This section covers three theorems of fundamental importance to the topic of differential calculus: The Extreme Value Theorem, Rolle's Theorem, and the Mean Value Theorem. It also discusses the relationship between differentiability and continuity.
Extreme Value Theorem
If is a function that is continuous on the closed interval , then has both a global minimum and a global maximum on . It is assumed that a and b are both finite.
The Extreme Value Theorem is a fundamental result of real analysis whose proof is beyond the scope of this text. However, the truth of the theorem allows us to talk about the maxima and minima of continuous functions on closed intervals without concerning ourselves with whether or not they exist. When dealing with functions that do not satisfy the premises of the theorem, we will need to worry about such things. For example, the unbounded function has no extrema whatsoever. If is restricted to the semi-closed interval , then has a minimum value of at , but it has no maximum value since, for any given value in , one can always find a larger value of for in , for example by forming , where is the average of with . The function has a discontinuity at . fails to have any extrema in any closed interval around since the function is unbounded below as one approaches from the left, and it is unbounded above as one approaches from the right. (In fact, the function is undefined for . However, the example is unaffected if is assigned any arbitrary value.)
The Extreme Value Theorem is an existence theorem. It tells us that global extrema exist if certain conditions are met, but it doesn't tell us how to find them. We will discuss how to determine the extrema of continuous functions in the section titled Extrema and Points of Inflection.
If a function, , is continuous on the closed interval , is differentiable on the open interval , and , then there exists at least one number c, in the interval such that .
Rolle's Theorem is important in proving the Mean Value Theorem. Intuitively it says that if you have a function that is continuous everywhere in an interval bounded by points where the function has the same value, and if the function is differentiable everywhere in the interval (except maybe at the endpoints themselves), then the function must have zero slope in at least one place in the interior of the interval.
If is constant on , then for every in , so the theorem is true. So for the remainder of the discussion we assume is not constant on .
Since satisfies the conditions of the Extreme Value Theorem, must attain its maximum and minimum values on . Since is not constant on , the endpoints cannot be both maxima and minima. Thus, at least one extremum exists in . We can suppose without loss of generality that this extremum is a maximum because, if it were a minimum, we could consider the function instead. Let with in be a maximum. It remains to be shown that .
By the definition of derivative, . By substituting , this is equivalent to . Note that for all in since is the maximum on .
since it has non-positive numerator and negative denominator.
since it has non-positive numerator and positive denominator.
The limits from the left and right must be equal since the function is differentiable at , so .
If is continuous on the closed interval , where ,and differentiable on the open interval , there exists at least one in the open interval such that
The Mean Value Theorem is an important theorem of differential calculus. It basically says that for a differentiable function defined on an interval, there is some point on the interval whose instantaneous slope is equal to the average slope of the interval. Note that Rolle's Theorem is the special case of the Mean Value Theorem when .
In order to prove the Mean Value Theorem, we will prove a more general statement, of which the Mean Value Theorem is a special case. The statement is Cauchy's Mean Value Theorem, also known as the Extended Mean Value Theorem.
If , are continuous on the closed interval and differentiable on the open interval , then there exists a number, , in the open interval such that
If and , then this is equivalent to
To prove Cauchy's Mean Value Theorem, consider the function . Since both and are continuous on and differentiable on , so is . . Since (see the exercises), Rolle's Theorem tells us that there exists some number in such that . This implies that , which is what was to be shown.
If exists then is continuous at . To see this, note that . But
This imples that or , which shows that is continuous at .
The converse, however, is not true. Take , for example. is continuous at 0 since and and , but it is not differentiable at 0 since but .