Calculus

Applications of the Derivative

Lagrange’s Mean Value Theorem

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Problems 1-2
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Problems 3-6

Lagrange’s mean value theorem 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 \(x = \xi\) on this interval, such that
\[f\left( b \right) – f\left( a \right) = f’\left( \xi \right)\left( {b – a} \right).\] 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
\[
{F\left( x \right) = f\left( x \right) }
{- \frac{{f\left( b \right) – f\left( a \right)}}{{b – a}}x.}
\]

French mathematician Joseph-Louis Lagrange (1736-1813)

Fig.1 Joseph-Louis Lagrange
(1736-1813)

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 \(\xi\) in the interval \(\left( {a,b} \right)\) such that
\[F’\left( \xi \right) = 0.\] It follows that
\[f’\left( \xi \right) – \frac{{f\left( b \right) – f\left( a \right)}}{{b – a}} = 0\] or
\[f\left( b \right) – f\left( a \right) = f’\left( \xi \right)\left( {b – a} \right).\]

Geometrical meaning of Lagrange’s mean value theorem

Figure 2.

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

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 \(\xi,\) 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 \(\xi \in \left( {a,b} \right)\) such that
\[
{f’\left( \xi \right) }
= {\frac{{f\left( b \right) – f\left( a \right)}}{{b – a}} = 0,}
\] 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
\(\xi \in \left( {a,b} \right)\) such that
\[
{f\left( {{x_2}} \right) – f\left( {{x_1}} \right) }
= {f’\left( \xi \right)\left( {{x_2} – {x_1}} \right) }
= {0 \cdot \left( {{x_2} – {x_1}} \right) = 0.}
\] Consequently,
\[f\left( {{x_1}} \right) = f\left( {{x_2}} \right).\]

Solved Problems

Click on problem description to see solution.

 Example 1

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 \(\xi\) satisfying the conditions of the theorem.

 Example 2

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 3

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 \(3\)).

 Example 4

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 5

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 6

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 1.

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 \(\xi\) 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 \(\xi:\)
\[
{f’\left( \xi \right) = \frac{{f\left( b \right) – f\left( a \right)}}{{b – a}},\;\;}\Rightarrow
{2\xi – 3 = \frac{{\left( {{4^2} – 3 \cdot 4 + 5} \right) – \left( {{1^2} – 3 \cdot 1 + 5} \right)}}{{4 – 1}},\;\;}\Rightarrow
{2\xi – 3 = \frac{{9 – 3}}{3} = 2,\;\;}\Rightarrow
{2\xi = 5,\;\;}\Rightarrow
{\xi = 2,5.}
\] It can be seen that the point \(\xi = 2,5\) lies in the interval \(\left( {1,4} \right).\)

Example 2.

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:
\[
{f’\left( \xi \right) = \frac{{f\left( b \right) – f\left( a \right)}}{{b – a}},\;\;}\Rightarrow
{ – \frac{2}{{{{\left( {\xi – 3} \right)}^2}}} = \frac{{f\left( 5 \right) – f\left( 4 \right)}}{{5 – 4}}.}
\] 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
\[
{- \frac{2}{{{{\left( {\xi – 3} \right)}^2}}} = \frac{{2 – 3}}{{5 – 4}},\;\;}\Rightarrow
{ – \frac{2}{{{{\left( {\xi – 3} \right)}^2}}} = – 1,\;\;}\Rightarrow
{{\left( {\xi – 3} \right)^2} = 2.}
\] In this case only the positive square root is valid. Hence,
\[
{\xi – 3 = \sqrt 2 ,\;\;}\Rightarrow
{\xi = 3 + \sqrt 2 \approx 4,41.}
\] 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 \(\xi = 3 + \sqrt 2 \approx 4,41.\)

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Problems 1-2
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Problems 3-6