Calculus/Some Important Theorems

← Derivatives of Exponential and Logarithm Functions Calculus Differentiation/Basics of Differentiation/Exercises →
Some Important Theorems

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 edit

Classification of Extrema edit

We start out with some definitions.

Global Maximum

A global maximum (also called an absolute maximum) of a function   on a closed interval   is a value   such that   for all   .

Global Minimum

A global minimum (also called an absolute minimum) of a function   on a closed interval   is a value   such that   for all   .

Maxima and minima are collectively known as extrema.

The Extreme Value Theorem edit

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   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 0 at   , but it has no maximum value since, for any given value   , one can always find a larger value of   for   , for example by forming   , where   is the average of   with 1. 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 0 from the left, and it is unbounded above as one approaches 0 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.

Rolle's Theorem edit

 
Rolle's Theorem

If a function   is

  • continuous on the closed interval  
  • differentiable on the open interval  
  •  

then there exists at least one number   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.

Proof of Rolle's Theorem edit

If   is constant on   , then   for every   , 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   be a maximum. It remains to be shown that   .

By the definition of derivative,   . By substituting   , this is equivalent to   . Note that   for all   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   .

Exercise edit

1. Show that Rolle's Theorem holds true between the x-intercepts of the function   .
1: The question wishes for us to use the  -intercepts as the endpoints of our interval.
Factor the expression to obtain   .   are our two endpoints. We know that   and   are the same, thus that satisfies the first part of Rolle's theorem ( ).

2: Now by Rolle's Theorem, we know that somewhere between these points, the slope will be 0. Where? Easy: Take the derivative.

 
Thus, at   , we have a spot with a slope of 0. We know that   (or 1.5) is between 0 and 3. Thus, Rolle's Theorem is true for this (as it is for all cases).
1: The question wishes for us to use the  -intercepts as the endpoints of our interval.
Factor the expression to obtain   .   are our two endpoints. We know that   and   are the same, thus that satisfies the first part of Rolle's theorem ( ).

2: Now by Rolle's Theorem, we know that somewhere between these points, the slope will be 0. Where? Easy: Take the derivative.

 
Thus, at   , we have a spot with a slope of 0. We know that   (or 1.5) is between 0 and 3. Thus, Rolle's Theorem is true for this (as it is for all cases).

Mean Value Theorem edit

 
Mean Value Theorem

If a function   is

  • continuous on the closed interval  
  • differentiable on the open interval  

then there exists at least one   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.

Cauchy's Mean Value Theorem edit

Cauchy's Mean Value Theorem

If   are

  • continuous on the closed interval  
  • differentiable on the open interval  

then there exists a number   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   such that   . This implies that

 

which is what was to be shown.

Exercises edit

2. Show that   , where   is the function that was defined in the proof of Cauchy's Mean Value Theorem.
   
   
3. Show that the Mean Value Theorem follows from Cauchy's Mean Value Theorem.
Let   . Then   and   , which is non-zero if   . Then
  simplifies to   , which is the Mean Value Theorem.
Let   . Then   and   , which is non-zero if   . Then
  simplifies to   , which is the Mean Value Theorem.
4. Find the   that satisfies the Mean Value Theorem for the function   with endpoints   and  .
 
 
5. Find the point that satisifies the mean value theorem on the function   and the interval   .
 
 

Solutions

Differentiability Implies Continuity edit

If   exists then   is continuous at   . To see this, note that   . But

 

This implies 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   .

← Derivatives of Exponential and Logarithm Functions Calculus Differentiation/Basics of Differentiation/Exercises →
Some Important Theorems