The mean value theorem (MVT), also known as Lagrange's mean value theorem (LMVT), provides a formal framework for a fairly intuitive statement relating change in a function to the behavior of its derivative. The theorem states that the derivative of a continuous and differentiable function must attain the function's average rate of change (in a given interval). For instance, if a car travels 100 miles in 2 hours, then it must have had the exact speed of 50 mph at some point in time.
Mean Value Theorem
Suppose that a function is
- continuous on the closed interval and
- differentiable on the open interval
Then, there is a number such that and
Simple-sounding as it is, the mean value theorem actually lies at the heart of the proof of the fundamental theorem of calculus, and is itself based ultimately on properties of the real numbers. There is a slight generalization known as Cauchy's mean value theorem; for a generalization to higher derivatives, see Taylor's theorem.
The statement seems reasonable upon inspection of an example or two. Below, is the slope of the tangent lines in the interval , and is the slope of the secant line joining the two endpoints and .
Note that the mean value theorem does not restrict to only one value, nor does it tell us where is (other than inside the interval).
Note also that is not required to be differentiable at the end points. Two examples suffice to illustrate it: on and on .
The mean value theorem is best understood by first studying the restricted case known as Rolle's theorem.
Suppose that a function is continuous on , differentiable on , and that . Then, there is a number such that and .
In other words, if a function has the same value at two points, then it must "level" somewhere between those points. By considering whether the function is increasing or decreasing immediately after the first point, it becomes clear that neither option can continue indefinitely if the function is to return to the same value; therefore, there must be a local maximum or minimum before the next point occurs.
Rolle's theorem quickly turns into the mean value theorem by simply skewing the graph of the function.
Let all be as in the theorem statement above.
Define a new function as the difference between and the line passing through points and . This line has equation
Therefore, the function has equation
Next, Rolle's theorem is useful. The function satisfies the conditions of the theorem:
- The function is continuous on because it is the sum of and a first-degree polynomial, both of which are continuous.
The function is differentiable on because both and the first-degree polynomial are differentiable. In fact, we can compute it directly:
By Rolle's theorem, there exists a value in such that . Therefore,
Determine all the numbers that satisfy the conclusion of the mean value theorem for on the interval .
Observe that is a polynomial which is continuous and differentiable over any interval, so the mean value theorem applies. Indeed, we can compute that , and the average rate of change is
We want to find the values of such that
Note that is one of the endpoints, so the mean value theorem actually guarantees another in the interior, and indeed we have .
A car starts from rest and drives a distance of in . Use the mean value theorem to show that the car attains a speed of at some point(s) during the interval.
The mean value theorem says that the average speed of the car (the slope of the secant line) is equal to the instantaneous speed (slope of the tangent line) at some point(s) in the interval.
The average velocity is
From the mean value theorem, we can find in the interval such that
Suppose that is a differentiable function for all . If for all and , what is the maximum value of
Since is differentiable on all intervals, we can choose any two points. So from the mean value theorem, we have
So the maximum possible value of is .
Given that is an arbitrary quadratic polynomial:
show that the point whose existence is guaranteed by the mean value theorem is the mid-point of the interval .
From the mean value theorem, we have
Does there exist a function such that , and for all
If such a function exists, then from the mean value theorem there is a number such that and
But this is impossible because of the assumption . Therefore, such a function does not exist.
The name of the mean value theorem may require a little explanation. For a function on an interval , not necessarily continuous, one may define its mean value, or the "average", by the formula
provided, of course, that the (Riemann) integral exists. The reason is simple: the integral gives the area under the curve , and dividing by the length of the interval yields the "average" height of the function between and . In picture, we must have parts of the curve go above the mean value, and parts go below. If is continuous, then it must cross the mean value at some point. That is the "original" mean value theorem.
Mean Value Theorem for Integrals
If is continuous on , then there exists a point between and such that
What does this have to do with the (actual) mean value theorem, other than the semblance of the indeterminate ? Well, if is an antiderivative of ), the fundamental theorem of calculus (Newton-Leibniz formula) makes the left-hand side equal to , so that we have
This is precisely the mean value theorem, but for .
However, that does not count as a real proof of the mean value theorem. First of all, the fundamental theorem of calculus actually relies on the mean value theorem in its proof. Secondly, here is indeed continuous on and differentiable on , but its derivative furthermore is required to be continuous for the Riemann integral above to exist. One can construct (though somewhat ad hoc) functions that are differentiable, but its derivative fails to be continuous at one point, in a way that makes it not (Riemann) integrable.
We can also use the mean value theorem to prove certain inequalities.
Use the mean value theorem to prove that for
Suppose that our function was . Note that is just a dummy variable as we will be using in our interval of choice. Now, given that is defined, continuous, and differentiable over the interval , by the mean value theorem, we see that
By observation, we know that so it follows that . Thus, .
This tells us that for