# Solutions To Mathematics Textbooks/Calculus (3rd) (0521867444)/Chapter 1

## Contents

# Question 1[edit]

## i[edit]

## ii[edit]

## iii[edit]

, then

Either , which means or , which means that .

## iv[edit]

## v[edit]

Note that the two middle terms do no appear to match, but if you follow the logical development in each term, you will see that there is a matching additive inverse for each term on both sides.

## vi[edit]

Use the same method as **iv**, expand the expression and cancel.

# Question 2[edit]

implies that and thus . Step 4 requires division by and thus is an invalid step.

# Question 4[edit]

## ii[edit]

All values of x satisfy the inequality, since it can be rewritten as , and for all x.

## iii[edit]

If , then .

Thus, , or .

## v[edit]

cannot be factored in its current form, so we first turn a part of the expression into a perfect square:

We then complete the square on , so now we have the expression:

- , which is positive for all values of x.

## vi[edit]

If , then .

Thus , or .

## viii[edit]

Complete the square:

Thus, .

## ix[edit]

Solve for first, then consider the third factor for both cases, giving us , or

## x[edit]

, or

## xi[edit]

If , then taking the base 2 logarithm on both sides:

## xii[edit]

## xiii[edit]

NOTE: The answer in the 3rd Edition provides or . Plugging in values , e.g. 10, gives us , so I think this is a misprint or an incorrect answer.

If , then , which is only positive when since .

# Question 5[edit]

## i[edit]

, which is true since

## ii[edit]

## iv[edit]

, therefore .

, therefore .

## v[edit]

, therefore .

## viii[edit]

If or are 0, then by definition .

Otherwise, we have already proved that for , therefore if , is true.

## ix[edit]

If , then since .

If , then . Since , we have .