# Solutions To Mathematics Textbooks/Principles of Mathematical Analysis (3rd edition) (ISBN 0070856133)/Chapter 5

< Solutions To Mathematics Textbooks | Principles of Mathematical Analysis (3rd edition) (ISBN 0070856133)## Contents

# Chapter 5Edit

## 6Edit

We will compute the derivative and show that it is always . The derivative of is given by the quotient rule to be The denominator is clearly always positive as we know it is the square of some real number , so in order to show that the entire derivative is always positive we need to show that the numerator is positive.

However, and we know that, by the mean value theorem there is some such that since . Since is monotonically increasing it must be that .

Thus, and so the derivative of is positive, which implies that is monotonically increasing.

## 11Edit

Since the denominator of goes to zero as we can employ L'Hopital's rule. Using this we get Since the denominator is again zero we apply L'Hopital's rule again

An example of a function that is not differentiable, but where this limit exists is The second derivative of this function does not exist at zero, as however we can compute the above limit at zero by

## 14Edit

A function is said to be convex if

First we shall prove () that monotonically increasing implies is convex.

Assume is monotonically increasing. Let for . The mean value theorem implies that such that . Similarly such that . Thus, since is assumed to be monotonically increasing we have Because we know that for some . Moreover, we have We note here that the direction of the inequality is preserved since we have and and (*) reducing yields but since the initial choice of was arbitrary (up to ordering) this holds for any . We must still address the case corresponding to . This is easy though.

For let . Then by what we have proven already now let so it follows from the above that

Thus if is monotonically increasing for ,

Now we prove () that convex implies is monotonically increasing.

The proof of this direction follows similarly to the previous. Let's assume is convex.

Then for (Note: we must have by (*)) and , . Following the previous reasoning in reverse we see that for .

So in the limit we have and in the limit we have thus, combining these statements we have so is monotonically increasing as desired.

Now it only remains to prove that (assuming exists for ) is convex if and only if for .

is convex is monotonically increasing by the previous proof. Moreover, is monotonically increasing (proved in class).

So we have shown that (assuming exists for ) is convex if and only if for as desired.