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

Question 1 i ii iii , then Either , which means or , which means that . iv v 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 Use the same method as iv, expand the expression and cancel. Question 2 implies that and thus . Step 4 requires division by and thus is an invalid step. Question 4 ii All values of x satisfy the inequality, since it can be rewritten as , and for all x. iii If , then . Thus, , or . v 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 If , then . Thus , or . viii Complete the square: Thus, . ix Solve for first, then consider the third factor for both cases, giving us , or x , or xi If , then taking the base 2 logarithm on both sides: xii xiii 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 i , which is true since ii iv , therefore . , therefore . vi , therefore . viii If or are 0, then by definition . Otherwise, we have already proved that for , therefore if , is true. ix If , then since . If , then . Since , we have . Question 6