We apply the Mean Value Theorem.
1 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 the existence of a critical number for a differentiable function on and interval in which the function is equal at the endpoints.
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 states that there exists a number between and such that . That is, has a critical number in the interval . Moreover, we can find by solving the equation This equation has infinitely many solutions: where is any integer. The solution (corresponding to ) is in the interval .
The critical number is
For the next example we need to recall that a root of a polynomial, , is a value , such that . We also need the fact that a quadratic polynomial has at most two roots, since if then by the quadratic formula
Let be a cubic polynomial, i.e., We will argue by contradiction to demonstrate that can have at most three roots. Consider a cubic polynomial with four roots. Label these roots in ascending order: . By definition of root, we have We apply Rolle’s Theorem to on each of the three intervals and to conclude that there are values in these respective intervals such that The above equations are telling us that has three distinct roots. However, since the degree of is , the Power Rule tells us that the degree of must be . The quadratic formula tells us that a quadratic polynomial can have at most two roots. We see that the assumption that the cubic polynomial has four roots has led to the absurdity that the quadratic polynomial has three roots. This contradiction implies that a cubic polynomial having four roots cannot exist. From this we conclude that a cubic polynomial can have at most three roots.
See the figure below which shows polynomial with four roots and the corresponding values, and given by Rolle’s Theorem.
2 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.
The Mean Value Theorem has both geometric and conceptual interpretations. Geometrically, the equation asserts that the tangent line at and the secant line on are parallel, as shown in the figure below. Conceptually, the equation tells us that at the point , the instantaneous rate of change of is equal to the average rate of change of over the interval .
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 . As an example of the conceptual interpretation of the theorem, consider a car that averages a speed of, say, 47.4 miles per hour on a long trip. The Mean Value Theorem tells us that the car must have been traveling at exactly 47.4 miles per hour at some time during the trip.
The Mean Value Theorem is an existence theorem because it asserts that there exists at least one value inside the interval that satisfies the equation In the examples and problems below, we will determine this special value.
Since is a polynomial, it is continuous and differentiable everywhere. Therefore, is continuous on the closed interval and differentiable on the open interval . Thus, satisfies the hypotheses of the MVT on . The conclusion of the theorem states 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 : Next, we compute the derivative, , and set it equal to this value: 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 . Furthermore, is the mid-point of the interval . When applying the MVT to a quadratic polynomial on any interval the value of that satisfies the theorem will always be the mid-point!
In the figure below, the secant line is in blue, and the tangent line at is in red. The Mean Value Theorem asserts that these lines are parallel, and the figure makes this clear.
The value of is: What do you notice about this value of in relation to the given interval? Is this always the case?
The value of is: What do you notice about this value of in relation to the given interval? Is this always the case?
The value of is: What do you notice about this value of in relation to the given interval? Is this always the case?
The values of in ascending order are: and are these values in the interval ?
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 . To find the values of which satisfy the conclusion of the theorem, we first compute Next, we compute Finally, we solve the equation Take the reciprocal of both sides: Divide both sides by : Squaring both sides yields the unique solution Note that the value is in the interval .
In the figure below, the secant line is in blue, and the tangent line at is in red. The Mean Value Theorem asserts that these lines are parallel, and the figure makes this clear.
The value of is: Is this value in the interval ?
The value of is: Is this value in the interval ?
Given that the function is continuous on the closed interval and differentiable on the open interval , find the values of which satisfy the conclusion of the Mean Value Theorem for on the interval .
The value of is: Is this value in the interval ?
The function is continuous and differentiable on the interval . Hence, is continuous on the closed interval and differentiable on the open interval , and the hypotheses of the theorem are satisfied. To find the values of that satisfy the conclusion of the Mean Value Theorem, we compute and, we compute We then set these equal and solve the resulting equation for : This gives the unique solution
Note that the value is in the interval .
In the figure below, the secant line is in blue, and the tangent line at is in red. The Mean Value Theorem asserts that these lines are parallel, and the figure makes this clear.
The value of is: Is this value in the interval ?
The function is continuous and differentiable on the interval . Hence, it is continuous on the closed interval and differentiable on the open interval and the hypotheses of the theorem are satisfied. To find the values of that satisfy the conclusion of the Mean Value Theorem, we compute Next, we compute Finally, we solve the equation Taking the reciprocal of both sides yields the unique solution
Note that the value is in the interval .
In the figure below, the secant line is in blue, and the tangent line at is in red. The Mean Value Theorem asserts that these lines are parallel, and the figure makes this clear.
The value of is: Is this value in the interval ?
The values of are (in ascending order): , and Are these values in the interval ?
Is the function continuous on the closed interval and differentiable on the open interval ? Why or why not? Determine if there are any values of that satisfy the conclusion of the Mean Value Theorem for on the interval .
Were there any values of ? Does this contradict the Mean Value Theorem?