Calculus

Applications of the Derivative

Applications of Derivative Logo

Lagrange’s Mean Value Theorem

Solved Problems

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Example 7

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

Example 8

Find the point C(ξ, η) on the curve \[y = {x^3},\] where the tangent is parallel to the chord connecting the points O(0, 0) and A(2, 8) (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 \[f\left( x \right) = \frac{{x - 2}}{{x + 2}}.\] 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 7.

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

Solution.

The hyperbolic function \(f\left( x \right) = \frac{1}{x}\) is continuous and differentiable on the interval \(\left[ {1,4} \right].\) Therefore, we can apply the MVT.

First, we write the values of the function at the endpoints:

\[f\left( 1 \right) = 1,\;\;f\left( 4 \right) = \frac{1}{4}.\]

Take the derivative and compose the relation given by the MVT:

\[f^\prime\left( x \right) = \left( {\frac{1}{x}} \right)^\prime = - \frac{1}{{{x^2}}}.\]
\[f^\prime\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}},\;\; \Rightarrow - \frac{1}{{{c^2}}} = \frac{{f\left( 4 \right) - f\left( 1 \right)}}{{4 - 1}},\;\; \Rightarrow - \frac{1}{{{c^2}}} = \frac{{\frac{1}{4} - 1}}{3},\;\; \Rightarrow - \frac{1}{{{c^2}}} = \frac{{ - \frac{3}{4}}}{3},\;\; \Rightarrow {c^2} = 4,\;\; \Rightarrow {c_{1,2}} = \pm 2.\]

We see that only one root \(c=2\) belongs to the open interval \(\left( {1,4} \right).\) Hence, the answer is \(c=2.\)

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 \(4\)).

Solution.

This function is continuous and differentiable for all \(x \in \mathbb{R}.\) Therefore, we can use the Lagrange formula:

\[f'\left( \xi \right) = \frac{{{y_A} - {y_O}}}{{{x_A} - {x_O}}},\]

where \(\xi\) is the abscissa of the point \(C,\) in which the tangent is parallel to the chord \(OA.\)

Application of Lagrange's mean value theorem to the function y=x^3
Figure 4.

Substituting the derivative

\[f'\left( x \right) = \left( {{x^3}} \right)^\prime = 3{x^2}\]

and the coordinates of the ends of the chord, we obtain:

\[3{\xi ^2} = \frac{{8 - 0}}{{2 - 0}},\;\; \Rightarrow 3{\xi ^2} = 4,\;\; \Rightarrow {\xi ^2} = \frac{4}{3},\;\; \Rightarrow \xi = \pm \frac{2}{{\sqrt 3 }}.\]

Obviously, only the positive square root is relevant here. Then the coordinates of the point \(C\) are given by

\[\xi = \frac{2}{{\sqrt 3 }} \approx 1,15;\;\;\eta = {\xi ^3} = {\left( {\frac{2}{{\sqrt 3 }}} \right)^3} = \frac{8}{{3\sqrt 3 }} \approx 1,54.\]

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 \(f\left( 5 \right).\)

Solution.

\(1.\) As the function has a finite derivative for all \(x,\) it is continuous and differentiable on \(\mathbb{R}.\) Hence, the Mean Value Theorem is applicable to this function.

According to the MVT, we can write the following relationship

\[f\left( b \right) - f\left( a \right) = f^\prime\left( c \right)\left( {b - a} \right),\]
\[ \Rightarrow f\left( 2 \right) - f\left( { - 2} \right) = f^\prime\left( c \right)\left( {2 - \left( { - 2} \right)} \right),\]
\[\Rightarrow f\left( { - 2} \right) = f\left( 2 \right) - 4f^\prime\left( c \right).\]

Since \(f^\prime\left( x \right) \le 5,\) we get the inequality.

\[f\left( { - 2} \right) \ge f\left( 2 \right) - 4 \cdot 5 = 1 - 20 = - 19.\]

Thus, the lower bound (the minimum possible value) of the function at \(x=-2\) is

\[{f_{\min}}\left( { - 2} \right) = - 19.\]

\(2.\) To estimate an upper bound for \(f\left( 5 \right)\) we again use the MVT in the form

\[f\left( b \right) - f\left( a \right) = f^\prime\left( c \right)\left( {b - a} \right),\]

Substitute the known boundary values:

\[ f\left( 5 \right) - f\left( 2 \right) = f^\prime\left( c \right)\left( {5 - 2} \right),\]
\[\Rightarrow f\left( 5 \right) = f\left( 2 \right) + 3f^\prime\left( c \right).\]

Taking into account that \(f^\prime\left( x \right) \le 5,\) we have

\[f\left( 5 \right) \le f\left( 2 \right) + 3 \cdot 5 = 1 + 15 = 16.\]

The upper bound (the maximum possible value) for \(f\left( 5 \right)\) is equal to

\[{f_{\max}}\left( {5} \right) = 16.\]

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 \(\left[ { - 3,3} \right].\)

Solution.

The function is continuous and differentiable everywhere in \(\mathbb{R}.\) Therefore, we can apply the MVT to this function. Calculate the values at the endpoints of the interval \(\left[ { - 3,3} \right]:\)

\[f\left( { - 3} \right) = {\left( { - 3} \right)^3} - \left( { - 3} \right) = - 24,\]
\[f\left( 3 \right) = {3^3} - 3 = 24.\]

The derivative is given by

\[f^\prime\left( x \right) = \left( {{x^3} - x} \right)^\prime = 3{x^2} - 1.\]

Solve the equation to determine the values of \(c:\)

\[f^\prime\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}},\;\; \Rightarrow 3{c^2} - 1 = \frac{{24 - \left( { - 24} \right)}}{{3 - \left( { - 3} \right)}},\;\; \Rightarrow 3{c^2} - 1 = \frac{{48}}{6} = 8,\;\; \Rightarrow 3{c^2} = 9,\;\; \Rightarrow {c^2} = 3,\;\; \Rightarrow {c_{1,2}} = \pm \sqrt 3 .\]

We obtained two values of \(c\) that lie in the given open interval and satisfy the MVT: \({c_1} = - \sqrt 3 ,\) \({c_2} = \sqrt 3.\)

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 \(\left[ { - 2,2} \right].\)

Solution.

Note that this function is a cubic polynomial and, hence, it is everywhere continuous and differentiable. Therefore, the MVT is applicable to this function.

Evaluate the function at the endpoints of the interval:

\[f\left( { - 2} \right) = {\left( { - 2} \right)^3} - 2 \cdot {\left( { - 2} \right)^2} - \left( { - 2} \right) + 1 = - 8 - 8 + 2 + 1 = - 13,\]
\[f\left( 2 \right) = {2^3} - 2 \cdot {2^2} - 2 + 1 = 8 - 8 - 2 + 1 = - 1.\]

Find the derivative:

\[f^\prime\left( x \right) = \left( {{x^3} - 2{x^2} - x + 1} \right)^\prime = 3{x^2} - 4x - 1.\]

Solve the equation given by the MVT:

\[f^\prime\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}},\;\; \Rightarrow 3{c^2} - 4c - 1 = \frac{{ - 1 - \left( { - 13} \right)}}{{2 - \left( { - 2} \right)}},\;\; \Rightarrow 3{c^2} - 4c - 1 = 3,\;\; \Rightarrow 3{c^2} - 4c - 4 = 0.\]

Using the quadratic formula, we get

\[D = {\left( { - 4} \right)^2} - 4 \cdot 3 \cdot \left( { - 4} \right) = 64,\]
\[{c_{1,2}} = \frac{{4 \pm \sqrt {64} }}{6} = \frac{{4 \pm 8}}{6} = - \frac{2}{3},2.\]

Thus, solving gives two values of \(c.\) Notice that the MVT requires the number \(c\) to lie in the open interval. Therefore, only one root \({c_1} = - \frac{2}{3}\) belongs to the given open interval \(\left( { - \frac{2}{3},2} \right).\)

Answer: \(c = - \frac{2}{3}.\)

Example 12.

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

Solution.

The function has a discontinuity at \(x = - 2,\) but it is continuous on the closed interval \(\left[ { - 1,2} \right]\) and differentiable on the open interval \(\left( { - 1,2} \right).\) Hence, the MVT is applicable to this function on the given interval.

First, we evaluate the function at the endpoints:

\[f\left( { - 1} \right) = \frac{{ - 1 - 2}}{{ - 1 + 2}} = - 3,\]
\[f\left( 2 \right) = \frac{{2 - 2}}{{2 + 2}} = 0.\]

Next, we find the derivative by the quotient rule:

\[f^\prime\left( x \right) = \left( {\frac{{x - 2}}{{x + 2}}} \right)^\prime = \frac{{\left( {x - 2} \right)^\prime\left( {x + 2} \right) - \left( {x - 2} \right)\left( {x + 2} \right)^\prime}}{{{{\left( {x + 2} \right)}^2}}} = \frac{{\cancel{x} + 2 - \cancel{x} + 2}}{{{{\left( {x + 2} \right)}^2}}} = \frac{4}{{{{\left( {x + 2} \right)}^2}}}.\]

Write the formula given by the MVT:

\[f^\prime\left( c \right) = \frac{{f\left( b \right) - f\left( a \right)}}{{b - a}},\;\; \Rightarrow \frac{4}{{{{\left( {c + 2} \right)}^2}}} = \frac{{0 - \left( { - 3} \right)}}{{2 - \left( { - 1} \right)}},\;\; \Rightarrow \frac{4}{{{{\left( {c + 2} \right)}^2}}} = \frac{3}{3} = 1,\;\; \Rightarrow {\left( {c + 2} \right)^2} = 4,\;\; \Rightarrow {c^2} + 4c = 0,\;\; \Rightarrow c\left( {c + 4} \right) = 0,\;\; \Rightarrow {c_1} = 0,\,{c_2} = - 4.\]

As you can see, only the first root \({c_1} = 0\) lies in the open interval \(\left( { - 1,2} \right).\)

Answer: \(c=0.\)

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 \(\Delta x.\)

Solution.

Using Lagrange's mean value theorem, we can calculate the value of the function at the point \(x + \Delta x,\) if we know the value of the function \(f\left( x \right)\) at the point \(x\) and the value of the derivative \(f'\left( \xi \right)\) at some intermediate point \(\xi.\) We write the formula in the following form:

\[f\left( {x + \Delta x} \right) = f\left( x \right) + f'\left( \xi \right)\Delta x.\]

Now we find the coordinate \(\xi\) (where the derivative is calculated) for the quadratic function. The derivative is given by

\[f'\left( x \right) = \left( {a{x^2} + bx + c} \right)^\prime = 2ax + b.\]

Substituting the expressions for the function \(f\left( x \right)\) and its derivative in the Lagrange formula, we get:

\[a{\left( {x + \Delta x} \right)^2} + b\left( {x + \Delta x} \right) + c = a{x^2} + bx + c + \left( {2a\xi + b} \right)\Delta x,\;\; \Rightarrow { \cancel{\color{blue}{a{x^2}}} + 2ax\Delta x + a{\left( {\Delta x} \right)^2}} + {\cancel{\color{red}{bx}} + \cancel{\color{green}{b\Delta x}} + \cancel{\color{maroon}{c}}} = {\cancel{\color{blue}{a{x^2}}} + \cancel{\color{red}{bx}}} + {\cancel{\color{maroon}{c}} + 2a\xi \Delta x + \cancel{\color{green}{b\Delta x}},} \Rightarrow {2ax\Delta x + a{\left( {\Delta x} \right)^2} = 2a\xi \Delta x,}\;\; \Rightarrow a\Delta x\left( {2x + \Delta x} \right) = 2a\xi \Delta x,\;\; \Rightarrow 2\xi = 2x + \Delta x,\;\; \Rightarrow 2\xi = 2\left( {x + \frac{{\Delta x}}{2}} \right),\;\; \Rightarrow \xi = x + \frac{{\Delta x}}{2}.\]

Hence, in this case, the derivative must be calculated at the point \(\xi = x + {\frac{{\Delta x}}{2}},\) i.e. shifted with respect to \(x\) by the half of the increment \(\Delta x.\) This result holds for any quadratic function with arbitrary values of \(x\) and \(\Delta x.\) Thus, the final formula for the quadratic function has the form

\[f\left( {x + \Delta x} \right) = f\left( x \right) + f'\left( {x + \frac{{\Delta x}}{2}} \right)\Delta x.\]

According to Lagrange's mean value theorem, the tangent drawn at the point \(\xi = x + {\frac{{\Delta x}}{2}}\) will be parallel to the chord joining the points \(A\left( {x,f\left( x \right)} \right)\) and \(B\left( {x + \Delta x,f\left( {x + \Delta x} \right)} \right)\) (see Figure \(5\)).

Lagrange formula for the quadratic function y=ax^2+bx+c
Figure 5.

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.

Solution.

We denote the roots of the function as \(a\) and \(b.\) By the condition, the function \({f\left( x \right)}\) is continuous and differentiable on the interval \(\left[ {a,b} \right].\) Consequently, we can apply Lagrange's mean value theorem:

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

where \(c\) is a point lying in the open interval \(\left( {a,b} \right).\)

Since \({f\left( a \right) - f\left( b \right)} = 0,\) then

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

Thus, the derivative \({f'\left( x \right)}\) has at least one root.

We emphasize that the derivative may have more than one root (for example, you can consider the function \(f\left( x \right) = \sin x\) on the interval \(\left[ {0,2\pi} \right],\) where the derivative has two roots). Lagrange's mean value theorem allows to prove the existence of at least one root.

It is clear that this scheme can be generalized to the case of \(n\) roots and derivatives of the \(\left( {n - 1} \right)\)th order. If a function has three real roots, then the first derivative will have (at least) two roots. Respectively, the second derivative will have at least one root. In general, if a function has \(n\) real roots, the derivative of the \(\left( {n - 1} \right)\)th order will have 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 \(x = 10.\)

Solution.

To estimate the value of \({f\left( {10} \right)},\) we use the Lagrange formula, which is written as

\[f\left( {10} \right) - f\left( 2 \right) = f'\left( c \right)\left( {10 - 2} \right),\]

where \(c\) is a point lying in the interval \(\left( {2,10} \right).\)

We rewrite this formula in the form

\[f\left( {10} \right) = f\left( 2 \right) + 8f'\left( c \right).\]

The maximum possible value of the derivative on the given interval is \(f'\left( x \right) = 4.\) Hence,

\[f\left( {10} \right) \le f\left( 2 \right) + 8 \cdot 4 = 4 + 32 = 40.\]

Thus, the maximum possible value of the function at the right boundary of the interval is \(40.\)

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 \(x=6.\)

Solution.

We use the Mean Value Theorem in the form

\[f\left( b \right) - f\left( a \right) = f^\prime\left( c \right)\left( {b - a} \right),\]

where \(c\) is a point in the interval \(\left( { - 2,6} \right).\)

Substituting the known values, we write the formula as follows:

\[f\left( 6 \right) - f\left( { - 2} \right) = f^\prime\left( c \right)\left( {6 - \left( { - 2} \right)} \right)\]

or

\[f\left( 6 \right) = f\left( { - 2} \right) + 8f'\left( c \right).\]

The maximum value of the derivative on the given interval is \(f'\left( c \right) = 3.\) Consequently,

\[f\left( 6 \right) \le f\left( { - 2} \right) + 8 \cdot 3 = 4 + 24 = 28.\]

Thus, the upper bound of the function at the right endpoint of the interval is \({f_{\max }}\left( 6 \right) = 28.\)

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