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

< Solutions To Mathematics Textbooks‎ | Calculus (3rd) (0521867444)

Contents

Question 1Edit

iEdit

 

 

 

 

 

 


iiEdit

 

 

 

 


iiiEdit

 , then

 

Either  , which means   or  , which means that  .


ivEdit

 

 

 

 

vEdit

 

 

 

 

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.


viEdit

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


Question 2Edit

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


Question 4Edit

iiEdit

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

iiiEdit

If  , then  .


Thus,  , or  .


vEdit

  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.


viEdit

If  , then  .


 


Thus  , or  .


viiiEdit

Complete the square:


 


Thus,  .

ixEdit

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


xEdit

 , or  


xiEdit

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


 


xiiEdit

 


xiiiEdit

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 5Edit

iEdit

 


 

 , which is true since  


iiEdit

 


 

 

 


ivEdit

 , therefore  .

 , therefore  .


vEdit

 , therefore  .

 

 

 


viiiEdit

If   or   are 0, then by definition  .

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


ixEdit

If  , then   since  .

If  , then  . Since  , we have  .

Question 6Edit