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

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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 .

Question 6[edit]