\sin x > 0\) for all \( x \in \left[\frac{\pi}{4}, \frac{3\pi}{4}\right] \).

• Clearly, it is differentiable on the open interval \(\left(\frac{\pi}{4}, \frac{3\pi}{4}\right)\), as both \(\log x\) and \(\sin x\) are differentiable functions on their respective domains.

• Evaluating at endpoints:

  \[ f\left(\frac{\pi}{4}\right) = \log\left(\sin\frac{\pi}{4}\right) = \log\left(\frac{1}{\sqrt{2}}\right), \quad f\left(\frac{3\pi}{4}\right) = \log\left(\sin\frac{3\pi}{4}\right) = \log\left(\frac{1}{\sqrt{2}}\right). \]

Thus, we observe that \( f\left(\frac{\pi}{4}\right) = f\left(\frac{3\pi}{4}\right) \).

Since all three hypotheses of Rolle’s theorem are satisfied, we conclude that there exists some \( c \in \left(\frac{\pi}{4}, \frac{3\pi}{4}\right) \) such that

\[ f'(c) = 0. \]

To explicitly verify, we compute the derivative:

\[ f'(x) = \frac{\cos x}{\sin x} = \cot x. \]

Setting \( f'(c) = 0 \), we get

\[ \cot c = 0 \implies c = \frac{\pi}{2}, \]

which indeed lies in the interval \(\left(\frac{\pi}{4}, \frac{3\pi}{4}\right)\).

Hence, as predicted by Rolle’s theorem, such a point \( c = \frac{\pi}{2} \) exists within the interval.

Example

Consider the function

\[ f(x) = e^x\sin x \]

defined on the interval \([0,\pi]\). Let us verify that Rolle's theorem is applicable:

First, note that \(f\) is a combination of exponential and trigonometric functions, both of which are infinitely differentiable. Hence, \(f\) is clearly continuous on \([0,\pi]\) and differentiable on \((0,\pi)\).

Evaluating at endpoints:

\[ f(0) = e^0\sin 0 = 0,\quad f(\pi) = e^{\pi}\sin\pi = 0. \]

Thus, \(f(0)=f(\pi)=0\). All the conditions of Rolle's theorem are satisfied.

Hence, we conclude there exists at least one \(c\in(0,\pi)\) such that:

\[ f'(c) = 0. \]

For completeness, let us find this \(c\) explicitly. Differentiating, we have:

\[ f'(x) = e^x(\sin x + \cos x). \]

Setting this equal to zero:

\[ e^c(\sin c + \cos c)=0 \implies \sin c + \cos c = 0 \implies \tan c = -1. \]

Solving this gives:

\[ c = \frac{3\pi}{4} \quad(\text{since } c\in(0,\pi)). \]

Thus, Rolle’s theorem guarantees—and we have explicitly found—the existence of the point:

\[ c=\frac{3\pi}{4}, \]

where the derivative vanishes.

Roots of an equation using Rolle's Theorem

In certain problems, Rolle's theorem can be used cleverly to establish that an equation has at least one root within a given interval. The central idea involves recognizing that if we have an equation of the form \(f(x)=0\), it may be advantageous to identify a function whose derivative is \(f(x)\). If such a function can be constructed, then Rolle's theorem can be applied to the integral (or antiderivative) to establish the existence of the desired root.

Consider the following example:

Problem:

Show that the equation

\[ e^x(\sin x + \cos x) = 0 \]

has at least one root in the interval \(\left[0,\pi\right]\).

Solution (Using Rolle's theorem):

We are asked to prove that the function

\[ f(x) = e^x(\sin x + \cos x) \]

has at least one root in the given interval. To approach this via Rolle’s theorem, consider defining a suitable auxiliary function, obtained by integration of the given expression:

Define a new function \(F(x)\) as follows:

\[ F(x) = \int e^x(\sin x + \cos x)\,dx. \]

Upon integrating, we have

\[ F(x) = e^x\sin x + C. \]

Choose the simplest possible antiderivative (taking \(C=0\)), thus defining explicitly:

\[ F(x)=e^x\sin x. \]

Clearly, \(F(x)\) is continuous on \(\left[0,\pi\right]\) and differentiable on \((0,\pi)\), being the product of exponential and trigonometric functions.

Now, observe that:

\[ F(0) = e^0\sin 0 = 0,\quad F(\pi)=e^{\pi}\sin\pi=0. \]

Thus, the hypotheses of Rolle’s theorem are satisfied by \(F(x)\). Hence, Rolle’s theorem guarantees the existence of some number \(c\in(0,\pi)\) such that:

\[ F'(c)=0. \]

But notice carefully:

\[ F'(x)=e^x(\sin x+\cos x). \]

Thus, there must exist at least one point \(c\in(0,\pi)\) for which

\[ F'(c)=e^c(\sin c+\cos c)=0. \]

Thus, Rolle's theorem has elegantly provided the required conclusion.

Constructing an Auxiliary Function form a given function for Rolle’s Theorem

A powerful approach when using Rolle’s theorem involves constructing an auxiliary function based on a given function \( f \). Given a function \( f \) which is continuous on the closed interval \([a,b]\) and differentiable on the open interval \((a,b)\), we can strategically define an auxiliary function \( g(x) \) that meets all the hypotheses of Rolle’s theorem. Importantly, the auxiliary function \( g(x) \) is not unique; multiple choices may be possible, provided all required conditions remain satisfied and what we want to achive. We can derive some powerful results from this process.

To illustrate this method concretely, suppose \( f \) is continuous on the interval \([a,b]\) and differentiable on the interval \((a,b)\). We introduce the following auxiliary function:

\[ g(x) = (x - a)(x - b)f(x). \]

This function \( g(x) \) explicitly satisfies all the hypotheses of Rolle’s theorem. Specifically, observe that it is continuous on the interval \([a,b]\), as it is the product of continuous functions. It is also differentiable on the interval \((a,b)\). Moreover, at the endpoints, we clearly have:

\[ g(a) = (a - a)(a - b)f(a) = 0, \quad\text{and}\quad g(b) = (b - a)(b - b)f(b) = 0, \]

ensuring the crucial condition \( g(a) = g(b) \).

Thus, by Rolle’s theorem, there must exist at least one point \( c \in (a,b) \) at which the derivative vanishes:

\[ g'(c) = 0. \]

To extract meaningful insight from this result, we now explicitly compute the derivative \( g'(x) \):

\[\begin{align} g'(x) &= f'(x)(x-a)(x-b) + f(x)(x-b) + f(x)(x-a) \\ &= (x-a)(x-b) f'(x) + f(x)(x-a) + f(x)(x-b). \end{align}\]

Therefore, the condition \( g'(x)=0 \) leads directly to:

\[ (x-a)(x-b) f'(x) + f(x)(x-a) + f(x)(x-b) = 0. \]

Provided \( x \neq a,b \), it is convenient to divide both sides by \((x-a)(x-b)\) to reveal a more elegant form:

\[ f'(x) + f(x) \left( \frac{1}{x-a} + \frac{1}{x-b} \right) = 0. \]

After rearranging, we obtain the insightful identity:

\[ f'(x) = -f(x) \left( \frac{1}{x-a} + \frac{1}{x-b} \right). \]

Thus, Rolle’s theorem rigorously guarantees the existence of at least one point \( c \in (a,b) \) for which the following important identity holds:

\[ f'(c) = - f(c) \left( \frac{1}{c-a} + \frac{1}{c-b} \right). \]

This demonstrates explicitly how Rolle’s theorem, combined with a carefully chosen auxiliary function, provides a powerful tool to derive nontrivial and useful relationships involving the original function \( f(x) \) and its derivative \( f'(x) \).

Applying the same reasoning, we now consider the auxiliary function

\[ g(x) = (f(x) - f(a))(f(x) - f(b)). \]

Since \( g(a) = g(b) = 0 \), Rolle’s theorem ensures the existence of some \( c \in (a,b) \) such that \( g'(c) = 0 \).

Expanding \( g(x) \):

\[ g(x) = f^2(x) - (f(a) + f(b)) f(x) + f(a) f(b). \]

Differentiating,

\[ g'(x) = 2f(x) f'(x) - (f(a) + f(b)). \]

Setting \( g'(c) = 0 \),

\[ 2 f(c) f'(c) = f(a) + f(b). \]

Rearranging, we obtain the compact and elegant identity:

\[ f'(c) f(c) = \frac{f(a) + f(b)}{2}. \]

This equation reveals that at some interior point \( c \), the product of \( f(c) \) and its derivative is simply the arithmetic mean of the function values at the endpoints.

Example

Consider a function \( f \) that is continuous and differentiable on \([0,a]\), where \( a > 0 \) and \( f(a) = 0 \). We define the auxiliary function

\[ g(x) = x f(x). \]

Since \( g(x) \) is a product of continuous and differentiable functions, it is itself continuous on \([0,a]\) and differentiable on \( (0,a) \). Evaluating \( g(x) \) at the endpoints:

\[ g(0) = 0 \cdot f(0) = 0, \quad g(a) = a f(a) = 0. \]

Since \( g(0) = g(a) \), Rolle’s theorem guarantees the existence of some \( c \in (0,a) \) such that \( g'(c) = 0 \).

Differentiating \( g(x) \),

\[ g'(x) = f(x) + x f'(x). \]

Setting \( g'(c) = 0 \), we obtain

\[ f(c) + c f'(c) = 0. \]

Rearranging,

\[ f'(c) = -\frac{f(c)}{c}. \]

This shows that at some interior point \( c \), the derivative of \( f \) is precisely the negative of the function value divided by \( c \), revealing a fundamental property of such functions.

This problem in some exam may be posed as follows:

"Given a continuous and differentiable function \( f \) on \( [0,a] \) such that \( f(a) = 0 \), prove that there exists \( c \in (0,a) \) such that"

\[ f'(c) = -\frac{f(c)}{c}. \]

To prove this, we attempt to derive the result by working backwards. Assume that the given identity holds at some \( c \):

\[ f'(c) = -\frac{f(c)}{c}. \]

This suggests a differential equation of the form

\[ f'(x) = -\frac{f(x)}{x}. \]

Rearrange this equation to express it in a form suitable for integration:

\[ \frac{d}{dx} \big( x f(x) \big) = 0. \]

Now, moving forward, we construct the function

\[ g(x) = x f(x). \]

Clearly, \( g(x) \) is continuous and differentiable on \( [0,a] \). Evaluating at the endpoints:

\[ g(0) = 0 \cdot f(0) = 0, \quad g(a) = a f(a) = 0. \]

Since \( g(0) = g(a) \), Rolle’s theorem applies, guaranteeing the existence of some \( c \in (0,a) \) such that \( g'(c) = 0 \). Computing \( g'(x) \),

\[ g'(x) = f(x) + x f'(x). \]

Setting this equal to zero at \( x = c \),

\[ f(c) + c f'(c) = 0. \]

Rearranging,

\[ f'(c) = -\frac{f(c)}{c}. \]

Rolle’s Theorem and Roots of a Function

Assume that a function \( f \) always satisfies the first two hypotheses of Rolle’s theorem, meaning it is continuous on any closed interval and differentiable on any open interval. Then we can make the following claims about the roots of \( f \):

I. Between any two consecutive roots of \( f \), there exists at least one root of \( f' \).

To see why this must be true, let \( \alpha \) and \( \beta \) be two consecutive roots of \( f \), meaning

\[ f(\alpha) = 0 \quad \text{and} \quad f(\beta) = 0. \]

Since \( f \) is continuous on \([ \alpha, \beta ]\) and differentiable on \( (\alpha, \beta) \), the hypotheses of Rolle’s theorem are satisfied.

By Rolle’s theorem, there exists some \( c \in (\alpha, \beta) \) such that

\[ f'(c) = 0. \]

This shows that between any two consecutive roots of \( f \), there is at least one root of \( f' \).

II. If \( f \) has \( n \) real roots, then \( f' \) has at least \( n - 1 \) real roots

Assume that \( f \) is a function that satisfies the hypotheses of Rolle’s theorem: it is continuous and differentiable wherever necessary. Suppose \( f \) has \( n \) distinct real roots, say

\[ \alpha_1 < \alpha_2 < \dots < \alpha_n. \]

Since \( f(\alpha_i) = 0 \) for each \( i \), we can consider consecutive pairs of these roots:

\[ (\alpha_1, \alpha_2), (\alpha_2, \alpha_3), \dots, (\alpha_{n-1}, \alpha_n). \]

By Rolle’s theorem, within each interval \( (\alpha_i, \alpha_{i+1}) \), there exists at least one point \( c_i \) such that

\[ f'(c_i) = 0. \]

Since there are \( n-1 \) such intervals, we obtain at least \( n-1 \) distinct roots of \( f' \), namely

\[ c_1, c_2, \dots, c_{n-1}. \]

Thus, we conclude that if \( f \) has \( n \) real roots, then \( f' \) has at least \( n-1 \) real roots.

III. If \( f \) is a polynomial of degree \( n \) with exactly \( n \) real roots, then \( f' \) has exactly \( n-1 \) real roots, all distinct from the roots of \( f \).

We establish this result in two steps:

1. Lower Bound on the Number of Roots of \( f' \):

   As shown earlier, if \( f \) has \( n \) distinct real roots, say

   \[ \alpha_1 < \alpha_2 < \dots < \alpha_n, \]

   then by Rolle’s theorem, \( f' \) must have at least one root in each interval \( (\alpha_i, \alpha_{i+1}) \), for \( i = 1, 2, \dots, n-1 \).
   Thus, \( f' \) must have at least \( n-1 \) real roots.

2. Upper Bound on the Number of Roots of \( f' \):

   Since \( f(x) \) is a polynomial of degree \( n \), its derivative \( f'(x) \) is a polynomial of degree \( n-1 \). A polynomial of degree \( n-1 \) can have at most \( n-1 \) real roots. Since we have already shown that \( f' \) has at least \( n-1 \) real roots, it follows that these are all the roots \( f' \) can have.

From Rolle’s theorem, we see that each root of \( f' \) lies in an interval between two consecutive roots of \( f \). This ensures that the roots of \( f' \) cannot coincide with the roots of \( f \), meaning all the roots of \( f' \) are distinct from those of \( f \).

Thus, if \( f \) has exactly \( n \) distinct real roots, then \( f' \) has exactly \( n-1 \) real roots, and none of these roots coincide with the roots of \( f \).

IV. If \( f' \) has \( m \) real roots, then \( f \) cannot have more than \( m+1 \) real roots.

We justify this result as follows:

1. Suppose \( f' \) has exactly \( m \) real roots, say

   \[ \beta_1 < \beta_2 < \dots < \beta_m. \]

   These are the critical points of \( f \), where \( f' \) changes sign.

2. The function \( f(x) \) can have at most one local extremum in each interval determined by these critical points. Thus, in the intervals

   \[ (-\infty, \beta_1), (\beta_1, \beta_2), \dots, (\beta_{m-1}, \beta_m), (\beta_m, \infty), \]

   \( f(x) \) can change direction at most \( m \) times.

3. A continuous function that changes direction at most \( m \) times can have at most \( m+1 \) real roots. If it had more than \( m+1 \) real roots, there would be more than \( m \) critical points by Rolle’s theorem, contradicting the assumption that \( f' \) has exactly \( m \) real roots.

Thus, we conclude that if \( f' \) has \( m \) real roots, then \( f \) cannot have more than \( m+1 \) real roots.

Lagrange’s Mean Value Theorem (MVT)

Lagrange’s Mean Value Theorem is a fundamental result in differential calculus that generalizes Rolle’s Theorem by removing the requirement that the function takes equal values at the endpoints. It provides a precise way to quantify the rate of change of a function over an interval.

Statement:

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

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

Then, there exists some \( c \in (a, b) \) such that

\[ f'(c) = \frac{f(b) - f(a)}{b - a}. \]

Geometrically, the theorem guarantees the existence of a point \( c \in (a, b) \) where the tangent to the curve \( y = f(x) \) is parallel to the secant line joining the points \( (a, f(a)) \) and \( (b, f(b)) \).

The secant line passing through \( (a, f(a)) \) and \( (b, f(b)) \) has the equation

\[ y - f(a) = \frac{f(b) - f(a)}{b - a} (x - a). \]

The slope of this secant line is

\[ \frac{f(b) - f(a)}{b - a}. \]

Since \( f(x) \) is differentiable on \( (a, b) \), it has a well-defined tangent line at every point in \( (a, b) \). The theorem asserts that there exists some \( c \in (a, b) \) such that the tangent line at \( x = c \) has the same slope as the secant line, meaning

\[ f'(c) = \frac{f(b) - f(a)}{b - a}. \]

Thus, there exists at least one point \( c \) where the curve \( y = f(x) \) has a tangent parallel to the secant line connecting \( (a, f(a)) \) and \( (b, f(b)) \).

LMVT and Rate of Change

Lagrange’s Mean Value Theorem provides an essential link between the average rate of change of a function over an interval and its instantaneous rate of change at some interior point.

Given a function \( f \) that is continuous on the closed interval \([a, b]\) and differentiable on the open interval \( (a, b) \), the theorem guarantees the existence of a point \( c \in (a, b) \) such that

\[ f'(c) = \frac{f(b) - f(a)}{b - a}. \]

This equation states that the instantaneous rate of change of \( f(x) \) at \( x = c \), given by \( f'(c) \), is equal to the average rate of change of \( f(x) \) over the interval \( [a, b] \).

• The average rate of change of \( f(x) \) from \( x = a \) to \( x = b \) is

  \[ \frac{f(b) - f(a)}{b - a}. \]

  This represents the change in the function value per unit change in \( x \) over the entire interval.

• The instantaneous rate of change of \( f(x) \) at \( x = c \) is given by \( f'(c) \), which represents how \( f(x) \) is changing at that specific point.

The theorem asserts that at some point \( c \) in \( (a, b) \), the instantaneous rate of change matches the average rate of change over the entire interval.

Motion along a Straight Line

In the context of motion along a straight line, the function \( x(t) \) represents the position of an object at time \( t \). The average velocity of the object over the time interval \([t_1, t_2]\) is given by

\[ \frac{x(t_2) - x(t_1)}{t_2 - t_1}. \]

This quantity represents the total displacement divided by the total time taken. It describes how the position of the object changes on average over the interval \([t_1, t_2]\).

The instantaneous velocity at any time \( t \) is given by the derivative \( x'(t) \), which represents the rate of change of position at that specific instant.

Lagrange’s Mean Value Theorem states that if \( x(t) \) is continuous on \([t_1, t_2]\) and differentiable on \( (t_1, t_2) \), then there exists some \( t \in (t_1, t_2) \) such that

\[ x'(t) = \frac{x(t_2) - x(t_1)}{t_2 - t_1}. \]

This means that at some instant \( t \) within the interval, the instantaneous velocity \( x'(t) \) is exactly equal to the average velocity over the entire interval. In other words, at least once during the motion, the object must be moving at a speed that matches the overall average speed computed over the interval.

Example

Let \( f(x) = ax^2 + bx + c \). Apply the Lagrange Mean Value Theorem (LMVT) on the interval \( [\alpha, \beta] \) and find the value of \( c \).

Solution:

Since \( f(x) = ax^2 + bx + c \) is a polynomial, it is continuous and differentiable everywhere, so the Mean Value Theorem applies on \( [\alpha, \beta] \). By LMVT, there exists some \( c \in (\alpha, \beta) \) such that

\[ f'(c) = \frac{f(\beta) - f(\alpha)}{\beta - \alpha}. \]

Differentiating \( f(x) \), we get \( f'(x) = 2ax + b \), so at \( x = c \),

\[ 2ac + b = \frac{f(\beta) - f(\alpha)}{\beta - \alpha}. \]

Expanding \( f(\beta) - f(\alpha) \),

\[ f(\beta) - f(\alpha) = (a\beta^2 + b\beta + c) - (a\alpha^2 + b\alpha + c) = a(\beta^2 - \alpha^2) + b(\beta - \alpha). \]

Using the identity \( \beta^2 - \alpha^2 = (\beta - \alpha)(\beta + \alpha) \),

\[ f(\beta) - f(\alpha) = (\beta - \alpha)(a(\beta + \alpha) + b). \]

Dividing by \( \beta - \alpha \),

\[ \frac{f(\beta) - f(\alpha)}{\beta - \alpha} = a(\beta + \alpha) + b. \]

Since \( 2ac + b = a(\beta + \alpha) + b \), subtracting \( b \) from both sides gives

\[ 2ac = a(\beta + \alpha), \]

which simplifies to

\[ c = \frac{(\beta + \alpha)}{2}. \]

Interestingly, \(c\) is a the midpoint of the interval \([\alpha, \beta]\).

Proof of Lagrange’s Mean Value Theorem

Let \( f \) be a function that is continuous on \( [a,b] \) and differentiable on \( (a,b) \). To apply Rolle’s theorem, we construct an auxiliary function

\[ g(x) = f(x) + \lambda x, \]

where \( \lambda \) is chosen such that \( g(a) = g(b) \). Expanding this condition,

\[ f(a) + \lambda a = f(b) + \lambda b. \]

Rearranging,

\[ \lambda (b - a) = f(b) - f(a), \]

which gives

\[ \lambda = -\frac{f(b) - f(a)}{b - a}. \]

Substituting back, we get

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

Since \( f \) is continuous on \( [a,b] \) and differentiable on \( (a,b) \), the same holds for \( g(x) \). From the choice of \( \lambda \), it follows that \( g(a) = g(b) \), so Rolle’s theorem applies on this function in \([a, b]\), ensuring the existence of some \( c \in (a,b) \) such that

\[ g'(c) = 0. \]

Differentiating,

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

Setting \( g'(c) = 0 \), we obtain

\[ f'(c) = \frac{f(b) - f(a)}{b - a}. \]

Thus, we have established the existence of a point \( c \in (a, b) \) where the derivative of \( f \) equals the slope of the secant line joining \( (a, f(a)) \) and \( (b, f(b)) \), completing the proof. \(\blacksquare\)

The proof once again shows how using an auxiliary function can help establish mathematical results in a precise and elegant way. By carefully defining a function \( g(x) \) that includes the given function \( f(x) \) along with a well-chosen linear term, we transformed the problem into a form where Rolle’s theorem could be applied.

This allowed us to find a point \( c \) where the derivative of \( f(x) \) is equal to the slope of the secant line, without needing to study \( f(x) \) directly. This method demonstrates a clever mathematical technique—by modifying a function in a structured way, we fit it into a well-known theorem, allowing us to prove an important result with clarity.

Using LMVT to Find Bounds on Values of Functions

Let \( f \) be continuous on \( [a, b] \) and differentiable on \( (a, b) \). Suppose the derivative \( f' \) is bounded on \( (a, b) \), meaning there exist constants \( \alpha \) and \( \beta \) such that

\[ \alpha \leq f'(x) \leq \beta, \quad \text{for all } x \in (a, b). \]

Given that \( f(a) \) is known, we aim to find the possible range of values for \( f(b) \).

Since the hypotheses of the Mean Value Theorem are satisfied, there exists some \( c \in (a, b) \) such that

\[ f'(c) = \frac{f(b) - f(a)}{b - a}. \]

From the given bound on \( f'(x) \), we know that

\[ \alpha \leq f'(c) \leq \beta. \]

Substituting \( f'(c) = \frac{f(b) - f(a)}{b - a} \), we obtain

\[ \alpha \leq \frac{f(b) - f(a)}{b - a} \leq \beta. \]

Multiplying throughout by \( (b-a) \), which is positive, we get

\[ \alpha (b-a) \leq f(b) - f(a) \leq \beta (b-a). \]

Rearranging to isolate \( f(b) \),

\[ f(a) + \alpha (b-a) \leq f(b) \leq f(a) + \beta (b-a). \]

Thus, the possible values of \( f(b) \) lie in the range

\[ \big[ f(a) + \alpha (b-a), \, f(a) + \beta (b-a) \big]. \]

This result gives precise upper and lower bounds on \( f(b) \) based on the given bounds on \( f'(x) \).

Example

Let \( f \) be a function that is continuous on \( [1,4] \) and differentiable on \( (1,4) \). Suppose that the derivative of \( f \) satisfies

\[ f'(x) \geq 2 \quad \text{for all } x \in (1,4), \]

and that

\[ f(1) = 3. \]

Find the possible values of \( f(4) \).

Solution:

Since \( f \) is continuous on \( [1,4] \) and differentiable on \( (1,4) \), the Mean Value Theorem applies. Thus, there exists some \( c \in (1,4) \) such that

\[ f'(c) = \frac{f(4) - f(1)}{4 - 1}. \]

Substituting \( f(1) = 3 \),

\[ f'(c) = \frac{f(4) - 3}{3}. \]

Given that \( f'(x) \geq 2 \) for all \( x \), we also have \( f'(c) \geq 2 \), so

\[ \frac{f(4) - 3}{3} \geq 2. \]

Multiplying by 3,

\[ f(4) - 3 \geq 6. \]

Adding 3 to both sides,

\[ f(4) \geq 9. \]

Thus, the function value at \( x = 4 \) must satisfy the inequality

\[ f(4) \geq 9. \]

This gives the minimum possible value of \( f(4) \), but since no upper bound on \( f'(x) \) is given, \( f(4) \) can be arbitrarily large.

Cauchy’s Mean Value Theorem (Generalized Mean Value Theorem)

Let \( f(x) \) and \( g(x) \) be functions that satisfy the following conditions on the closed interval \( [a, b] \):

1. \( f(x) \) and \( g(x) \) are continuous on \( [a, b] \).
2. \( f(x) \) and \( g(x) \) are differentiable on \( (a, b) \).
3. \( g'(x) \neq 0 \) for all \( x \in (a, b) \).

Then, there exists some \( c \in (a, b) \) such that

\[ \frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}. \]

This theorem generalizes Lagrange’s Mean Value Theorem by applying it to two functions instead of one. If we take \( g(x) = x \), then \( g'(x) = 1 \), and the theorem reduces to the standard Mean Value Theorem.

Proof of Cauchy’s Mean Value Theorem

To prove the theorem, we define an auxiliary function

\[ h(x) = f(x) + \lambda g(x), \]

where \( \lambda \) is chosen so that \( h(a) = h(b) \). Expanding this condition,

\[ f(a) + \lambda g(a) = f(b) + \lambda g(b). \]

Rearranging,

\[ \lambda (g(b) - g(a)) = f(b) - f(a). \]

Solving for \( \lambda \),

\[ \lambda = -\frac{f(b) - f(a)}{g(b) - g(a)}. \]

Thus, we redefine

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

Since \( f \) and \( g \) are continuous on \( [a,b] \) and differentiable on \( (a,b) \), the same holds for \( h(x) \). Also, by construction,

\[ h(a) = h(b), \]

so Rolle’s theorem applies, ensuring the existence of some \( c \in (a, b) \) such that

\[ h'(c) = 0. \]

Differentiating,

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

Setting \( h'(c) = 0 \), we obtain

\[ f'(c) = \frac{f(b) - f(a)}{g(b) - g(a)} g'(c). \]

Thus, we have established the existence of a point \( c \in (a, b) \) where

\[ \frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}. \]

This completes the proof of Cauchy’s Mean Value Theorem.