The value of is:
We apply the Mean Value Theorem.
Rolle’s Theorem
We begin with a special case of the Mean Value Theorem known as Rolle’s Theorem. This theorem uses the Extreme Value Theorem to guarantee a critical number of a differentiable function under certain circumstances.
Rolle’s Theorem Suppose the function is continuous on the closed interval and
differentiable on the open interval . If , then there exists a number between and
such that
critical number on the interval . At this value of our differentiable function, the derivative is zero.
(and hence a critical number)
Show that the function has a critical number in the interval .
The function is differentiable (and hence also continuous) on the interval and since we can apply Rolle’s Theorem to on the interval . The theorem says that there exists a number between and such that . That is, has a critical number in the interval . Moreover, we can find since and the equation has as a solution in the interval.
The function is differentiable (and hence also continuous) on the interval and since we can apply Rolle’s Theorem to on the interval . The theorem says that there exists a number between and such that . That is, has a critical number in the interval . Moreover, we can find since and the equation has as a solution in the interval.
Use Rolle’s Theorem to show that a cubic polynomial can have at most 3
roots.
Recall that a root of a polynomial, , is a value , such that . 
Recall that a root of a polynomial, , is a value , such that .
Also recall that a quadratic polynomial has at most two roots, since if then Let be a cubic polynomial, i.e., We will argue by contradiction to demonstrate that can have at most 3 roots.
Suppose that a cubic polynomial, , can have 4 roots. From smallest to largest, we label the 4 roots and . Then, by definition of root, We can now apply Rolle’s Theorem to on and to conclude that has a root on each of these intervals.
That is, there exist numbers and such that where: and .
This would mean that has 3 roots, which is not possible: is a quadratic polynomial, and so has at most 2 roots.
This contradiction implies that a cubic polynomial cannot have 4 roots. Thus, a cubic polynomial can have at most 3 roots.
In fact, a similar argument can be used to show that a polynomial of degree can have at most roots for any whole number, .
The Mean Value Theorem
The Mean Value Theorem is one of the most far-reaching theorems in calculus. It
states that for a continuous and differentiable function, the average rate of
change over an interval is attained as an instantaneous rate of change at
some point inside the interval. The precise mathematical statement is as
follows.
Mean Value Theorem Suppose that the function is continuous on the closed interval
and differentiable on the open interval . Then there exists a number between and
such that
Geometrically, the left-hand side of the conclusion of the MVT represents the slope of the tangent line to at Meanwhile, the right-hand side represents the slope of the secant line connecting the points and . Since their slopes are equal, these two lines are parallel. Conceptually, the left-hand side represents the instantaneous rate of change of at while the right-hand side represents the average rate of change of over the interval from to . Thus, the MVT says that at some point, the instantaneous rate of change will be equal to the average rate of change.
As an example of the conceptual interpretation of the theorem, consider a car that averages a speed of 57.4 miles per hour on a long trip. By the MVT the car must have been traveling at exactly 57.4 miles per hour at some instant during the trip.
The MVT is considered an existence theorem because it asserts that there exists at least one value inside the interval that satisfies the equation In our examples, we will determine the exact value of .
example 1 Verify that the function satisfies the hypotheses of the Mean Value
Theorem on the interval . Then find all values that satisfy the conclusion of the
theorem.
Since is a polynomial, it is continuous on the closed interval and differentiable on the open interval . Thus, satisfies the hypotheses of the MVT on . We now know that the equation has at least one solution in the interval . To find the solution(s), we first compute the value of the right-hand side using and : Since, , the conclusion of the MVT guarantees that the equation has at least one solution in the open interval . This is easily verified, since if , then which is in the open interval . Furthermore, note that is actually the mid-point of the interval . When applying the MVT to a quadratic polynomial on any interval the point will always be the mid-point! 
Since is a polynomial, it is continuous on the closed interval and differentiable on the open interval . Thus, satisfies the hypotheses of the MVT on . We now know that the equation has at least one solution in the interval . To find the solution(s), we first compute the value of the right-hand side using and : Since, , the conclusion of the MVT guarantees that the equation has at least one solution in the open interval . This is easily verified, since if , then which is in the open interval . Furthermore, note that is actually the mid-point of the interval . When applying the MVT to a quadratic polynomial on any interval the point will always be the mid-point!
In the above figure, the blue line in the secant line for on the interval , and the red line is the tangent line at . The Mean Value Theorem asserts that these lines are parallel.
(problem 1a) Given that the function satisfies the hypotheses of the MVT on the
interval , find the value of in the open interval which satisfies the conclusion of the
theorem.
(problem 1b) Given that the function satisfies the hypotheses of the MVT on the
interval , find the value of in the open interval which satisfies the conclusion of the
theorem.
The value of is:
(problem 1c) Given that the function satisfies the hypotheses of the MVT on the
interval , find the value of in the open interval which satisfies the conclusion of the
theorem.
The value of is:
(problem 1d) Given that the function satisfies the hypotheses of the MVT on the
interval , find the values of in the open interval which satisfy the conclusion of the
theorem.
The values of in ascending order are: and
example 2 We will verify that the function satisfies the hypotheses of the MVT on
the interval and we will find the special value of that satisfies the Mean Value
Equation. The function is continuous on the interval and differentiable on the
interval . Hence, it is continuous on the closed interval and differentiable on the
open interval . Then, by the MVT, the Mean Value Equation holds for some
number in the interval . To find , we first compute Next we compute Finally
we solve the equation which gives and so Note that the value is in the
interval and so the special value, , guaranteed to exist by the MVT, is in this
example.
In the above figure, the blue line in the secant line for on the interval , and the red line is the tangent line at . The Mean Value Equation asserts that these lines are parallel, and this is clear in the figure.
(problem 2a) Given that the function satisfies the hypotheses of the MVT on the
interval , find the value of in the open interval which satisfies the conclusion of the
theorem.
The value of is:
(problem 2b) Given that the function satisfies the hypotheses of the MVT on the
interval , find the value of in the open interval which satisfies the conclusion of the
theorem.
The value of is:
(problem 2c) Given that the function satisfies the hypotheses of the MVT on the
interval , find the value of in the open interval which satisfies the conclusion of the
theorem.
The value of is:
example 3 We will verify that the function satisfies the hypotheses of the MVT on
the interval and we will find the special value of that satisfies the Mean Value
Equation. The function is continuous and differentiable on the interval . Hence, it is
continuous on the closed interval and differentiable on the open interval . Then, by
the MVT, the Mean Value Equation holds for some number in the interval . To find ,
we first compute Next we compute Finally we solve the equation which gives
Note that the value is in the interval and so the special value, , guaranteed to exist by the MVT, is in this example.
In the above figure, the blue line in the secant line for on the interval , and the red line is the tangent line at . The Mean Value Equation asserts that these lines are parallel, and this is clear in the figure.
(problem 3) Given that the function satisfies the hypotheses of the MVT on the
interval , find the value of in the open interval which satisfies the conclusion of the
theorem.
The value of is:
example 4 We will verify that the function satisfies the hypotheses of the MVT on
the interval and we will find the special value of that satisfies the Mean Value
Equation. The function is continuous and differentiable on the interval . Hence, it is
continuous on the closed interval and differentiable on the open interval . Then, by
the MVT, the Mean Value Equation holds for some number in the interval . To find ,
we first compute Next we compute Finally we solve the equation which gives
Note that the value is in the interval and so the special value, , guaranteed to exist by the MVT, is in this example.
In the above figure, the blue line in the secant line for on the interval , and the red line is the tangent line at . The Mean Value Equation asserts that these lines are parallel, and this is clear in the figure.
(problem 4a) Given that the function satisfies the hypotheses of the MVT on the
interval , find the value of in the open interval which satisfies the conclusion of the
theorem.
The value of is:
(problem 4b) Given that the function satisfies the hypotheses of the MVT on the
interval , find the value of in the open interval which satisfies the conclusion of the
theorem.
The value of is:
Here is a detailed, lecture style video on the Mean Value Theorem:
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