CONQUERING CALCULUS POST-SECONDARY PREPARATION SOLUTIONS TO ALL EXERCISES. Andrijana Burazin and Miroslav Lovrić

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1 CONQUERING CALCULUS POST-SECONDARY PREPARATION SOLUTIONS TO ALL EXERCISES Andrijana Burazin and Miroslav Lovrić c 7 Miroslav Lovrić. This is an open-access document. You are allowed to download, print and share this document, or use it otherwise for noncommercial purposes. Making copies and distributing this document in an format) for profit is prohibited. Version Ma 7.

2 FOREWORD This file contains solutions to all eercises from the book Calculus... Fear No More [Review and Reference for First Year College and Universit Courses], Second Edition, written b Miroslav Lovrić, and published b Nelson. In general, all details of a solution are provided. Sometimes, related theoretical concepts, methods or formulas are recalled. Keep in mind that in man cases there are multiple correct was of solving a question and/or simplifing an answer); just because our answer looks different from the one given here does not mean that it s incorrect. Some eercises require that ou use a calculator. Do not read this manual! It is far more beneficial to tr to solve an eercise on our own. Start and see how far ou can go. If ou get stuck, identif the problem first tr to understand wh ou are having difficulties, and then look up the solution. This wa, ou will learn not onl what the problem is or what ou have problems with), but also how that particular problem has been resolved. If ou just read a solution, ou might not recognize the hard parts) - or, even worse, ou might miss the whole point of the eercise. Big Thank ou! to Andrijana Burazin for writing solutions to all eercises. We accept full responsibilit for errors and will be grateful to anbod who brings them to our attention. Your comments and suggestions will be greatl appreciated. Andrijana Burazin Ma 7 Miroslav Lovrić Department of Mathematics and Statistics McMaster Universit Hamilton, Ontario, Canada L8S K lovric@mcmaster.ca CONTENTS. Numbers and Operations.... Basic Algebra.... Basic Facts and Formulas from Geometr...[no eercises]. Equations and Inequalities Elements of Analtic Geometr Functions Trigonometric Functions Eponential and Logarithmic Functions Mathematical Language; Mathematical Thinking and Logic...59

3 Section. Numbers and Operations Section. Numbers and Operations. a) Note that 6 is a multiple of 7. Thus Alternativel, note that both the numerator and the denominator are multiples of 6: b) Factor, and then cancel: Note that it is not necessar to factor all terms so we did not factor ), nor to factor as far as possible i.e., all the wa to the prime factors; e.g., 6 suffices, and there was no need to write ). c) Factor, and then cancel b and b : 69 d) Factor, and then cancel b 7 and b b: ab 7bc 7 a b a a 7 b c c c Note that it is not alwas necessar to use a dot to indicate the product: for instance ab a b and a a. However, in some cases we must use the dot, as in 7 ; alternativel, we use brackets: 7)). e) Both 9 and are multiples of : 9z z z z f) Both 6 and are multiples of : 6 ) ) ) ) In the above, we cancelled b and b.. B es we mean equal to a b a b c d and b no we mean not equal to c d a) a + b a + b) a b c d c d c d ;es. b) a b d c a b d + c) a b c d ;no. c) b a b + a) d c d + c) b + a d + c a b c d ;es. d) a b d c a b c d) a b c d ;no. e) b a b a) b + a c d c d c d a b c d ;es. f) b a a b) d c c d) a b c d ;no.

4 Calculus: Fear No More Solutions to All Eercises g) a b d c h) a b c d ;no. a b d c) a b c d ;es.. B es we mean equal to )z t) a b and b no we mean not equal to )z t) a b )z t) a) ;no. a b )z t) )z t) )z t) b) ;es. b a b a) a b )z t) )z t) )z t) )z t) )z t) c) ;no. b a b a b a a b) a b )z t) )z t) )z t) d) ;es. b a a b) a b )t z) )t z) )[ t z)] )z t) e) ;es. a b a b a b a b )t z) [ )][ z t)] )z t) f) ;es. a b a b a b )t z) [ )][ z t)] )z t) g) ;no. a b a b a b )t z) [ )][ z t)] )z t) )z t) h) ;no. b a a b) a b) a b. a) Since all denominators are equal, we combine the numerators: b) Since all denominators are equal, we combine the numerators: a + b 7 a 7 + b 7 b a 7 a + b a + b b a) 7 a + b a + b b + a 7 a + b 7 c) Since both denominators are equal, we combine the numerators and then simplif: a + b) a b) a + b) a b) ab ab ab a +ab + b a ab + b ) ab a +ab + b a +ab b ab ab ab d) Since all denominators are equal, we combine the numerators: c c + d d c + d + c d c + d c d +c d c d c d) c + d c + d c + d e) Write as a fraction with denominator and then combine the three fractions:

5 Section. Numbers and Operations f) The common denominator is 8; thus )) + )) + 5 ) ) g) The common denominator is b)a) 6ab; thus a b + b aa) a ba) + bb) ab) b)a) b)a) a 6ab + b 6ab ab 6ab a +b ab 6ab h) The common denominator is ; thus )5) + )6) ) 5 5) 56) ) + ) 5) i) The common denominator is a b ;thus a +b a + b ) ab + a b a +b)a b ) a b a +)a b) ba b) a b +a b a b a) ab a) + a b + a b a b a a b + b a b a b +a b a b + a b) a +b a b a b +a b a b a b a +b a b 5. a) Multipl the fractions first, then cancel the resulting fraction and add to /c : a bc b ac + c b) Multipl the fractions and cancel b : + + z ab abc + c c + c c + + z)) + + z + ) + ) + b) a bb)

6 Calculus: Fear No More Solutions to All Eercises Note: it is not possible to cancel the final answer b +. However, the fraction can be rewritten in the following wa, b viewing + as a single term: + + z + c) Multipl and then cancel: + )+z z + + z + mn n m 7 mn n m 7 mn))m) 9m n )n)7) 9n m d) First add the terms in parentheses the common denominator is z), and then multipl b z: z z + ) ) z z z) + ) ) z) ) z z + z + z z z + ) + z Alternativel, multipl the terms in parentheses b z first, and then simplif: z z + ) z z z + z z e) Multipl and then cancel b + : z z + z z + + ) + ) + ) ) + ) ) + )) ) + ) ) 9 + ) + ) ) 9 Note: whenever needed as was the case above), we can write an epression A as a fraction: A A. f) Multipl first, cancel the resulting fractions, and then simplif: + + ) + ) ) + ) + g) Multipl and cancel: a +b a b 7a b) a +b) )7a b)) 7 7 a +b a b)a +b) ) We cancelled b a b, a +b and.

7 Section. Numbers and Operations 5 6. a) To divide two fractions means to multipl the fraction in the numerator b the reciprocal of the fraction in the denominator: b) We write as a fraction, then divide the two fractions and cancel: c) We write 5 as a fraction 5, then divide the two fractions and cancel: d)asinb)andc), ) ) 7) 67) In combining the three fractions, we used the fact that the common denominator of, and 6 is. Note that both 87 and are multiples of How do we know? The sums of digits of the two numbers, and + 6, are divisible b ). Thus, we cancel: e) Write c c, divide the two fractions and simplif: ab ab c c c c ab c c ab c ab c g) Simplif the top and the bottom fractions first, and then divide: + z z + z z z) z) + z z) z) z z z + z z z z z+ z z z z + z z z z + z h) Simplif the top and the bottom fractions first, and then divide note that 5 7 ): a 7 b c 5 d a 7 b7) 7) c 5 d5) 5) a 7b 7 c 5d 5 a 7b 7 5 5a 7b) a 7b) c 5d 7c 5d) c 5d i) Simplif the epression on the top the common denominator is ab), and then divide the fractions: a b + b a a + b aa) ba) + bb) ab) a + b a ab + b ab a + b a +b ab a +b a + b ab a + b ab

8 6 Calculus: Fear No More Solutions to All Eercises j) Simplif the epression on the top the common denominator is ), and then divide the fractions: ) ) ) ) 7. Remember: we use round bracket or empt dot to sa that a point real number) is not included, and square bracket or filled dot when we wish to include a point real number).. / / 8. The leftmost interval includes all negative numbers up to and including -, so we denote it b, ], The middle interval contains all numbers between and, not including but including ; so, we write, ]. The rightmost interval is, ). 9. a) [ /9, ). To include /9 we used the square bracket. Remember to alwas use round brackets with infinit. b) [ 7,.). Remember: number included, use square bracket; number not included, use round bracket. c), 8) and, ) d), 5.5) and 5.5, ). Note that 5.5 does not belong to either interval. e) [,.7) and.7, ] f), ) and, ). Zero is not a negative number, so we do not include it.. First row: interval: /, 7/); inequalit: / <<7/; verbal: all real numbers between /and7/, not including / and not including 7/; number line: / 7/ Second row: interval:,π); inequalit: <<π; verbal: all real numbers between and π, not including and not including π; number line: π Third row: interval:.7, ); inequalit: >.7; verbal: all real numbers greater than.7, not including.7; number line:.7 Fourth row: interval: [.,.9); inequalit:. <.9; verbal: all real numbers between. and.9, including. and not including.9; number line:..9

9 Section. Numbers and Operations 7 Fifth row: interval: [, 7.]; inequalit: 7.; verbal: all real numbers between and 7., including both and 7.; number line: 7.. a) is correct. Both fractions are equal to. b) is correct. Zero divided b a non-zero number is zero. c).7.7 is correct, because the smbol means greater than or equal to. d).7 >.7 is incorrect; A > Brepresents a strict inequalit, i.e., it does not allow the two numbers A and B to be equal. e) is incorrect. Division b zero does not produce a real number. f).. is incorrect. On a number line,. istotherightof.7, so the correct relation is. >... a) We use the definition, and then simplif: +.9 if ) if +.9 < { +.9 if.9.9 if <.9 b) We use the definition, and then simplif: 7 if 7 7 7) if 7 < To solve the inequalit 7 within the condition, we wrote: if 7 +7 if< 7 7 Replacing with < we obtain the remaining inequalit. c) As in a), if { if ) if < + if < { + if if > d) We use the definition, and then simplif:. if. {. + if 5..) if. <. if >5 To solve the inequalit. within the condition, we wrote:...

10 8 Calculus: Fear No More Solutions to All Eercises and therefore 5. Replacing with < we obtain the remaining inequalit.. a) Since π is positive, π π. B definition, / /) /, and thus / /. Since. is negative,..).. Note: of course, we could have written right awa that / / or.., since absolute value returns positive numbers. We included intermediate steps to practice the definition. Wh is the definition of absolute value so mess if all it does is to convert a number into a positive number? The reason is that we want to do more than just do calculations with numbers. The definition, as we have seen alread, needs to be applied to algebraic epressions as well. b) Recall that the distance between two numbers on a number line is computed as the absolute value of their difference; so, the distance between and is ) +. c) Simplif the term involving the absolute value first: d) Start b simplifing the terms involving the absolute value: + ) a) As usual, we write out the definition first, and then simplif: if { if ) if < if < + if if > b) As in a), 6 5 if ) if 6 5 < { 6 5 if if6<5 6 5 if if< 5 6 c) Draw a number line, start at 5 and walk units to the right obtaining 5 + 7) and then start at 5 and walk units to the left obtaining 5 7). So the solutions are 7 and7. There is a longer alternative algebraic solution good for practice!). Note that we are asked to solve the equation 5. Recall that 5 if 5 { 5 if 5 5 5) if 5 < +5 if<5 Thus, the equation 5 splits into two cases: { 5 if 5 +5 if<5 B simplifing, we obtain two solutions: { 7 if 5 7 if <5

11 Section. Numbers and Operations 9 d) Draw a number line, start at and walk units to the right obtaining +9) and then start at and walk units to the left obtaining ). So the solutions are and 9. Alternativel, we can solve ) algebraicall. The equation + implies + and thus 9)or +), and thus and. e) Using the product of absolute values formula ab a b, wewrite f) Using the product of absolute values formula ab a b, wewrite. Thus, is not correct. Alternativel, all we need to do is to find one number for which the given formula does not hold. For instance: if,then ) isnotequalto. 5. a) ) ) ) ) ) 8. b) Removing the negative number in the eponent, we obtain 6. c) is undefined, since division b zero does not ield a real number. d) /) ) 8 8. Remember: it s the top times the reciprocal of the bottom.) 8 6. a) Using the definition) a a a a a we write.).).).).)., b) Note that the minus sign, unlike in a), is not raised to the fourth power. Thus. [.).).).)]., c) Remove the negative sign from the eponent and then proceed as in a):.).).).).).),. d) Write /. and then simplif: ).)..).).).),. Alternativel, we could use laws of eponents: ) ) ) ), or: ) ),,, e) Write /. and then simplif: ).)..).),,.

12 Calculus: Fear No More Solutions to All Eercises As in d), there are alternatives, using laws of eponents; for instance: ) ) ) 6,, f) Remove the negative sign from the eponent first: 5) 5) 5) 5) 5 g) Remove the negative sign from the eponent, and note that the minus sign in front of 5 is not under the power of : 5 5 5)5) 5 h) Remove the negative sign from the eponent, and note that the minus sign in front of 5 is not under the power of : 5 5 5)5)5) 5 i) Remove the negative sign from the eponent: 5) 5) 5) 5) 5) 5 5 j) Remove the negative sign from the eponent:.5).5).5).5).5) a) To multipl the powers of a, add the eponents: a a a a a + )++ ) a + a b) To divide the powers of a, subtract the eponents: a a a ) a + a 6 c) B laws of eponents, + ) 9 Note that we used the fact that. d) Simplif using laws of eponents, and cancel: ) e) Using laws of eponents, 5 ) ) 5)) 5 ) ) 5 f) Simplif the fraction within the brackets first: ) )) 5 ) 5))

13 Section. Numbers and Operations Alternativel, using ) a n b a n b : n ) ) ) )) )) 8 8) g) Simplif the fraction within the brackets first: ) )) 5 ) Alternativel, using ) a n b a n b : n ) ) ) ) 8 ) ) ) 8 ) h) As in f) and g), ) c 5 c ) 5 c 5 ) 5 c 5 c i) Simplif the epression inside the brackets first: a ) b a b a b )) a b +) ab ) a b ) a b ab a b j) Using laws of eponents, ) a 8 ) b 8 ) a 8 ) b 8 b a b ) a 6 a ) b b6 a a 6 b b 6 a a 8 b 8 b8 a 8 k) Using laws of eponents, ) 7 ) ) 7 ) ) ) ) 7 ) l) As in j), a ) ) a b a a8 b b 8 b a b a b ab) m) Using laws of eponents, a + b c) a + b c) ) a + b c) 7) a + b c) a + b c) n) Using laws of eponents, + ) + ) ) + ) 6 + ) + ) + ) ) ) 7 ) 8 b a 8. a) 6 is not defined since no real number raised to the power of gives a negative number. b) Recall that 5 ;thus 5 ) 5 5 ) 5

14 Calculus: Fear No More Solutions to All Eercises Note: it is a good idea to remember powers of up to, sa, 6 6, and the first few powers of, sa, up to 8. c) We use the fact that 5 and therefore ) 5 ;thus 5 ) 5 ) 5 ) 5 d) Using laws of eponents, ans recalling that 6 and 8, we write ) ) ) ) e) Use the fact that : ) ) f) Using the law of radicals for fractions, we write Alternativel, there is a longer calculation: ) ) ) g) 5 because 5. h) Recalling the cancellation formula n a n a when n is even, we write ) Alternativel, ) 6 i) Recalling the cancellation formula n a n a when n is even, we write Alternativel, or 6 ) j) Recall the cancellation formula n a n a when n is odd. Thus ) Alternativel, or ) ) ) ) ) 8, because ) 8. k) / is not defined, because no real number squared gives a negative number.

15 Section. Numbers and Operations l) Write the square root as a rational eponent: 6 6) 6 ) 6 ) m) 6 is not defined. We can start the calculation b writing 6 but this is as far as we can go. No real number squared gives 6. n) Using the law of radicals for fractions, we write Alternativel, ) o) Using laws of radicals, we write 6 6 One, of several alternatives: 6 6 6) 6 ) ) ) ) ) ) a) Since 6, we get 6 6 b) Since 5 9 5, we write c) Write ; then d) Since 7 6, we write e) Write 6 8 wh? because 8);then f) Write 8 wh? because 8);then 8 8 g) Write 8 7 wh? because 7 ); then h) Write wh? because ); then

16 Calculus: Fear No More Solutions to All Eercises i) Write 6 wh? because 6 ); then 6 6 j) Write wh? because ); then. a) To multipl, we add eponents: b) Convert radicals to fractional eponents and simplif: 5 ) 5 ) ) 5 ) c) Using the power to another power law of eponents, we write ) 5 d) Using the power to another power law of eponents, we write ) e) We write 9 9 f) Convert the radical to a fractional eponent and then multipl: g) We can keep radicals ) note that ) ), or convert to rational eponents: ) ) h) Convert to rational eponents and then multipl when multipling, keep in mind that ): ) ) i) Convert to rational eponents and then multipl: ++ j) Convert to rational eponents and use laws of eponents: ab6 c ab 6 c ) a b 6 ) c ) a b c a b c k) Multipling each term in the first bracket with each term in the second bracket, we obtain a + b) a b) a a + b a a b b b a ) b ) a b

17 Section. Numbers and Operations 5 Note that b a a b.) Instead of working with radicals, we can use rational eponents: a + b) a b)a + b )a b )a a a b + b a b b a b a b Note that we can use the difference of squares formula + ) ) which is reviewed in the net section): a + b) a b) a ) ) b a b l) Multipling each term in the first bracket with each term in the second bracket, we obtain a ) a ) b +5 b a a b a + a b 5 b b a ) ) ab + ab 5 b a +9 ab 5b We can use rational eponents as well: a ) a ) b +5 b a b )a +5b ) a a +a b b a 5b b a +a b a b 5b a +9b b 5b m) Multipl the bracket b and simplif: ) ) ) Alternativel, )) ) ) n) Multipling each term in the first bracket with each term in the second bracket, we obtain a + b) a ab + b )a + b )a ab) + b ) a + b )a a b + b ) a a a a b + a ab + b a b a b + b b a a b + a b + b a a b + b a + b Note: after reviewing the difference of cubes formula, tr this question again!). a) The law of radicals for products implies that is correct. b) The formula is incorrect. For instance, when, then, whereas. Or, when, then is not defined, whereas. c) B the law of radicals for fractions, 6 The formula is correct. 6

18 6 Calculus: Fear No More Solutions to All Eercises d) The formula ) 7 is incorrect. The correct formula is obtained from the laws of eponents: ) e) The formula ) is correct, since ) f) The formula + 5 is incorrect. For instance, when, the left side is and the right side is. g) The formula a a is incorrect. The correct formula is obtained from the laws of eponents: a a a a a a h) The formula a b) a b is correct, since a b) a ) b) a ) b a b i) The formula a + b) a + b is incorrect. For eample, if a andb thena+b) /, whereas a + b a) ) / ) 5)/ 5) / 6 / ) / 5) /) /) 5) 5. For the record: the formulas that we used in the calculation above, from first to last equals sign: n A A /n with n ),A/B) n A n /B n with n /); net, we recall that 6 and 5 5, so that 5 5) ;thena m ) n A mn with m andn /), and finall we simplif eponents and use the fact that A A. Note: there are other was to simplif this epression; for instance 5 6 5) 5 ) 5 ; in the last step we used the formula n A n A which holds for all odd eponents n in our case n ). Tr to find other was of correctl simplifing this! b) One wa to simplif is:. If needed, we rationalize the denominator and write Alternativel, we could simplif given epression as.. In the first and the last steps we used the formula A/B A/ B. ) / c) / 9 9 ) / /) / ) / /) 7 8.

19 Section. Numbers and Operations 7 As usual, there are other was of simplifing; for instance: ) / ) / ) ) / ) /) 9 ) ) ) 8 7 d) Likewise, 6 / 9 / ) / ) / /) /) a) Combine the cube roots: b)ithelpstowriteas : 5 + ) 5 + ) ) c) Because the denominator of the power is, we write the numbers involved as squares: ) ) d) Same idea as in c): write 8 as a cube and. as a square: 8 +. ) +. ) +. + Alternativel, e) Write both numbers as squares: 8 +. ) + ) +.9) +.). ) +. ) To simplif., we wrote: Alternativel,.. 5) ) +.) 9 + ) + ) + ) ) f) Divide the two roots and rewrite 75 using a smaller number under the square root: ) 7 8.

20 8 Calculus: Fear No More Solutions to All Eercises g) Rewrite the second and the third terms using a smaller number under the square root: h) Multipl and then comine the square roots: 8+7 ) ) 8 Alternativel: 8+7 ) ) +7 ) ). a) 7 7 7) /) ) ) b) a + b a + b a + b ) ) c) 5 5 5/) ) ) ) d) z z z ) z ) e) ) ) ) 6 6 ) ) f) ) ) ) 8 8 ) ) 5. a) 7 7 7) /) b) ) ) /) ) ) ) c) ) ) ) d) a b 6 c 8 a b 6 c 8 a b 6 c 8 ) a b c 8 e) ) ) ) ) 6 6 ) f) Using ), we write ). ) 6. a) Using laws of eponents,..5 ) ) ).9.9.)).9 6.) First equals sign is due to the formula m n m+n, and the third one due to m / n m n. We computed 6.).9 using m ) n mn. ) / b) / /) 6) / / /) ) / /) / 9/.

21 Section. Numbers and Operations 9 In the first step we simplified the fraction ) 6. The second equals sign is due to the formulas AB) n A n B n, here applied with A,B 6 and n /, and m ) n mn : 6) / ) / 6 ) / /) 6 /). Finall, in the last step we use the fact that multipling two powers with the same base results in the addition of their eponents. c) We get a) a) a) a) ) a) a) a) a, since a) a and a). 7. a) Using laws of eponents,. ).).6 b) Using laws of eponents,. z. ).6.6.).6 z. ).6.6.).6) z.).6).6.8 z z.66 c) Divide inside the brackets first:..7 )...7)).. )..).).5.5 d) Using laws of eponents, a b ) a ) b ) a b a 6 b 6 e)workoneachfractionfirst,andthenmultiplithelpstowriteas ): c ) ).6 d..6 c ).6 d. ).6 d c d c c.6.6 d.6.6) d.).6) c.6) c.6.6 d.6. d. c.6 c.6.6 d.6. d. c.6 c.6+.6 d c. d..8 c..8 d.

22 Calculus: Fear No More Solutions to All Eercises f) Simplif the fraction first: ) 7. + ). + ) ) ) ) ) a) The formula for the product of fractions a b c d ac bd applied from right to left so we are undoing the product) implies that is correct. b) The formula + is incorrect. If and, then the left side is, whereas + the right side is /. Another eample: if and, then the left side is not defined, whereas the right side is /. c) The formula + is incorrect. If and, then the left side is, whereas the right side is. d) The calculation recall the wa we divide fractions!) proves that is incorrect. e) The multiplication law for radicals n a n b n ab implies that The given formula is correct. f) The formula is incorrect. For instance, if and 5, then 5 5 whereas 5 )5) 5. Alternativel, b computing the common denominator we obtain the correct formula: g) The formula 7 is incorrect; b computing the common denominator we obtain h) We do not square a binomial, or a trinomial just b squaring each term. Thus, the formula + ) + is incorrect. It s eas to find values for and to prove this. For instance, if and then + ) ), whereas +. i) The formula a a is incorrect. If a, then the left side is a, whereas the right side is a.

23 Section. Numbers and Operations Note: If a 7, then a 7) 7and a 7) 7. This, however, does not prove that the formula is correct nor would an number of calculations such as this one). The formula needs to hold for all real numbers a. If it does not hold for a single value of a, then the formula is not correct. j) The formula k + k k is incorrect: if, then the left side is and the right side is. k) Since k k k+k k, the given formula is correct. l) Using a ) b ab we compute k ) k k. Thus, k ) k k is incorrect. Alternativel, Take andk ;then k ) k ) 8 5, whereas k ) 6 6.

24 Calculus: Fear No More Solutions to All Eercises Section. Basic Algebra. a) m + b.. So, m andb.. b) m + b 6 6. So, m 6 andb. c) m + b 7.So,m andb 7. d) m + b 5 5. So, m 5 and b. e) Think of as + ). Comparing m + b + ) we see that m and b. f) From m + b +wegetm andb. g) From m + b +wegetm andb. h) m + b So,m and b 6. i) m + b 5 5.So,m 5 and b. j) m + b ) So,m andb 8. k) m + b So,m 5 and b 5.. a) a + b + c So,a, b 9,andc. b) a + b + c So,a, b 7,andc. c) a + b + c So,a, b,andc. d) From a + b + c ) + +weseethata, b,c e) Comparing a + b + c + ) +wegeta,b, c. f) Comparing a + b + c + +wegeta,bandc.. a) m + n + p +. So, m,n,andp. b) m + n + p So,m,n 6,andp. c) m + n + p + +. So,m, n,andp. d) m + n + p ) So,m,n 6, and p.. a) Multipl each with each, to get + ) ) )) )+)+ ) ) ) + ) b) Using the formula for the square of the difference, we compute.) ) ).) +.). +. c) We multipl the first two terms first or the last two difference of squares!) ) +) ) +) ) ) )

25 Section. Basic Algebra d) Using the cube of the difference formula, we get ) ) ) ) + )) ) 8 ) e) Using the formula for the square of the difference, we compute 5 ) ) ) Alternativel, using EWE, 5 ) 5 ) )) ) ) ) ) ) 5 )+ 5 ) ) f) Simplif and use the square of the sum formula: ) ) 5 + 5))+) g) Using the formula for the square of the difference, we compute..).).).) +.) Alternativel, using EWE,..)..)..).).).).).).) +.).) h) Using EWE, + ) ) )) )) )) + )) + )) ))

26 Calculus: Fear No More Solutions to All Eercises i) Recognizing the epression as a difference of squares, we find 7) +7) ) 7) 9 Alternativel, using EWE, 7) +7) ))+)7) 7)) 7)7) j) Recognizing the epression as a difference of squares, we find ) + ) ) ) k) Multipl the two brackets using the difference of squares formula) and then multipl b : 7+ ) 7 ) [ 7) ) ] [ 9 8] Alternativel, without using the difference of squares formula: 7 + )7 ) [7)7) 7) )+ )7) ) )] [ ] l) Using the difference of squares a b)a + b) a b with a andb, we obtain ) + ) ) ) Alternativel, using EWE, ) + ) ) ) ) )+ )) ) ) + + m) Using the difference of squares formula, we obtain ) ) ) + ) 9 Using EWE, ) ) ) ) ) ) + + ) ) )) n) Using the cube of the sum formula, +) + ) + )+ ) )

27 Section. Basic Algebra 5 If we cannot recall the formula we need, we can still do it using EWE: +) +) +) +) +) + + +) +) + +) ) ) o) Using the cube of the difference formula, ) ) ) ) ) + ) 8 ) + 6) As in n), if we are not sure about the formula, we use EWE: ) ) ) ) ) ) +6 ) ) a) a + b + c ) +. So,a, b, and c. b) From a + b + c ) 7) it follows that a, b, and c. c) From ) a + b + c 5) we conclude that a, b /, and c 5/. d) From a + b + c..).5 +.5) it follows that a.5, b.85, and c.6. e) From ) a + b + c ) it follows that a /, b /, and c /8.

28 6 Calculus: Fear No More Solutions to All Eercises f) From a + b + c.8.)..) we obtain a 6.6, b., and c.. 6. a) Using EWE, +) +) +) Using the square of the sum formula, +) ) + ) b) Using EWE, + ) + ) Note that in the above calculation we used ) + ). c) We find ) ) d) Using EWE, +) ) Using the difference of squares formula, +) ) ) ) 9 e) Using EWE, ) ) ) Alternativel, converting to fractional powers, ) ) ) + ) )) + ) ) +) ) )) )) 9 Using the difference of squares formula, ) + ) ) 9 f) Using EWE, + ) + ) + ) Using the square of the sum formula, + ) ) + ) g) Using EWE, 5 ) 5 ) 5 ) Using the square of the sum formula, 5 ) ) ) 5)

29 Section. Basic Algebra 7 h) Using EWE, + ) + ) + ) +) Alternativel, using the square of the difference formula, + ) + ) + + ) i) Using EWE, ) ) ) ) ) +) ) Using the formula for the cube of the difference, ) ) j) Using the formula for the cube of the sum, ) ) + / + ) ) + ) ) /) + /) ) + /) )+ / / +7 k) Using the difference of squares formula, + ) 7 ++ ) 7) +) Look at the formulas a + b) a +ab + b and a b) a ab + b. Both epansions have three terms, two of which are squares, and the sign of the third term is determined b the sign within the binomial being squared. In this eercise we think in the opposite direction: we know the three terms which are the result of squaring a binomial, and we need to guess that binomial. a) Looking at we identif two squares: and 6 6. The sign of the remaining term is +, so our guess is +6). Check: +6) +)6) Works. b) +9 7) c) In a +ab +b we identif the squares a and b b). Thus, we tr a +b). Check: a +b) a +a)b)+b) a +ab +b.

30 8 Calculus: Fear No More Solutions to All Eercises d)notethat6 ).Thus ). e)notethat9 ) and.thus9 + ). f) Note that 6a a) and 5b )5b). Thus 6a ab +5b a 5b). Check: a 5b) a) a)5b)+5b) 6a ab +5b. g) ) h) Note that ). Thus ) Check: + ) +) ) i) Note that 9 ). Thus ). Check: 9 ) j) Note that 9 ). Thus ) ) ) ) a) We add and subtract b/) 6/) 6, to get ) ) 6 +8) 6 Keep in mind that the three terms in the bracket are equal to + ) b ; i.e., ). b) We add and subtract b/) 6/) 6, to get ) ) ) 55 c) We add and subtract b/) 7/) 9/, to get ) d) We add and subtract b/) 7/) 9/, to get e) Factor out first [ ] ) 9 + ) ) ) [ +) ] [ ) ] ) f) Factor out of the terms involving and then complete the square within the square brackets: +5 [ ] +5 [ +) ] +5 [ ) ] +5 ) 7

31 Section. Basic Algebra 9 g) We factor out the coefficient of and then complete the square: 9 [ 9 ] [ ) 8 ] 6 6 [ 9 ) ] 8 9 ) h) We factor out the coefficient of and then complete the square within the square brackets: [ + 9 ) 9 ] [ ] + [ ) ] 9 + ) 7 + ) 9 i) We factor out / frist: +6 [ + ] [ + + ) ] [ + ) ] + ) 6 + ) 7 j) The coefficient of is b /7; thus, b/ /, and b/) /96: + [ ) ] + ) k) Factor / out of the terms involving : [ ] + 9 [ ) ] ) [ + ) ] ) + l) Factor. out of the terms involving : ).5.[ + +) ].5. +)..5. +).6 9. a) ) b) Group the first two and the last two terms: + +) +) +) ) Alternativel, group the first and the third, and the second and the fourth terms: + + ) + ) +) ) c) Group the first two and the last two terms or in some other wa, as in b)): + + ) + ) + ) )

32 Calculus: Fear No More Solutions to All Eercises d) Group the first two and the last two terms: a 6ab +ab 6b aa b)+b a b) a b)a +b )a b)a + b ) e) Two numbers whose sum is and whose product is 8 are 7 and. Thus, ) +) f) Two numbers whose sum is and whose product is 8 are 7 and. Thus, ) ) g) Factor out first, and then proceed b finding numbers whose sum is and whose product is. Thus ) +5) ) h) Factor 5 out, and then recall the difference of squares formula: ) 5 +) ) i) Factor out the common term, and then proceed as usual sum is 5, product is ): ) +) +). a) We write + a) + b) and check all combinations of integers a and b so that ab. In this case, there are onl two possibilities, a andb, and a andb. B tring +) ) and ) +), we see that the first one works. Thus, +) ) b) As in a), we write + + a) + b) and check the two cases of integers a and b so that ab. We find + ) +) c) Write a) + b) and check all pairs of values for a and b such that ab : a andb i.e., 5+)+)), a andb i.e., 5 ) )), a andb i.e., 5 +) + )), a andb i.e., 5 ) )), a andb i.e., 5 +) + )), a andb i.e., 5 ) )). It turns out that 5 +5 ) ) d) Write +7 + a) + b) and check the pairs of values for a and b such that ab 7: a andb 7,a7andb,a andb 7, and a 7 andb. It turns out that +7 ) 7) e) There are two cases: a) + b) and a) + b). Because ab, possible values are a andb i.e., 6 +) +)and + ) + )), a and b i.e., 6 +) +) and + ) + )), a andb i.e., 6 ) ) and ) )), and a andb i.e., 6 ) ) and ) )). It turns out that ) +)

33 Section. Basic Algebra f) There are two cases: a) + b) and6 + a) + b). Because ab, possible values are a andb i.e., 6 +) ) and + ) )), a and b i.e., 6 ) +)and ) + )), a andb i.e., 6 ) +)and ) + )), and a andb i.e., 6 +) ) and + ) )). We find that 6 + ) ) g) There are two cases: + a) + b) and + a) + b). As in e) and f), we have check pairs of integers a and b such that ab. Sooner or later, we discover that + ) ) h) There are two cases: a) + b), a) + b). As in e) and f), we have check pairs of integers a and b such that ab. We find ) ). a) Write. ).) and use a b a + b)a b) witha and b. to get. ).).) +.). b) We write 8 ) ) + ) c) Even though the numbers might not look like it, the are squares: 5 ) 5 ) ) ) d) Factor 5 out first: ) 5 ) 5) ) 5 5) +5) e) Factor 5 out first: ) 5 ) ) ) 5 ) + ) f) a b c a) bc ) ) a bc a + bc ) g) Start with ) ) to get ) + ). Now + cannot be factored an further, but can: ) ) ) + ). Thus ) + ) ) + ) + ). h) Recall that ). Thus 6 ) ) ) + )

34 Calculus: Fear No More Solutions to All Eercises i) Write ) and 5 5. Thus ) ) ) ) j) Using a b a + b)a b) witha +andb, we obtain +) ) [ +) )] [ +)+ )] +) + ). a) Using the sum of the cubes formula a + b a + b)a ab + b )witha and b, we obtain +7) +) +) + ) +) +9) b) Using the sum of the cubes a + b a + b)a ab + b )witha and b, we obtain +6) +) +) + ) +) + 6) c) Start with a difference of squares 6 ) ) +) ) Each binomial can further be factored; thus 6 +) ) +) +) ) + +) Alternativel, if we start with a difference of cubes, we get 6 ) ) ) ) + + ) +) ) + +) It is not obvious how to factor + +. It involves a trick we add and subtract : ) + ++ ) + +) +) Thus is now the difference of squares, so +) ++) + ) and our answer agrees with the one we obtained earlier). d) We write + /) +) ) / + /) ) / ) + ) ) ) / + / / + e) Write + z 6 ) +z ) and use the sum of the cubes formula a +b a+b)a ab+b ) with a and b z : + z 6 ) +z ) + z )) )z )+z ) ) + z ) z + z ) f) Factor out ab first: a b ab 7 ab a b ) ab a b)a + ab + b ). a) We group the terms and then factor: + + ) +) + ) + ) + ) ).

35 Section. Basic Algebra Alternativel, we group the first and the third terms and the second and the fourth terms: + )+ ) ) + ) ) + ). b) Factor out and then use the difference of squares: ) ) +). c) Factor out: ) and use the difference of squares: 9 5 ) ) 5) ) 5) +5). d) Using the difference of squares twice, we get 5 ) ) +) ) +) +). e) 8 5 5) ) 5) ) 5) +5) f) Again, it s the difference of squares: 6 ) ) 9 7 ) 7 + ) 7 g) ) ) + ). h) We factor the given epression in the form +6 + a) + b) and +6 + a) + b) and tr combinations of a and b for which ab 6. We get +6 ) ). i) We factor 8 + a) + b) and check integer values of a and b for which ab. We get 8 +) ) j) We hope to factor the given epression in the form + + a) + b); thus, we need to find two numbers whose sum is and product is. We find that 5 and satisfthese requirements, and so + ) +5). k) Start with +7 + a) + b) and tr combinations of a and b whose product is. Tr a,b : +) ) 8+ 7, which does not work. Choosing a andb,weget ) +) works! Had it not worked, we would have tried a,b ora, b, etc. l) Start b factoring out: + + ). Now, we are looking for two numbers whose sum is and whose product is. The numbers and work, and so + + ) ) +). m) Factor out and then use the same strateg as in a) or c): 5 5) ) +5). n) Because 6 ) ), we use the difference of squares 6 ) +).

36 Calculus: Fear No More Solutions to All Eercises The first factor is the difference of squares again, and thus 6 ) +) +). o) Combine the first two and the last two terms: a b ac + ab c bc a b ac)+ab c bc )aab c)+bcab c) a + bc)ab c). a) ) b) Note that each term contains a : a + a + a + a a a + a a a a +a) c) z z + z z z + z) d) Factor out +: a +) c +) a c) +) e) Rewrite the last two terms to identif a common factor: a b) a + b a b) a b) )a b) f) Combine the first two and the last two terms: ac + bc ad bd ca + b) da + b) a + b)c d) g) Combine the first two and the last two terms: + +) +) ) +) h) Rewrite ) in the second term as : a ) ab ) a )+ab ) a + ab) ) a + b) ) Alternativel, factor out a ): a ) ab ) a )[ b] a )[ ) + b)] a ) + b) i) Combine the first two and the last two terms: ab b a + ba ) a ) b )a ) j) Difference of squares: k) Difference of squares: a 9c a) c) a c)a +c) 9 ) 7) 7) + 7) l) Factor out, then use the difference of squares formula: ) ) +) m) Difference of cubes: a b 7 ab) ) ab )ab) +ab + )ab )a b +ab +9)

37 Section. Basic Algebra 5 n) Difference of squares: + ) + ) [ + )][ + + )] ) + + ) o) Factor out, then use the difference of squares formula: ) ) ) ) + ) p) Group the first two terms using the difference of squares: +) +) ) +) +) +) +)) +) ) +) q) Combine the first two and the last two terms: )+ + ) +) + ) r) Factor out a b and then use the difference of cubes formula: a 5 b a b 5 a b a b )a b a b)a + ab + b ) s) Difference of squares: a b) c [a b) c][a b)+c] a b c)a b + c) t) Start with the difference of squares formula: ) +) [ ) +)] [ )+ +)] ) + +) 5) +) 5) + ) u) Use the difference of squares and the difference of cubes formulas: a b a + b a b ) a b ) a b)a + b) a b)a + ab + b ) a b)[a + b) b + ab + a )] a b)a + b b ab a ) 5. a) Cancel b + : b) Cancel b 7: c) Cancel b a + b) : ) + ) + ) 7 + ) + ) a + b) a + b) a + b) a + b

38 6 Calculus: Fear No More Solutions to All Eercises d) The onl wa to cancel 6 and is to split the fraction: e) Factor, and then cancel: 9 +) ) ) +) + f) Factor, and then cancel: ) + ) ) + ) + ) Note that we factored out the minus sign: +) ). g) Factor, and then cancel: + ) + ) ) + 6 h) Cancel b +) : ) +) ) +) + i) Factor using the techniques we have covered: ) +) ) +) + j) Factor, and then cancel b : ) 5 +6 ) ) ) +) ) +) ) +) +) ) ) ) ) k) Factor and then cancel: a a a) + a) a) + a + a ) + a + a + a l) We can factor the denominator, but there is nothing to cancel: + a + a + a + a) a + a ) m) Factor out and cancel: ) n) Factor the difference of squares in the denominator and cancel b c d: c d) d c c d) d c)d + c) c d) c d)d + c) c d d + c) c d c + d In order to cancel, we factored out the minus sign: d c d + c) c d). o) Factor the numerator using the difference of squares, and factor out the minus sign in the denominator: ) + ) + )

39 Section. Basic Algebra 7 Alternativel, p) Factor and cancel: ) ) + ) + ) q) Factor and cancel: ) ) ) +) +) ) ) ) ) r) Factor out the minus sign in the denominator no need to factor anthing else): a b)c d ) d c a b)c d ) c d a b) ) s) Rearrange the numerator and then factor: 6 a a 6 a 6) a a 6 a 6 a )a +a + 6) a +a +6 a )a +) a + t) Factor out the minus signs and then cancel: )z t) )t z) )[ t z) ] )t z) t z) [ )] t z) )t z) Note that z t) [ t z)] t z). 6. a) Factor out 5 in the numerator and then cancel: ) b) Factor and then cancel b : 9 + 9) ) +) ) +) +) +) +) c) Factor b grouping and then cancel: a + b a b a + b) a + b) )a + b) a + b + a + b a + b)+a + b) + )a + b) + d) a b + ab a b a b aba + b) a ba b) a + b aa b) e) Factor 5 out of both numerator and denominator: ) 5 5 ) f) Keep in mind that + cannot be factored: ) + ) + ) + ) g) ) 9 9 ) ) ) + ) +

40 8 Calculus: Fear No More Solutions to All Eercises h) Factor and cancel: a +b a b a + b) a + b) )a + b) a a a ) a a + b ) a 7. a) Factor and cancel b a: a a 5 + a a aa a +) a a a + a b) Keep in mind that + a cannot be factored. a + a 6 a + a a + a ) a + a ) a + a ) +a c) Factor and cancel: a + a b + ab a + a) b + a) a b d) Factor and cancel: a b + ab a b + a b ab a + b) a bb + a) b a e) Factor a in the numerator: a + a + a + a + a a a a + a ) a a +a a a f) To cancel with the denominator, factor out a in the numerator: a + a + a a a + a ) a a + a a + a ) Alternativel: a + a + a a + a ) a a +a a +a )a g) Factor and cancel: a + a + a + a + a + a a a + a a a + a) a + a) a a h) Factor a in the denominator: a a a + a a a a a a a) a 8. a) The fraction is not defined for that s wh we sa > and not ): if > { if > if < if < b) In the first line, we sa > to eclude. ) if > ) if < { if > if <

41 Section. Basic Algebra 9 c) We eclude the value 5/ for which the denominator is zero. To cancel, rewrite the denominator as 5 5+) 5): 5 if 5 > d) As in c), ) 5 ) ) ) if 5 < if > if < 5 5) if > 5 5) 5) if< 5 if < if > 9. a) The common denominator of and is their product; therefore ) ) ) ) ) ) ) +) ) ) + ) ) b) The least common multiple of, and ) is ). Thus + ) ) ) ) ) ) + +) ) + ) ) +5 + ) ) c) Since 5 +6 ) ) we see that the common denominator is ) ): ) ) ) ) + ) ) ) + ++ ) ) + ) ). a) The common denominator is ) + ): ) ) + ) + ) + ) ) + )+ ) + ) ) ) ) + ) )

42 Calculus: Fear No More Solutions to All Eercises b) Since ab + b ba + b), the common denominator is ba + b): a + b b a + b + b a + b)a + b) ab + b ba + b) a + b)a + b) b + b ba + b) a + b)a + b) a + b ba + b) b b) a + b)b) + b ba + b) c) Since a b a +b)a b), the common denominator is a +b)a b): a +b a b a b a b) a +b)a b) a +b) a b)a +b) a b)a +b) a b) a +b) a b)a +b) a b a 6b a b)a +b) a b a b)a +b) a +b + a b)a +b) d) The common denominator is z: + z + z + z + z)z z) +)) z) + ) z) + z)z +))+) z z z + z + z e) The common denominator is ) + ): ) + ) ) ) + + ) + ) + ) ) + ) + )+ ) ) ) f) Factor the denominators to figure out the common denominator: +) ) +) +) ) + )) ) +) ) )) +) )) +) ) +) +) 6 ) +) ) +)

43 Section. Basic Algebra ) +) 6 +6 ) +) +6 6) ) +) +8) ) +8) ) +) +) g) Because b +b 8b +)b ), the common denominator is b +)b ): + b b + + b b +b 8 + b b + + b b +)b ) b +)b ) b + )) + b +)b ) b )b + )) b )) b +)b )) + b) b +)b )) b +)b ) + b +) b ) + b b +)b ) b +b 8+b +6 b ++b b +)b ) b +6b + b +)b ) h)notethatbecause ) the first two fractions add up to zero. Thus + ) + ) + ) + ) ) + ) ) + ) )+ + + ) ) i) The common denominator is m ;thus m m m m m m j) The common denominator is a + ;thus a + a) a+ a a) + a + a a + + a + a+ a + k) The common denominator is a + ;thus a a + a) a a a) a + a ) a a ) a a + a + a a a a+ a + l) Factoring using the difference of cubes and the difference of squares formulas and then canceling the fraction, we get 7 9 ) ) ) ) ) + +9) + +9 ) +) +

44 Calculus: Fear No More Solutions to All Eercises m) Factor, and then cancel: + + ) +) +) ) +) +) +) ) + In the denominator, we factored b writing a) + b) and then checking integer values of a and b such that ab. n) Factor and cancel first: ab + b ) a b a + b + b ba + b) a b a + b + b ba b)+b ab Alternativel, compute the common denominator: ab + b ) a b a + b + b ab + b ) a b a + b + a + b b a + b ab + b )a b)+b a + b) a + b a b + ab ab b + ab + b a + b a b + ab aba + b) ab a + b a + b o) Factor using the difference of squares formula, and then cancel: p) Factor and cancel: a + b) a b a b)a + b) + ) 6 +) a + b)a + b) ) + ) a + b a b) + ) ) +) q) Simplif the terms in the parentheses first: a ) a + b) b a b ab a + b) a b ab r) Compute the common denominator, factor, and cancel: a + ) a b b b a b + a a b ab b a)b + a) a b b a s) Recall: the top times the reciprocal of the bottom: t) Simplif the numerator first: b a b 5 ab 5 a b b a a a + b ab a + b a b ab u) Simplifing the double fraction, a b a + b a b a b 5 5 ab a b a b)a + b) a + b ab b a + a b b b a a b a b a a + b a b ab

45 Section. Basic Algebra v) Start b calculating common denominators: In the last step, we simplified using the top times the reciprocal of the bottom rule. Net, we cancel the fraction 6 +9 ) w) Start b calculating common denominators, then simplif the double fraction and cancel: + +. a) How do we factor 6 out of or an other term)? We need to generate 6 first, and then factor it out. Multipl and divide b 6, thus not changing the value of the epression: We do the same with the remaining terms some are obvious, such as 6 ): ) 6 b) We need to identif in each term keep in mind the formula a+b a b ): ) We can check our answer b epanding the right side b multipling b ). c) We need to identif in each term: ) d) We need to identif / in each term keep in mind the formula a+b a b ): + / 5/ + / / / / / / 5/ + / / /) e) We need to identif in each term: ) + + ) f) We need to identif / in each term: + / 9/ + / 7/ / 5/ / / / 9/ + 7/ 5/ / ) g) We write / 5) + 7 / ) / 5) + 7 / ) / [ 5) + 7 ) ]

46 Calculus: Fear No More Solutions to All Eercises h) We write / 5) + 5 / 5) / 5) + 5 / 5) [ / 5) 5) + ] [ 7 5 / 5) 5 5 ] i) As in h), / 5) + / 5) / 5) + / 5) 5) j) We need to identif in each term: [ 5) + ] 5) [5 )] [ + ] 7 + k) We need to identif / in each term: / 5/ + / / 7 / / + / / / / / [ 5/ + / 7 / + / / ] l) We need to identif / in each term: 5+ / 5+ / Note that / /. / / / 5 / / + / / / / / 5 / + / ) 5 + ). a) ) 6) + 5 5) 57 Thus,

47 Section. Basic Algebra 5 b) ) ) 8 Thus, c) ) 5 Thus, d) + + ) ) Thus, e) ) ) Thus, f) g) 6 6) ) 7 7 9) ) Thus, Thus,

48 6 Calculus: Fear No More Solutions to All Eercises h) + + ) Thus, + +. a) Subtract and add 6: ) Check: see Eercise f). b) Subtract and add : + ) Check: ) c) Subtract and add 5, to get ) Check: ) 5 d) Add and subtract, thus getting ) Check: + + ) e) Subtract and add : + ) + + +

49 Section. Basic Algebra 7 Check: ) f) Write 5 + 6: Check: ) 6 +5) Wetrtoidentifarootof + b guessing. Recall that, the root must divide the free coefficient. Substituting, we get ) +) ), soisnota root. Substituting, we get ) + ) )! So, isarootofthe given polnomial, i.e., + is its factor. Thus + +)q) where the unknown polnomial q) is identified using long division: ) + ) ) We write i.e., ) + ). Now we factor + +) ) and so ) +) ) Of course, we could have initiall guessed some other root or ). Therewouldbeno change, however we follow the same procedure and arrive at the same factorization.

50 8 Calculus: Fear No More Solutions to All Eercises 5. a) Multipling and dividing b 7+, we get ) 7) ) b) Multipl and divide b the conjugate of the denominator: ) ) c) Multipl and divide b the conjugate of the denominator: + +) + ) ) +) d) As above, + + ) ) ) ) e) Multipl and divide b the conjugate of the denominator: ) + ) ) ++) +) ++) ++ f) Multipl and divide b +6: ) ) 6) g) We multipl and divide b + h) Multipl and divide b + : ) ) ) + ) ) ++ + ) ) ) + + ) )

51 Section. Equations and Inequalities 9 Section. Equations and Inequalities. a) Substituting into the given equation, we compute + + ) So, is not a solution. isnotasolutionbecause + + ) is not zero. Substitute : + + ) + + So, is a solution of the given equation. b) We are checking whether or not the left side 5 +isequalto : : ) 5) :) 5) :) 5) + : ) 5 ) Thus, isasolution. c) We are checking whether or not the left side + + is equal to zero: : ) + ) ) ) : ) + ) ) ) : ) +) ) ) + : ) +) ) ) Thus, and are the solutions.. a) Simplif.5) + +) + 6 gather like terms together Dividing b 5, we get. b) Multipling the equation b 5 5 is the lowest common denominator), we get

52 5 Calculus: Fear No More Solutions to All Eercises c) Simplif and gather like terms together. 6.. ) d) Gather like terms together, divide and simplif In the last step we rationalized the denominator which might, or might not be required): Note that we could have canceled the fraction b instead.. a) Multipling the equation b is the lowest common denominator), we get ) 5 ) b) The lowest common denominator of and+ is their product )+). Multipling the given equation b ) +), we get ) +)+ ) +) + +)+ ) c) We multipl the given equation b + ) or, cross-multipl!) ) +) + 5 +) 5

53 Section. Equations and Inequalities a) The lowest common denominator is 6. Multipling the given equation b 6 ields v Dividing b, we get 5/. b) We cross multipl to get rid of fractions: ) ) +) c) The common denominator is ); thus + ) + ) ) + ) d) Simplif each side first, and then combine the terms:.6.) ) e) Multipl both sides b the common denominator ): + ) + ) ) ) + ))

54 5 Calculus: Fear No More Solutions to All Eercises f) The common denominator is the product of the two denominators; thus, multipling the equation b + ) ), we obtain + ) +) g) Simplif each side and then combine the terms: 5 +) ) ) a) Computing the square root of both sides, we obtain ± ± ) ± ± b) Computing the square root of both sides of weobtain ± ± 6. Note that ) / /) / /)/) /6 c) The square of an real number is zero or positive. Thus, the given equation has no solutions. d) Computing the square root of both sides, we obtain +) +± ±

55 Section. Equations and Inequalities 5 ± So, the solutions are )/ / and )/. e) Compute the square root of both sides and then solve for : 5) 5± ± +5 ± +5 5 ± 6. a) Since the square of a real number cannot be negative, the given equation has no solutions. b) Computing the square root of both sides, we get 6± ; i.e., 6±. Recall that can be simplified as follows and so 6±. c) We are asked to solve + +. Factoring, we get + + +) +6) and so and 6. d) We have to solve. To factor, we think of writing + a) + b) and tr those a and b for which ab a andb, a andb,aandb, and a andb ). We find that ) + ) Thus, the solutions are from ) / andfrom + ). Alternativel, using the quadratic formula, we get ± +8 ± 9 ± Thus, the solutions are 8/6 / and 6/6. 7. a) We add and subtract the square of one half of the coefficient of, which is /) /, and write +/ /+5, i.e., + ) +5 Net, we recall that the square term involves one half of the coefficient of i.e., one half of ) ) 9 Since the square of ever number is zero or positive, the above equation has no solutions. b) First, divide the equation b so that the coefficient of is 7 7