The Mean Value Theorem, MVT, with Solved Examples

Statement of Mean Value Theorem (MVT) : Let a function $f$ be

contineous on $\left[a\;,\;b\right]$ .
differentiable on $\rbrack a,b\lbrack$ .

then there exist a point $c\in\rbrack a,b\lbrack$ such that

$\frac{f(b)-f(a)}{b-a}=\;f'(c)$

MVT Proof: Define a new function $F$ by

$F(x)= Ax + f(x)$

where A is a constant to determined such that $F(a)=F(b)$

The functiom $Ax$ is contineous on $\left[a\;,\;b\right]$ and differentiable on $\rbrack a,b\lbrack$ . Hence $F(x)$ is contineous and differentiable respecctively on $\left[a\;,\;b\right]$ and $\rbrack a,b\lbrack$ . The function $F$ satisfies all the conditions of Rolle’s theorem therefore, there exists a point $c\in\rbrack a,b\lbrack$ such that

$F'(c)=0\;\;\;\; i.e,\;\;\; A + f'(c)=0$

or $f'(c)=-A$

From $F(a)=F(b)$ , we have

$Aa+f(a)=Ab +f(b)$

or $-A(b-a) =f(b)- f(a)$

or $-A=\frac{f(b)-f(a)}{b-c}$

substituting this value of $A$ into $(1)$ , we get

$\frac{f(b)-f(a)}{b-c}=f'(c)$ as required.

Geometrical Interpretation of Mean Value Theorem, MVT

The mean value theorem MVT has a simple geometrical interpretation

Let the graph of f on $\left[a\;,\;b\right]$ be represented by the curve APB. Suppose the chord AB makes an angle of measure $\beta$ with the x axis

Then

$\tan\;\beta\;=\frac{BM}{AM}=\frac{f(b)-f(a)}{b-a}……………(1)$

by the mean value theorem, we have

$\frac{f(b)-f(a)}{b-a}=f'(c)……………(2)$

where $c\in\rbrack a,b\lbrack$

From $(1)$ and $(2)$ we get $\tan\;\beta\;=f'(c)$

Thus at the point $P (c, f(c) )$ , The tangent line ton the graph of $y=f(x)$ is parallel to the chord of $AB$ .

For physical interpretation of MVT , Suppose $s= f(t)$ denotes the distance S that a particle has travelled at time $t$ . If $f$ contineous on $\left[a\;,\;b\right]$ . and differentiable on $\rbrack a,b\lbrack$ then by mean value theorem (mvt) there exists a point $t_0\in\rbrack a,b\lbrack$ such that

$\frac{f(b)-f(a)}{b-c}=f'(t_0)$

$i.e,$ the velocity $f'(t_0)$ equal to the average velocity $\frac{f(b)-f(a)}{b-a}$ from $a$ to $b$ .

For Example, If a car travels 400km in 5 hours then at some time the car has the speed of 80km per hour.

Also called LaGrange’s Mean Value Theorem after the name of French Mathematician J.L Lagrange(1736-1813)

Questions and their solution related to the Mean Value Theorem

Question 1: Verify MVT for $f\left(x\right)=x^2$ on the interval $\lbrack1,3\rbrack$ .

Solution:

Check Continuity and Differentiability: First, ensure $f\left(x\right)=x^2$ is continuous on $\lbrack1,3\rbrack$ and differentiable on $(1,3)$ Since $f(x)$ is a polynomial function, it is continuous and differentiable everywhere.

Calculate the Average Rate of Change: The average rate of change of $<em>f</em>$ on $\lbrack1,3\rbrack$ .

$f^'\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}=\frac{9-1}2=4$ .

Find c That Satisfies MVT:

We need to find c such that

$f^'\left(c\right)=4$

Since

$f'\left(x\right)=2x$

put $x=c$ .

$f'\left(c\right)=2c$ .

$4=2c$

$c=2$

This $<em>c</em>=2$ lies in the interval $\lbrack1,3\rbrack$ and satisfies the theorem.

Question 2: Determine if MVT applies to f\left(x\right)=\frac1x on [1,4] and find the value of c if it does.

Solution:

Check Continuity and Differentiability: The function f\left(x\right)=\frac1x is continuous and differentiable on [1,4] except at x=0, which is not in the interval.

Calculate the Average Rate of Change:

The average rate of change of f on [1,4] is

f’\left(c\right)=\frac{f\left(4\right)-f\left(1\right)}{4-1}=\frac{{\displaystyle\frac14}-1}3=-\frac14

Find �c That Satisfies MVT:

We find c such that

f’\left(c\right)=-\frac14

f’\left(c\right)=-\frac1{x^2}

-\frac14=-\frac1{c^2}

\Rightarrow\;\;\;\;c^2=4

so c=2 and c=-2

onlu c=2 is within the interval [1,4].

Question 3: Prove MVT does not apply to f(x)=∣x∣ on [−1,1][−1,1].

Solution:

Check Continuity and Differentiability: The function f\left(x\right)=\left|x\right| is continuous on [−1,1]. However, it is not differentiable at x=0 because the left-hand and right-hand derivatives do not match.
MVT Application: Since �(�)=∣�∣f(x)=∣x∣ fails the differentiability criterion at �=0x=0, the Mean Value Theorem does not apply to it on the interval [−1,1][−1,1].

These examples illustrate how the Mean Value Theorem can be applied or why it might not apply in certain situations.

Uses of Mean Value Theorem

The Mean Value Theorem (MVT) is a cornerstone of calculus that has numerous applications in various fields such as mathematics, physics, engineering, and economics. Its significance lies in providing a formal way to understand how functions behave over an interval. Here are some key uses of the Mean Value Theorem:

Determining Function Behavior: MVT can help determine if a function is increasing or decreasing over an interval. If the derivative �′(�)f′(c) is positive for all �c in the interval, the function is increasing. Conversely, if �′(�)f′(c) is negative, the function is decreasing.

Proving the Existence of Roots: MVT can be used in conjunction with Rolle’s Theorem to prove the existence of at least one root in a function within a given interval. If a continuous function �f has equal values at the endpoints of an interval, there exists at least one point �c in the interval where �′(�)=0f′(c)=0.

Establishing the Uniqueness of Solutions: In differential equations, MVT can be used to prove the uniqueness of solutions under certain conditions. This is crucial in ensuring that the solutions derived from mathematical models are reliable and consistent.

Approximation of Functions: The theorem is foundational for developing numerical methods, such as the Newton-Raphson method, which approximates the roots of a function. It provides a theoretical basis for the convergence of these methods.

Analysis of Error Bounds: In numerical analysis, MVT can help estimate the error in numerical differentiation and integration. Knowing the bounds of the function’s derivative allows for estimating the maximum error in calculations.

Optimization Problems: MVT plays a role in solving optimization problems by identifying critical points where the derivative equals zero. This information can be used to find local maxima and minima of functions, which is essential in various optimization scenarios.

Economics and Finance: In economics, MVT can be used to model the behavior of cost functions and to analyze the marginal cost at a certain level of production. It’s also applied in finance to model and predict the behavior of financial instruments over time.

Physics and Engineering: MVT is used in physics to derive equations related to motion and in engineering to analyze the stress and strain on materials. The theorem helps in understanding the distribution and impact of forces within structures.

Informing Integral Calculus Theories: The Fundamental Theorem of Calculus and the Mean Value Theorem for Integrals are closely related to MVT. These relationships form the basis for much of the theory behind integral calculus, including techniques for evaluating definite integrals.

Verifying Lipschitz Continuity: MVT is instrumental in proving that a function satisfies Lipschitz continuity, a property important in the study of differential equations and numerical analysis for ensuring stability and convergence of solutions.

Geometrical Interpretation of Mean Value Theorem, MVT

Questions and their solution related to the Mean Value Theorem

Question 1: Verify MVT for on the interval .

Solution:

Question 2: Determine if MVT applies to f\left(x\right)=\frac1x​ on [1,4] and find the value of c if it does.

Solution:

Question 3: Prove MVT does not apply to f(x)=∣x∣ on [−1,1][−1,1].

Solution:

Uses of Mean Value Theorem

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Question 1: Verify MVT for $f\left(x\right)=x^2$ on the interval $\lbrack1,3\rbrack$ .

Question 2: Determine if MVT applies to f\left(x\right)=\frac1x on [1,4] and find the value of c if it does.