# MATH 409 Lecture 16

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## Mean Value Theorem

### Points of Local Extremum

Let be a function defined on .

We say that attains a **local maximum** at a point if there exists such that for all . Note local because we restrict the function to the -neighborhood of .

Similarly, attains a **local minimum** at a point if there exists such that for all .

### Fermat's Theorem

not little or last...

**Theorem. [Fermat.]** If a function is differentiable at a point of local extremum (maximum or minimum), then .

*Proof.* Assume without loss of generality that is a point of local minimum. Since is defined on an open interval containing , there exists such that for all , .

- In the case , this implies
- In the case , this implies

By antisymmetry (in particular, and ) we conclude that .

### Rolle's Theorem

**Theorem.** Suppose that with . If a function is continuous on the interval , differentiable on , and if , then for some .

*Proof.* By the Extreme Value Theorem, the function attains its (absolute) maximum and minimum on . We consider two cases:

- In the case , at least one of the extrema is attained at a point in . Then by Fermat's theorem.
- In the case , the function is constant on . Then for all .

**Corollary.** If a polynomial has distinct real roots, then the polynomial has at least distinct real roots.

*Proof.* Let be distinct real roots of ordered so that . By Rolle's Theorem, the derivative has a root in each of intervals , , ..., .

### Intermediate Value Theorem for Derivatives

**Theorem.** Suppose that a function is differentiable on an interval with . If is a real number between and , then for some .

*Proof.* First consider the case when , , and . Since is differentiable on , it must be continuous on . By the extreme value theorem, attains its absolute minimum on at some point . Since , we have for sufficiently small . Hence . Similarly, implies that for sufficiently small . Hence . We obtain that . Then due to Fermat's theorem.

Now consider , , and . Then the function is differentiable on with and . By the previous case, for some . Then .

In the general case when , consider a function . It is differentiable on and for all . It follows that lies between and . By the above, for some . Then .

**Note:**We had a similar theorem about continuous functions. Now although is continuous on , its derviative need not be continuous.

**Corollary.** If a function is differentiable on an open interval , then the derivative has no jump discontinuities in .

If a derivative has a jump discontinuity, then there are many values for which there can be no such that . This means that not every function can be a derivative.

### Mean Value Theorem

**Theorem.** If a function is continuous on and differentiable on , then there is a such that .

In other words, there exists a point between two endpoints and such that the derivative (i.e. slope of the tangent line) is equivalent to the slope of the chord line through and .

*Proof.* Let for .

We observe that the function is differentiable. Moreover, for all . By construction, and .

It follows that the function is continuous on , differentiable on , and satisfies . By Rolle's Theorem, for some . We have , thus or equivalently, .

#### Monotone Functions (Revisited)

**Theorem.** Suppose that a function is continuous on an interval and differentiable on .

- is increasing on if and only if on .
- is decreasing on if and only if on .
- is strictly increasing on
*if*(not biconditional) . - is strictly decreasing on
*if*(not biconditional) . - is constant on if and only if on .

*Proof.* Let . By the mean value theorem, for some . Obviously if and only if . Likewise, if and only if . This proves statements 3 and 4, and the (⇒) part of statements 1 and 2. The (⇐) parts of 1 and 2 follows from the Comparison Theorem (by taking ) Finally, statement 5 follows from antisymmetry between 1 and 2.

#### Generalized Mean Value Theorem

**Theorem.** If and are continuous on and differentiable on , then there exists a such that

If either or is the identity function, we get exactly the mean value theorem.

*Proof.* For any , let . This is simply a linear combination of and . Observe that the function is continuous on and differentiable on . Further, and . Hence . By Rolle's Theorem, for some . It remains to notice that .

Therefore

## Taylor's Formula

**Theorem.** Let and be an open interval. If a function is times differentiable on , then for each pair of points , there is a point between and such that

*Proof.* Let us fix and define

The function is infinitely differentiable on . The function is defined and differentiable on . By the Generalized Mean Value Theorem, for every , , there exists a point between and such that . Note that the latter follows both in the case and in the case (both sides of the equation get negated, so it remains equivalent). Clearly . Further

Summing up over from to , we obtain that . Finally, so that for . It follows that , which implies Taylor's formula.

**Note:**The function is a polynomial of degree at most . It is called the

**Taylor polynomial**of order generated by centered at . One can check that and for .