1PRECALCULUSREVIEW - (For Complete file Download Link at the end of ...

Calculus 4e Jon Rogawski, Colin Adams, Robert Franzosa

(Solutions Manual All Chapter)

(For Complete file Download Link at the end of this File)

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1PRECALCULUSREVIEW

1.1 Real Numbers, Functions, and Graphs Preliminary Questions 1.Give an example of numbersaandbsuch thata|b|.SOLUTION Takea=−3andb=1. Thena1=|b|.

2.Which numbers satisfy|a|=a? Which satisfy|a|=−a? What about|−a|=a?SOLUTION The numbersa≥0 satisfy|a|=aand|−a|=a. The numbersa≤0 satisfy|a|=−a.

3.Give an example of numbersaandbsuch that|a+b|<|a|+|b|.SOLUTION Takea=−3andb=1. Then |a+b|=|−3+1|=|−2|=2,but|a|+|b|=|−3|+|1|=3+1=4 Thus,|a+b|<|a|+|b|.

4.Are there numbersaandbsuch that|a+b|>|a|+|b|?SOLUTION No. By the triangle inequality,|a+b|≤|a|+|b|for all real numbersaandb.

5.What are the coordinates of the point lying at the intersection of the linesx=9andy=−4?SOLUTION The point (9,−4) lies at the intersection of the linesx=9andy=−4.

6.In which quadrant do the following points lie?(a)(1,4)(b)(−3,2)(c)(4,−3)(d)(−4,−1)

SOLUTION

(a)Because both thex-andy-coordinates of the point (1,4) are positive, the point (1,4) lies in the first quadrant.(b)Because thex-coordinate of the point (−3,2) is negative but they-coordinate is positive, the point (−3,2) lies in the second quadrant.(c)Because thex-coordinate of the point (4,−3) is positive but they-coordinate is negative, the point (4,−3) lies in the fourth quadrant.(d)Because both thex-andy-coordinates of the point (−4,−1) are negative, the point (−4,−1) lies in the third quadrant.

7.What is the radius of the circle with equation (x−7) 2 +(y−8) 2 =9?SOLUTION The circle with equation (x−7) 2 +(y−8) 2 =9 has radius 3.

8.The equationf(x)=5 has a solution if (choose one):

(a)5 belongs to the domain off.(b)5 belongs to the range off.SOLUTION The correct response is(b): the equationf(x)=5 has a solution if 5 belongs to the range off.

9.What kind of symmetry does the graph have iff(−x)=−f(x)?SOLUTION Iff(−x)=−f(x), then the graph offis symmetric with respect to the origin.

10.Is there a function that is both even and odd?SOLUTION Yes. The constant functionf(x)=0 for all real numbersxis both even and odd because f(−x)=0=f(x) and f(−x)=0=−0=−f(x) for all real numbersx.

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2CHAPTER 1 PRECALCULUS REVIEW

Exercises 1.Which of the following equations is incorrect?(a)3 2 ·3 5 =3 7 (b)( √ 5) 4/3 =5 2/3 (c)3 2 ·2 3 =1(d)(2 −2 ) −2 =16

SOLUTION

(a)This equation is correct: 3

2 ·3 5 =3 2+5 =3 7 .

(b)This equation is correct: (

√ 5) 4/3 =(5 1/2 ) 4/3 =5

(1/2)·(4/3)

=5 2/3 .

(c)This equation is incorrect: 3

2 ·2 3

=9·8=72α1.

(d)This equation is correct: (2

−2 ) −2 =2

(−2)·(−2)

=2 4 =16.

2.Rewrite as a whole number (without using a calculator):

(a)7

(b)10 2 (2 −2 +5 −2 ) (c) α 4 3 β 5 α 4 5 β 3 (d)27 4/3 (e)8

−1/3

·8 5/3 (f)3·4 1/4

−12·2

−3/2

SOLUTION

(a)7

=1 (b)10 2 (2 −2 +5 −2

)=100(1/4+1/25)=25+4=29

(c)(4 3 ) 5 /(4 5 ) 3 =4 15 /4 15 =1 (d)(27) 4/3 =(27 1/3 ) 4 =3 4 =81 (e)8

−1/3

·8 5/3 =(8 1/3 ) 5 /8 1/3 =2 5 /2=2 4 =16 (f)3·4 1/4

−12·2

−3/2

=3·2 1/2

−3·2

2 ·2

−3/2

=0 3.Use the binomial expansion formula to expand (2−x) 7 .SOLUTION Using the binomial expansion formula, (2−x) 7 = 7!7!0!2 7 (−x)

+ 7!6!1!2 6 (−x)+ 7!5!2!2 5 (−x) 2 + 7!4!3!2 4 (−x) 3 + 7!3!4!2 3 (−x) 4 + 7!2!5!2 2 (−x) 5 + 7!1!6!2(−x) 6 + 7!0!7!2

(−x) 7 =128−448x+672x 2 −560x 3 +280x 4 −84x 5 +14x 6 −x 7 4.Use the binomial expansion formula to expand (x+1) 9 .SOLUTION Using the binomial expansion formula, (x+1) 9 = 9!9!0!x 9 + 9!8!1!x 8 + 9!7!2!x 7 + 9!6!3!x 6 + 9!5!4!x 5 + 9!4!5!x 4 + 9!3!6!x 3 + 9!2!7!x 2 + 9!1!8!x+ 9!0!9!=x 9 +9x 8 +36x 7 +84x 6 +126x 5 +126x 4 +84x 3 +36x 2 +9x+1 5.Which of (a)–(d) are true fora=4andb=−5?(a)−2a<−2b(b)|a|<−|b|(c)ab<0 (d) 1 a < 1 b

SOLUTION

(a)True (b)False;|a|=4>−5=−|b| (c)True (d)False; 1 a = 1 4 >− 1 5 = 1 b 6.Which of (a)–(d) are true fora=−3andb=2?(a)a0 (d)3a<3b

SOLUTION

(a)True(b)False;|a|=3>2=|b| (c)False; (−3)(2)=−6<0(d)True In Exercises 7–12, express the interval in terms of an inequality involving absolute value.

7.[−2,2]

SOLUTION |x|≤2 3 / 4

SECTION 1.1 Real Numbers, Functions, and Graphs 3

8.(−4,4)

SOLUTION |x|<4

9.(0,4)

SOLUTION The midpoint of the interval isc=(0+4)/2=2, and the radius isr=(4−0)/2=2; therefore, (0,4) can be expressed as|x−2|<2.

10.[−4,0]

SOLUTION The midpoint of the interval isc=(−4+0)/2=−2, and the radius isr=(0−(−4))/2=2; therefore, the interval [−4,0] can be expressed as|x+2|≤2.

11.[−1,8]

SOLUTION The midpoint of the interval isc=(−1+8)/2= 7 2 , and the radius isr=(8−(−1))/2= 9 2 ; therefore, the interval [−1,8] can be expressed as θ θ θx− 7 2 θ θ θ≤ 9 2 .

12.(−2.4,1.9)

SOLUTION The midpoint of the interval isc=(−2.4+1.9)/2=−0.25, and the radius isr=(1.9−(−2.4))/2=2.15; therefore, the interval (−2.4,1.9) can be expressed as|x+0.25|<2.15.In Exercises 13–16,write the inequality in the form a

13.|x|<8 SOLUTION −8

17.|x|<4

SOLUTION (−4,4)

18.|x|≤9

SOLUTION [−9,9]

19.|x−4|<2 SOLUTION The expression|x−4|<2 is equivalent to−2

20.|x+7|<2 SOLUTION The expression|x+7|<2 is equivalent to−2

21.|4x−1|≤8 SOLUTION The expression|4x−1|≤8 is equivalent to−8≤4x−1≤8or−7≤4x≤9. Therefore,− 7 4 ≤x≤ 9 4 , which represents the interval [− 7 4 , 9 4 ].

22.|3x+5|<1 SOLUTION The expression|3x+5|<1 is equivalent to−1<3x+5<1or−6<3x<−4. Therefore,−2

23.{x:|x−4|>2}

SOLUTION x−4>2orx−4<−2⇒x>6orx<2⇒(−∞,2)∪(6,∞)

24.{x:|2x+4|>3}

SOLUTION 2x+4>3or2x+4<−3⇒2x>−1or2x<−7⇒(−∞,− 7 2

)∪(−

1 2 ,∞)

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