CHAPTER 1Section 1.1Section 1.2Section 1.5A.1. a. Domain isc.CHAPTER 2Section 2.1Section 2.222.Section 2.32. a.Section 2.4Section 2.5A.Sections 2.1 and 2.6B.Section 2.7SlopePointSection 2.8Section 2.8CHAPTER 3Section 3.1Section 3.2Section 3.3A.12.a. , ,Section 2.728. a. is the rate of growth f the bacteria population when t=5 hours. Its units are bacteria per hour.Section 3.4Section 3.5C.8.Section 3.6 Part ISection 3.6 Part IISection 3.7CHAPTER 4Section 4.1Section 1.7Section 4.2Section 4.3Section 4.4Section 4.6Section 4.7Section 4.8Section 4.9CHAPTER 5Section 5.1Section 5.2Section 5.4Section 5.3MATH 171: HOMEWORK ANSWERS Answers to the Additional Homework Questions and even textbook exercisesCHAPTER 1Section 1.1 A. See page 11.B.28. Domain is ( ) ( ) ( ), 2 2, 1 1,- �- � - - � - �Section 1.2 A. a. (i)b. (iii)c. (iv)d. (ii)e. (vi)f. (v)Section 1.5 A.1. a. Domain is ( ),- ��b. Range is ( )0,�c. (i) a > 1 (ii) 0 < a < 1Section 1.632.a. The natural logarithm is the log with base e, denoted ln(x)b. The common logarithm is the log with base 10, denoted log(x)c.CHAPTER 2Section 2.1 1. a. What is the slope of a function at a given point?b. What is the instantaneous velocity of a moving particle at an instant in time?2. a. To find the slope of a function at a given point P we find the slope of the secant line connecting P and a nearby point Q. Then, we let Q move closer to P and continue to find the slope of the secant line as a better and better approximation to the slope of the function at P. We then take the limit of these slopes and that will be the slope of the function at P.b. To find the instantaneous velocity at time t0, we find the average velocity on the time interval from t0 to t1. We then get better and better approximations by letting the time interval get smaller and smaller. The limit of the average velocities is the instantaneous velocity at time t0.3. In both of these methods, we start with something known (slope of a secant, average velocity) to findan approximation to something unknown (slope ofa tangent, instantaneous velocity). We then improve the approximation and this leads to a limitSection 2.2 A.1. a. 6b. 4c. does not existd. 6e. 2f. 2g. 2h. 22. a. Trueb. Falsec. Falsed. Truee. Truef. Trueg. TrueB.2. As x approaches 1 from the left, f(x) approaches 3; as x approaches 1 from the right, f(x) approaches 7.No, the limit does not exist because the left- and right-hand limits are different.4. a. ( )03limxf x�=b.( )34limxf x-�=131-4 -3 -2 -1 1 2 3-11234567xy-2 -1 1 2 3 4 5-11234567xyMATH 171: HOMEWORK ANSWERS c.( )32limxf x+�=d.( )3limxf x�does not exist because the limits to parts (b) and (c) are not equal.e.( )3 3f =22.6 2x xyx-=From the graph it appears that the 06 21.1limxx xx�-�As 0x-�As 0x+�x( )f xx( )f x-.0000047 1.098606 .0000043 1.098618-.0000024 1.098609 .0000021 1.098615-.0000012 1.098611 .0000011 1.098614-.000001 1.098611 .000001 1.098614From the table, it appears that the 06 21.1limxx xx�-�Therefore, based on graphical and numerical evidence, 06 21.1limxx xx�-�Section 2.32. a.( )( )( )( )2 2 22 0 2lim lim limx x xg xf x g x f x� � �+ = + =+ =� �� �d.Since ( )10limxg x�-=and g is in the denominator, thelimit does not exist.8. a. The left-hand side of the equation is not defined for x=2, but the right-hand side is.b.Since the equation holds for all x�2, it follows that both sides of the equation approach the same limit as 2x �. Remember that when finding( )limx af x� we never consider x=a.36.a.i.( )1 1 12 21 11 21 1lim lim limx x xx xxx x+ + +� � �- -= = + =- -ii.( )( )1 1 12 21 11 21 1lim lim limx x xx xxx x- - -� � �- -= = - + =-- - -b.No, ( )1limxF x�does not exist because( ) ( )1 1lim limx xF Fx x- +� ��c. f(x)Section 2.4 A.2. The graph of f has no hole, jump or vertical asymptote.6. One example is:B.1. a. 1.5b. 0.25c. does not existd. does not existe. 6f.1pp -g.93e--h. 22. a.( ) ( ), 2 2,-�ȥb.( )1,�c.( ),- ��3. a. f (a) = L . Because f is continuous,lim ( ) ( )x af x f a�=.b. Let f (0) = 1 because 0sinlim 1xxx�=c. None of the statements must be true Section 2.5 A.2.a. The graph of a function can approach a vertical asymptote but not cross it.The graph of a function can intersect a horizontalasymptote, and it can do so an infinite number oftimes.132MATH 171: HOMEWORK ANSWERS b. The graph of a function can have 0, 1 or 2 horizontal asymptotes.Sections 2.1 and 2.6 A. 1.432.(5)f=143. a.P Q xPQm(1, 3) (2, 8) 18 351-=(1, 3) (1.5, 5.25) 0.5 4.5(1, 3) (1.1, 3.41) 0.1 4.1(1, 3) (1.01, 3.0401) 0.01 4.01(1, 3) (1.001, 3.004001) 0.001 4.001(1, 3) (1.0001, 3.00040001) 0.0001 4.0001(1, 3) (1.00001,3.0000400001)0.00001 4.00001Best Guess: 4b.P Q xPQm(1, 3) (0, 0) –10 331-=-(1, 3) (0.5, 1.25) –0.5 3.5(1, 3) (0.9, 2.61) –0.1 3.9(1, 3) (0.99, 2.9601) –0.01 3.99(1, 3) (0.999, 2.996001) –0.001 3.999(1, 3) (0.9999, 2.99960001) –0.0001 3.9999(1, 3) (0.99999,2.9999600001)–0.00001 3.99999Best Guess: 44. Average velocity is change in position divided by change in time. In this case we need position at time t2 minus position at time t1, divided by t2 minus t1. Av velocity = 2 12 1( ) ( )s t s tst t t-D=D -5. The instantaneous velocity at t1 is the limit of the average velocities as t2 gets very, very close to t1. Inst velocity = 0 0lim ( av vel) limt tstD � D �D=D6. The average velocity is an "average" of how fast the particle moved during a given time interval whereas the instantaneous velocity is the velocity of the particle at an instant. Also, average velocity is a ratio of the change in position to the change intime. Instantaneous velocity at time t is the limit of the average velocity as the time interval shrinksto zero.B.2.a. Average velocity ( ) ( )f a h f ast h+ -D=D=b. Instantaneous velocity( ) ( )0limhf a h f ah�+ -=Section 2.7 A. 1.0( ) ( )( ) limhf a h f af ah�+ -�=2. Look at answers for 2.1at the beginning of Chapter 2 – #2a and 2b.3. First find f(1)Then find an expression for f(1 + h)Substitute these into the definition to get(1 ) (1)f h fh+ - and do the algebra needed to simplify the fraction. Finally, take the limit as0h �4.Slope Point–3 F–1 C0 A,E1 B2 D5. In order, from smallest to largest: e, c, b, d, f, aB.4.a.4 23y x= -b.( )144f�=Section 2.8 A. 1.0( ) ( )( ) limhf x h f xf xh�+ -�=2. f(x) is differentiable at a if ( )f a�exists. That is, the 0( ) ( )limhf a h f ah�+ -exists.3. a. –2