# Lagrange’s Mean Value Theorem

• ### The Mean Value Theorem (MVT)

Lagrange’s mean value theorem (MVT) states that if a function $$f\left( x \right)$$ is continuous on a closed interval $$\left[ {a,b} \right]$$ and differentiable on the open interval $$\left( {a,b} \right),$$ then there is at least one point $$x = c$$ on this interval, such that

$f\left( b \right) – f\left( a \right) = f’\left( c \right)\left( {b – a} \right).$

This theorem (also known as First Mean Value Theorem) allows to express the increment of a function on an interval through the value of the derivative at an intermediate point of the segment.

#### Proof.

Consider the auxiliary function

$F\left( x \right) = f\left( x \right) + \lambda x.$

We choose a number $$\lambda$$ such that the condition $$F\left( a \right) = F\left( b \right)$$ is satisfied. Then

${f\left( a \right) + \lambda a = f\left( b \right) + \lambda b,\;\;}\Rightarrow {f\left( b \right) – f\left( a \right) = \lambda \left( {a – b} \right),\;\;}\Rightarrow {\lambda = – \frac{{f\left( b \right) – f\left( a \right)}}{{b – a}}.}$

As a result, we have

${F\left( x \right) = f\left( x \right) } {- \frac{{f\left( b \right) – f\left( a \right)}}{{b – a}}x.}$

The function $$F\left( x \right)$$ is continuous on the closed interval $$\left[ {a,b} \right],$$ differentiable on the open interval $$\left( {a,b} \right)$$ and takes equal values at the endpoints of the interval. Therefore, it satisfies all the conditions of Rolle’s theorem. Then there is a point $$c$$ in the interval $$\left( {a,b} \right)$$ such that

$F’\left( c \right) = 0.$

It follows that

$f’\left( c \right) – \frac{{f\left( b \right) – f\left( a \right)}}{{b – a}} = 0$

or

$f\left( b \right) – f\left( a \right) = f’\left( c \right)\left( {b – a} \right).$

### Geometric interpretation

Lagrange’s mean value theorem has a simple geometrical meaning. The chord passing through the points of the graph corresponding to the ends of the segment $$a$$ and $$b$$ has the slope equal to

${k = \tan \alpha } = {\frac{{f\left( b \right) – f\left( a \right)}}{{b – a}}.}$

Then there is a point $$x = c$$ inside the interval $$\left[ {a,b} \right],$$ where the tangent to the graph is parallel to the chord (Figure $$2$$).

### Physical interpretation

The mean value theorem has also a clear physical interpretation. If we assume that $$f\left( t \right)$$ represents the position of a body moving along a line, depending on the time $$t,$$ then the ratio of

$\frac{{f\left( b \right) – f\left( a \right)}}{{b – a}}$

is the average velocity of the body in the period of time $$b – a.$$ Since $$f’\left( t \right)$$ is the instantaneous velocity, this theorem means that there exists a moment of time $$c,$$ in which the instantaneous speed is equal to the average speed.

Lagrange’s mean value theorem has many applications in mathematical analysis, computational mathematics and other fields. Let us further note two remarkable corollaries.

### Corollary $$1$$

In a particular case when the values of the function $$f\left( x \right)$$ at the endpoints of the segment $$\left[ {a,b} \right]$$ are equal, i.e. $$f\left( a \right) = f\left( b \right),$$ the mean value theorem implies that there is a point $$c \in \left( {a,b} \right)$$ such that

${f’\left( c \right) } = {\frac{{f\left( b \right) – f\left( a \right)}}{{b – a}} = 0,}$

that is, we get Rolle’s theorem, which can be considered as a special case of Lagrange’s mean value theorem.

### Corollary $$2$$

If the derivative $$f’\left( x \right)$$ is zero at all points of the interval $$\left[ {a,b} \right],$$ then the function $$f\left( x \right)$$ is constant on this interval. Indeed, for any two points $${x_1}$$ and $${x_2}$$ in the interval $$\left[ {a,b} \right],$$ there exists a point $$c \in \left( {a,b} \right)$$ such that

${f\left( {{x_2}} \right) – f\left( {{x_1}} \right) } = {f’\left( c \right)\left( {{x_2} – {x_1}} \right) } = {0 \cdot \left( {{x_2} – {x_1}} \right) = 0.}$

Consequently,

$f\left( {{x_1}} \right) = f\left( {{x_2}} \right).$

• ## Solved Problems

Click a problem to see the solution.

### Example 1

A thermometer was taken from a freezer and placed in the boiling water. It took 22 sec for the thermometer to rise from $$-10^\circ C$$ to $$100^\circ C.$$ Determine the average rate of temperature change.

### Example 2

Check the validity of Lagrange’s mean value theorem for the function $f\left( x \right) = {x^2} – 3x + 5$ on the interval $$\left[ {1,4} \right].$$ If the theorem holds, find a point $$c$$ satisfying the conditions of the theorem.

### Example 3

Let $$f\left( x \right) = \sqrt {x + 4} .$$ Find a point $$c$$ that satisfies the Mean Value Theorem for the function on the interval $$\left[ {0,5} \right].$$

### Example 4

The position of a particle is given by the function $$s\left( t \right) = 2{t^2} + 3t – 4.$$ Find the time $$t=c$$ in the interval $$0 \le t \le 4$$ when the instantaneous velocity of the particle equals to its average velocity in this interval.

### Example 5

Check the validity of Lagrange’s mean value theorem for the function $f\left( x \right) = \frac{{x – 1}}{{x – 3}}$ on the interval $$\left[ {4,5} \right].$$

### Example 6

Find all points $$c$$ satisfying the conditions of the MVT for the function $$f\left( x \right) = {x^3} – x$$ in the interval $$\left[ { – 2,1} \right].$$

### Example 7

Given the function $$f\left( x \right) = \large{\frac{1}{x}}\normalsize .$$ Find all values of $$c$$ that satisfy the Mean Value Theorem for $$f\left( x \right)$$ on the interval $$\left[ {1,4} \right].$$

### Example 8

Find the point $$C\left( {\xi ,\eta } \right)$$ on the curve $$y = {x^3},$$ where the tangent is parallel to the chord connecting the points $$O\left( {0,0} \right)$$ and $$A\left( {2,8} \right)$$ (Figure $$4$$).

### Example 9

Suppose that $$f\left( 2 \right) = 1$$ and $$f^\prime\left( x \right) \le 5$$ for all values of $$x.$$
1. Determine a lower bound for $$f\left( -2 \right).$$
2. Determine an upper bound for $$f\left( 5 \right).$$

### Example 10

Let $$f\left( x \right) = {x^3} – x.$$ Find all numbers $$c$$ that satisfy the Mean Value Theorem for $$f\left( x \right)$$ on the interval $$\left[ { – 3,3} \right].$$

### Example 11

Given the function $$f\left( x \right) = {x^3} – 2{x^2} – x + 1.$$ Find all points of $$c$$ satisfying the conditions of the Mean Value Theorem for the function on the interval $$\left[ { – 2,2} \right].$$

### Example 12

Let $$f\left( x \right) = \large{\frac{{x – 2}}{{x + 2}}}\normalsize .$$ Find all values of $$c$$ that satisfy the Mean Value Theorem for the function on the interval $$\left[ { – 1,2} \right].$$

### Example 13

Compose the Lagrange formula for the quadratic function $$f\left( x \right) = a{x^2} + bx + c$$ for arbitrary values of $$x$$ and $$\Delta x.$$

### Example 14

The function $${f\left( x \right)}$$ is everywhere continuous and differentiable. Prove that if the function $${f\left( x \right)}$$ has two real roots, then its derivative $${f’\left( x \right)}$$ has at least one root.

### Example 15

The function $${f\left( x \right)}$$ is continuous and differentiable on the interval $$\left[ {2,10} \right].$$ It is known that $$f\left( 2 \right) = 8$$ and the derivative on the given interval satisfies the condition $$f’\left( x \right) \le 4$$ for all $$x \in \left( {2,10} \right).$$ Determine the maximum possible value of the function at $$x = 10.$$

### Example 16

The function $$f\left( x \right)$$ is continuous and differentiable on the interval $$\left[ { – 2,6} \right].$$ It is known that $$f\left( { – 2} \right) = 4$$ and the derivative on the interval satisfies the condition $$f^\prime\left( x \right) \le 3$$ for all $$x \in \left( { – 2,6} \right).$$ Determine an upper bound of the function at the right endpoint $$x=6.$$

### Example 1.

A thermometer was taken from a freezer and placed in the boiling water. It took 22 sec for the thermometer to rise from $$-10^\circ C$$ to $$100^\circ C.$$ Determine the average rate of temperature change.

Solution.

The average rate of temperature change $$\large{\frac{{\Delta T}}{{\Delta t}}}\normalsize$$ is described by the right-hand side of the formula given by the Mean Value Theorem:

${\frac{{\Delta T}}{{\Delta t}} = \frac{{T\left( {{t_2}} \right) – T\left( {{t_1}} \right)}}{{{t_2} – {t_1}}} }={ \frac{{100 – \left( { – 10} \right)}}{{22}} }={ \frac{{110}}{{22}} }={ 5\,\frac{{^\circ C}}{{\sec }}}$

### Example 2.

Check the validity of Lagrange’s mean value theorem for the function $f\left( x \right) = {x^2} – 3x + 5$ on the interval $$\left[ {1,4} \right].$$ If the theorem holds, find a point $$c$$ satisfying the conditions of the theorem.

Solution.

The given quadratic function is continuous and differentiable on the entire set of real numbers. Hence, we can apply Lagrange’s mean value theorem. The derivative of the function has the form

${f’\left( x \right) = {\left( {{x^2} – 3x + 5} \right)^\prime } } = {2x – 3.}$

Find the coordinates of the point $$c:$$

${f’\left( c \right) = \frac{{f\left( b \right) – f\left( a \right)}}{{b – a}},\;\;}\Rightarrow {2c – 3 }={ \frac{{\left( {{4^2} – 3 \cdot 4 + 5} \right) – \left( {{1^2} – 3 \cdot 1 + 5} \right)}}{{4 – 1}},\;\;}\Rightarrow {2c – 3 = \frac{{9 – 3}}{3} = 2,\;\;}\Rightarrow {2c = 5,\;\;}\Rightarrow {c = 2,5.}$

You can see that the point $$c = 2,5$$ lies in the interval $$\left( {1,4} \right).$$

### Example 3.

Let $$f\left( x \right) = \sqrt {x + 4} .$$ Find a point $$c$$ that satisfies the Mean Value Theorem for the function on the interval $$\left[ {0,5} \right].$$

Solution.

The function is continuous on the closed interval $$\left[ {0,5} \right]$$ and differentiable on the open interval $$\left( {0,5} \right),$$ so the MVT is applicable to the function.

The derivative has the form

${f^\prime\left( x \right) = \left( {\sqrt {x + 4} } \right)^\prime }={ \frac{1}{{2\sqrt {x + 4} }}.}$

Find the coordinates of the point $$c:$$

${f^\prime\left( c \right) = \frac{{f\left( b \right) – f\left( a \right)}}{{b – a}},\;\;} \Rightarrow {\frac{1}{{2\sqrt {c + 4} }} = \frac{{\sqrt {5 + 4} – \sqrt {0 + 4} }}{{5 – 0}},\;\;} \Rightarrow {\frac{1}{{2\sqrt {c + 4} }} = \frac{1}{5},\;\;} \Rightarrow {\sqrt {c + 4} = \frac{5}{2},\;\;} \Rightarrow {c + 4 = \frac{{25}}{4},\;\;} \Rightarrow {c = \frac{9}{4} = 2.25}$

It can be seen that the point $$c = 2.25$$ belongs to the open interval $$\left( {0,5} \right).$$

### Example 4.

The position of a particle is given by the function $$s\left( t \right) = 2{t^2} + 3t – 4.$$ Find the time $$t=c$$ in the interval $$0 \le t \le 4$$ when the instantaneous velocity of the particle equals to its average velocity in this interval.

Solution.

The function $$s\left( t \right)$$ satisfies the conditions of the Mean Value Theorem, so we can write

$s^\prime\left( c \right) = \frac{{s\left( b \right) – s\left( a \right)}}{{b – a}},$

where $$a=0,$$ $$b=4.$$

Take the derivative:

${s^\prime\left( t \right) = \left( {2{t^2} + 3t – 4} \right)^\prime }={ 4t + 3.}$

Substituting this in the formula above, we get

${4c + 3 = \frac{{40 – \left( { – 4} \right)}}{4},\;\;} \Rightarrow {4c + 3 = 11,\;\;} \Rightarrow {4c = 8,\;\;} \Rightarrow {c = 2.}$

Answer: $$c=2.$$

### Example 5.

Check the validity of Lagrange’s mean value theorem for the function $f\left( x \right) = \frac{{x – 1}}{{x – 3}}$ on the interval $$\left[ {4,5} \right].$$

Solution.

This function has a discontinuity at $$x = 3,$$ but on the interval $$\left[ {4,5} \right]$$ it is continuous and differentiable. Therefore, the mean value theorem is applicable here. Find the derivative:

$\require{cancel} {f’\left( x \right) = {\left( {\frac{{x – 1}}{{x – 3}}} \right)^\prime } } = {\frac{{{{\left( {x – 1} \right)}^\prime }\left( {x – 3} \right) – \left( {x – 1} \right){{\left( {x – 3} \right)}^\prime }}}{{{{\left( {x – 3} \right)}^2}}} } = {\frac{{1 \cdot \left( {x – 3} \right) – \left( {x – 1} \right) \cdot 1}}{{{{\left( {x – 3} \right)}^2}}} } = {\frac{{\cancel{x} – 3 – \cancel{x} + 1}}{{{{\left( {x – 3} \right)}^2}}} } = { – \frac{2}{{{{\left( {x – 3} \right)}^2}}}.}$

Using the Lagrange formula, we get:

${f’\left( c \right) = \frac{{f\left( b \right) – f\left( a \right)}}{{b – a}},\;\;}\Rightarrow { – \frac{2}{{{{\left( {c – 3} \right)}^2}}} = \frac{{f\left( 5 \right) – f\left( 4 \right)}}{{5 – 4}}.}$

The values of the function at the endpoints are

${f\left( 4 \right) = \frac{{4 – 1}}{{4 – 3}} = 3,}\;\;\;\kern-0.3pt {f\left( 5 \right) = \frac{{5 – 1}}{{5 – 3}} = 2.}$

Then

${- \frac{2}{{{{\left( {c – 3} \right)}^2}}} = \frac{{2 – 3}}{{5 – 4}},\;\;}\Rightarrow { – \frac{2}{{{{\left( {c – 3} \right)}^2}}} = – 1,\;\;}\Rightarrow {{\left( {c – 3} \right)^2} = 2.}$

In this case only the positive square root is valid. Hence,

${c – 3 = \sqrt 2 ,\;\;}\Rightarrow {c = 3 + \sqrt 2 \approx 4,41.}$

Thus, the point at which the tangent to the graph is parallel to the chord lies in the interval $$\left( {4,5} \right)$$ and has the coordinate $$c = 3 + \sqrt 2 \approx 4,41.$$

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Problems 1-5
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Problems 6-16