math150 - CALCULUS: THE ANSWERS MATH 150: CALCULUS WITH...

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CALCULUS: THE ANSWERS MATH 150: CALCULUS WITH ANALYTIC GEOMETRY I VERSION 1.3 KEN KUNIYUKI and LALEH HOWARD SAN DIEGO MESA COLLEGE
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(Answers to Exercises for Chapter 1: Review) A.1.1. CHAPTER 1: REVIEW 1) f ± 4 () = 16 ; fa + h = a + h 2 = a 2 + 2 ah + h 2 2) Polynomial: Yes, Rational: Yes, Algebraic: Yes 3) Polynomial: No, Rational: Yes, Algebraic: Yes 4) Polynomial: No, Rational: No, Algebraic: Yes 5) Polynomial: No, Rational: No, Algebraic: No 6) a) ±² , ² , b) , ² , c) , ± 1 2 ³ ´ µ · ¸ ¹± 1 2 ,2 ³ ´ µ · ¸ ¹ 2, ² , d) ± 2, 5 ² ³ ) ´ 5, µ , e) ± 2, ² , f) ,7 ( ³ ´ , g) , ± 1 ( ³ ´ µ 3, ² · ) 7) ; Domain is 0, ± ² ³ ) ; Range is ± 2, ² ³ ´ ) 8) the y -axis; the function is even 9) the origin; the function is odd 10) the function is even 11) the function is even 12) the function is neither even nor odd 13) gx = x 4 + x , fu = u 8 ; there are other possibilities 14) gt = 1 t , = 4 ; there are other possibilities 15) gr = r 2 , = sin u ; there are other possibilities 16) a) undefined, b) ± 1 , c) 2 , d) ± 3 , e) ± 3 2 , f) ± 3 17) Hint on a): Use a Double-Angle ID; Hint on b): Use a Pythagorean ID 18) a) x ± ± x = ² ³ 6 + 2 n ,o r x = 7 6 + 2 n r x = 3 2 + 2 nn integer ´ µ · ¸ ¹ º · Note: ± ² 6 can be replaced by 11 ± 6 , 3 2 can be replaced by ± 2 , etc. b) x ± ± x 9 + 2 n 3 n integer ³ ´ µ · ¸ ¹ , or, equivalently, x ± ± x = 9 + 2 n 3 r x = 5 9 + 2 n 3 n integer ³ ´ µ · ¸ ¹
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(Answers to Exercises for Chapter 2: Limits and Continuity) A.2.1 CHAPTER 2: LIMITS AND CONTINUITY SECTION 2.1: AN INTRODUCTION TO LIMITS 1) 57 2) 11/7 3) 10 4) ± 2 5) a) f 1 () = 0.9999 , f 0.1 = 0.0999 , f 0.01 = 0.0099 ; b) ± 0.0001 , No 6) a) 11/7, b) 11/7; as a consequence, the answer to Exercise 2 is the same. 7) No; a counterexample: lim x ± 0 + x = 0 , while lim x ± 0 x does not exist (DNE). See also Example 8. 8) Yes 9) No; a counterexample: see Example 10 on hx = x + 3, x ± 3 7, x = 3 ² ³ ´ 10) No; a counterexample: see Example 9 on gx = x + x ± 3 . 11) a) b) 1, 0, DNE; c) 3, 3 , DNE, 2 12) a ) b ) ± 1 , 1, DNE 13) a) 0 (liters), which means that, if the gas’s temperature approaches absolute zero (from above), its volume approaches zero (liters). b) DNE, because temperatures cannot go below absolute zero. The domain of V does not include values of T below absolute zero.
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(Answers to Exercises for Chapter 2: Limits and Continuity) A.2.2 SECTION 2.2: PROPERTIES OF LIMITS and ALGEBRAIC FUNCTIONS 1) 39 + 18 3 16 , or 3 16 13 + () 2) a) DNE, b) 0, c) DNE 3) a) 0, b) 0, c) 0 4) a) 0, b) DNE, c) DNE, d) 0, e) , f) DNE, g) 0 5) a) DNE, b) 0, c) DNE, d) 0, e) DNE, f) DNE, g) 6 6) Yes, by linearity of the limit operator. 7) No; a counterexample: lim x ± 0 x 2 = 0 . See also Example 5 on lim x ±² 7 x + 7 2 . SECTION 2.3: LIMITS AND INFINITY I 1) a) b) 2, c) y = 2 2) a) 0; b) 0 or 1; c) 0, 1, or 2 3) a) 0 + , b) 0 ± , c) 0 + , d) , e) ± , f) ± , g) ± , h) 0 + , i) 0 4) Yes 5) No; some counterexamples: Example 5 on fx = sin x x ; also, = sin x + 2 sin x + 2 . 6) a) DNE, b) 0 or 0 + , c) 0 or 0 ± , d) 0 or 0 + , e) DNE, f) ± 7) a) ± b) i. lim x x 5 + 3 x 4 ³ 2 = lim x x 5 = ² ; ii. lim x x 5 + 3 x 4 ³ 2 = lim x x 5 ± 1 + 3 x ³ 2 x 5 ´ µ · ¸ ¹ ± 1 ²³ ´ ´µ ´´ = ² c) i. lim x ±²³ 2 x 3 ² 6 x 2 + x = lim x 2 x 3 = ; ii. lim x 2 x 3 ² 6 x 2 + x = lim x 2 x 3 ± 1 ² 3 x + 1 2 x 2 ´ µ · ¸ ¹ ± 1 ´ ´´ = d) lim w 5 w ³ 4 w 4 = lim w ³ 4 w 4 = ³²
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math150 - CALCULUS: THE ANSWERS MATH 150: CALCULUS WITH...

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