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

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Question 1

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, then

Either , which means or , which means that .


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.


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


Question 2

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


Question 4

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

If , then .


Thus, , or .


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.


If , then .



Thus , or .


Complete the square:



Thus, .

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


, or


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




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

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, which is true since




, therefore .

, therefore .


, therefore .

If or are 0, then by definition .

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


If , then since .

If , then . Since , we have .

Question 6

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