Lagrange’s Mean Value Theorem

Solved Problems

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:

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

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

Solution.

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

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

Substituting the derivative

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

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

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

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

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

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

Substitute the known boundary values:

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

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

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

The derivative is given by

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

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

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:

Find the derivative:

Solve the equation given by the MVT:

Using the quadratic formula, we get

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

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:

Next, we find the derivative by the quotient rule:

Write the formula given by the MVT:

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

Answer: \(c=0.\)

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:

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

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

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

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 \(5\)).

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:

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

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.

Solution.

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

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

We rewrite this formula in the form

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

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

Solution.

We use the Mean Value Theorem in the form

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

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

or

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

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