# 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

### Classification of Extrema

We start out with some definitions.

Global Maximum
A global maximum (also called an absolute maximum) of a function $f$  on a closed interval $[a,b]$  is a value $f(c)$  such that $f(c)\geq f(x)$  for all $x\in [a,b]$  .
Global Minimum
A global minimum (also called an absolute minimum) of a function $f$  on a closed interval $[a,b]$  is a value $f(c)$  such that $f(c)\leq f(x)$  for all $x\in [a,b]$  .

Maxima and minima are collectively known as extrema.

### The Extreme Value Theorem

Extreme Value Theorem
If $f$  is a function that is continuous on the closed interval $[a,b]$  , then $f$  has both a global minimum and a global maximum on $[a,b]$  . It is assumed that $a,b$  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 $f(x)=x$  has no extrema whatsoever. If $f(x)$  is restricted to the semi-closed interval $[0,1)$  , then $f$  has a minimum value of 0 at $x=0$  , but it has no maximum value since, for any given value $c\in [0,1)$  , one can always find a larger value of $f(x)$  for $x\in [0,1)$  , for example by forming $f(d)$  , where $d$  is the average of $c$  with 1. The function $g(x)={\frac {1}{x}}$  has a discontinuity at $x=0$  . $g(x)$  fails to have any extrema in any closed interval around $x=0$  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 $x=0$  . However, the example is unaffected if $g(0)$  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

Rolle's Theorem

If a function $f(x)$  is

• continuous on the closed interval $[a,b]$
• differentiable on the open interval $(a,b)$
• $f(a)=f(b)$

then there exists at least one number $c\in (a,b)$  such that

$f'(c)=0$

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

If $f$  is constant on $[a,b]$  , then $f'(x)=0$  for every $x\in [a,b]$  , so the theorem is true. So for the remainder of the discussion we assume $f$  is not constant on $[a,b]$  .

Since $f$  satisfies the conditions of the Extreme Value Theorem, $f$  must attain its maximum and minimum values on $[a,b]$  . Since $f$  is not constant on $[a,b]$  , the endpoints cannot be both maxima and minima. Thus, at least one extremum exists in $(a,b)$  . We can suppose without loss of generality that this extremum is a maximum because, if it were a minimum, we could consider the function $-f$  instead. Let $f(c)$  with $c\in (a,b)$  be a maximum. It remains to be shown that $f'(c)=0$  .

By the definition of derivative, $f'(c)=\lim _{h\to 0}{\frac {f(c+h)-f(c)}{h}}$  . By substituting $h=x-c$  , this is equivalent to $\lim _{x\to c}{\frac {f(x)-f(c)}{x-c}}$  . Note that $f(x)-f(c)\leq 0$  for all $x\in [a,b]$  since $f(c)$  is the maximum on $[a,b]$  .

$\lim _{x\to c^{-}}{\frac {f(x)-f(c)}{x-c}}\geq 0$  since it has non-positive numerator and negative denominator.

$\lim _{x\to c^{+}}{\frac {f(x)-f(c)}{x-c}}\leq 0$  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 $c$  , so $\lim _{x\to c}{\frac {f(x)-f(c)}{x-c}}=0=f'(c)$  .

### Exercise

1. Show that Rolle's Theorem holds true between the x-intercepts of the function $f(x)=x^{2}-3x$  .

1: The question wishes for us to use the $x$ -intercepts as the endpoints of our interval.

Factor the expression to obtain $x(x-3)=0$  . $x=0,x=3$  are our two endpoints. We know that $f(0)$  and $f(3)$  are the same, thus that satisfies the first part of Rolle's theorem ($f(a)=f(b)$ ).

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

${\frac {dy}{dx}}=2x-3$

Thus, at $x=1.5$  , we have a spot with a slope of 0. We know that ${\frac {3}{2}}$  (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

Mean Value Theorem

If a function $f(x)$  is

• continuous on the closed interval $[a,b]$
• differentiable on the open interval $(a,b)$

then there exists at least one $c\in (a,b)$  such that

$f'(c)={\frac {f(b)-f(a)}{b-a}}$

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 $f(a)=f(b)$  .

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

Cauchy's Mean Value Theorem

If $f(x),g(x)$  are

• continuous on the closed interval $[a,b]$
• differentiable on the open interval $(a,b)$

then there exists a number $c\in (a,b)$  such that

$f'(c)(g(b)-g(a))=g'(c)(f(b)-f(a))$

If $g(b)\neq g(a)$  and $g'(c)\neq 0$  , then this is equivalent to

${\frac {f'(c)}{g'(c)}}={\frac {f(b)-f(a)}{g(b)-g(a)}}$

To prove Cauchy's Mean Value Theorem, consider the function

$h(x)=f(x)(g(b)-g(a))-g(x)(f(b)-f(a))-f(a)g(b)+f(b)g(a)$

Since both $f$  and $g$  are continuous on $[a,b]$  and differentiable on $(a,b)$  , so is $h$  .

$h'(x)=f'(x)(g(b)-g(a))-g'(x)(f(b)-f(a))$

Since $h(a)=h(b)$  (see the exercises), Rolle's Theorem tells us that there exists some number $c\in (a,b)$  such that $h'(c)=0$  . This implies that

$f'(c)(g(b)-g(a))=g'(c)(f(b)-f(a))$

which is what was to be shown.

### Exercises

2. Show that $h(a)=h(b)$  , where $h(x)$  is the function that was defined in the proof of Cauchy's Mean Value Theorem.

{\begin{aligned}h(a)&=f(a)(g(b)-g(a))-g(a)(f(b)-f(a)-f(a)g(b)+f(b)g(a)\\&=f(a)g(b)-f(a)g(a)-g(a)f(b)-g(a)f(a)-f(a)g(b)+f(b)g(a)\\&=0\end{aligned}}  {\begin{aligned}h(b)&=f(b)(g(b)-g(a))-g(b)(f(b)-f(a))-f(a)g(b)+f(b)g(a)\\&=f(b)g(b)-f(b)g(a)-g(b)f(b)-g(b)f(a)-f(a)g(b)+f(b)g(a)\\&=0\end{aligned}}

3. Show that the Mean Value Theorem follows from Cauchy's Mean Value Theorem.

Let $g(x)=x$  . Then $g'(x)=1$  and $g(b)-g(a)=b-a$  , which is non-zero if $b\neq a$  . Then
${\frac {f'(c)}{g'(c)}}={\frac {f(b)-f(a)}{g(b)-g(a)}}$  simplifies to $f'(c)={\frac {f(b)-f(a)}{b-a}}$  , which is the Mean Value Theorem.

4. Find the $x=c$  that satisfies the Mean Value Theorem for the function $f(x)=x^{3}$  with endpoints $x=0$  and $x=2$ .

$x={\frac {2{\sqrt {3}}}{3}}$

5. Find the point that satisifies the mean value theorem on the function $f(x)=\sin(x)$  and the interval $[0,\pi ]$  .

$x={\frac {\pi }{2}}$

## Differentiability Implies Continuity

If $f'(x_{0})$  exists then $f$  is continuous at $x_{0}$  . To see this, note that $\lim _{x\to x_{0}}(x-x_{0})f'(x_{0})=0$  . But

{\begin{aligned}\lim _{x\to x_{0}}(x-x_{0})f'(x_{0})&=\lim _{x\to x_{0}}(x-x_{0}){\frac {f(x)-f(x_{0})}{x-x_{0}}}\\&=\lim _{x\to x_{0}}(f(x)-f(x_{0}))\\&=\lim _{x\to x_{0}}f(x)-f(x_{0})\end{aligned}}

This imples that $\lim _{x\to x_{0}}f(x)-f(x_{0})=0$  or $\lim _{x\to x_{0}}f(x)=f(x_{0})$  , which shows that $f$  is continuous at $x=x_{0}$  .

The converse, however, is not true. Take $f(x)=|x|$  , for example. $f$  is continuous at 0 since $\lim _{x\to 0^{-}}|x|=\lim _{x\to 0^{-}}-x=0$  and $\lim _{x\to 0^{+}}|x|=\lim _{x\to 0^{+}}x=0$  and $|0|=0$  , but it is not differentiable at 0 since $\lim _{h\to 0^{-}}{\frac {|0+h|-|0|}{h}}=\lim _{h\to 0^{-}}{\frac {-h}{h}}=-1$  but $\lim _{h\to 0^{+}}{\frac {|0+h|-|0|}{h}}=\lim _{h\to 0^{+}}{\frac {h}{h}}=1$  .

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