CHAPTER 1Section 1.1Section 1.2Section 1.3Section 1.5A.1. a. Domain isCHAPTER 2Section 2.1Section 2.2Section 2.4Sections 2.1 and 2.6Section 2.7SlopePointSection 2.8CHAPTER 3Section 3.1Section 3.2Section 3.4Section 3.5Section 3.6 Part ISection 3.6 Part IISection 3.7Section 1.7CHAPTER 4Section 4.1Section 4.2Section 4.3Section 4.4Section 4.6Section 4.7Section 4.8Section 4.9CHAPTER 5Section 5.1Section 5.2MATH 171: HOMEWORK ANSWERS Answers to the Additional Homework QuestionsCHAPTER 1Section 1.1 A.2. a.50 1000q p=- +b.12050p q-= +Section 1.2 A. a. (i)b. (iii)c. (iv)d. (ii)e. (vi)f. (v)Section 1.3 B.1. a. (iii)b. (iv)c. (v)d. (vi)e. (i)f. (ii) 2. a. ( ) 2sin 24xf x� �= +� �� �b.( ) 5cos3xf x� �=� �� �Section 1.5 A.1. a. Domain is 0y >b. Range is all real numbersc. (i) a > 1(ii)0 < a < 1CHAPTER 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. Calculus tries to estimate the area by dividing up the region into very thin rectangles. The sum of the areas of these rectangles is an approximation to the area of the region. We improve the approximation by increasing the number of rectangles. The actual area is the limit of the sum of the areas of these rectangles, as the rectangles become infinitely small.4. In both of these methods, we start with something known (slope of a secant, area of a rectangle) to find an approximation to something unknown (slope of a tangent, area under a curve). 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. Truec. False115-2 -1 1 2 3 4 5-11234567xy-4 -3 -2 -1 1 2 3-11234567xyd. Truee. Truef. Trueg. TrueSection 2.4 B.1. a. 1.5b. 0.25c. does not existd. does not existe. 6f.1pp -g.93e--h. 22. a. All real numbers except x = 2b. u > 1c. All real numbers.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 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.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 as 0h �116MATH 171: HOMEWORK ANSWERS 4.Slope Point–3 F–1 C0 E12A1 B2 D5. In order, from smallest to largest:e, c, b, d, a, fSection 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 (removable discontinuity), –1 (jump), 2 (infinite discontinuity) b. –2, –1 and 2 because function is discontinuous at these points. Also, 0 and 1 because of the sharp “corners” – the left hand derivative does not equal the right hand derivative.4. None.5. i., ii. and iii. Definition of continuity6. a. Sometimes false. For example, at a sharp “corner” function is continuous but not differentiable.b. True.7.( ) 4 3f x x�= +CHAPTER 3Section 3.1 A.1. a.5/ 6xb.1 5/ 225 3x x x- -+ -c.2xyd.22x x-C. 1. a.12-b. 02. a. 0b. 0.5c. –2d. does not existe. does not existf. 2Section 3.2B.1. 0( ) ( )( ) limhf x h f xf xh�+ -�=For an explanation, see the answers for 2.1 2a, 2b2.54 4xy = +Section 3.4 B.2. a. Falseb. Truec. Falsed. TrueSection 3.5 A.1. Different2. a. Pb. Cc. Cd. Ce. Pf. Cg. Ph. Ci. Cj. P3. b. x cosine square rootc. x sine exponentiald. x exponential sinef. x five times sineh. x sine fifth poweri. x square root log4. a. 12( )f x x=, ( ) 1g x x= -, 2( )h x x=. The twelfth power of (one minus x squared).b. ( ) sin( )f x x=, ( )xg x e=, ( ) tan( )h x x=. Sine of exponential of tan of xc. ( )xf x e=, ( ) sin( )g x x=, 2( )h x x=. Exponential of sine of x-squaredd. ( )f x x=, ( ) tan( )g x x=, 3( )h x x=. Square root of tan of x-cubed117B.1. 6( ) 6xf x e�=2. ( ) 2 cos 2f q q�=3. 2 3 / 2( 2)( )( 4 )tg tt t- +�=+4. 10sin 5y x�=-5. '( ) cosx xg x e e=6. sin( ) cosxh x e x�= �7. 23xyx�=+8. ( )2xef xx�=9. xy e-�=-10. 3 44 cos( )y x x�=11. 34sin cosy x x�= �12. 2'( 1)xxeye-=+ 13. 21 1cosyxx-� �=� �� �14. 2cos sinx x xyx-�=15. 2( ) 2 sin 5 5 cos 5g t t t t t�= +16.( ) ( cos 4 4 sin 4 )t tf t e t e t- -�=- + Section 3.6 Part I B.1. a. –1 m/sec b. 0 m/sec2. The graph of the derivative is:Section 3.6 Part II