Ch 8 Properties of Exponents Notes

C domain lnsinx is only defined for sinx 0 sinx 0 for

This preview shows page 12 - 16 out of 16 pages.

c. domain ln(sinx) is only defined for sinx > 0 sinx > 0 for (0, ), (2 , 3 ), etc at sinx = 1 n | n I 2 ln(1) = 0 as sin x 0 ln x d. critical points cotx = 0 when cosx = 0 x = 5 3 , , 2 2 2 7 , 2 not in domain increase: 5 0, , 2 , 2 2 decrease: 5 , , ,3 2 2 local max at 5 ,0 , ,0 2 2 2 0 + 4.a 0
Image of page 12

Subscribe to view the full document.

13 3. Given ln y 2 + 3y = ln x, find d x y. 2ln y + 3y = ln x / 1 2 (y ) y + 3y / = 1 x * multiply by xy 2xy / + 3xyy / = y y / = y 2x 3xy 4. Graph y = ln(x 2 1) a. domain y x 2 1 > 0 (x + 1)(x 1) > 0 ( , 1) (1, ) b. x int(s) 0 = ln(x 2 1) e 0 = x 2 1 1 = x 2 1 x 2 = 2 x = 2 c. symmetry ln[( x) 2 1] = ln(x 2 1) even function d. y / = 2 2 1 2x 2x x 1 x 1 no max or min because x = 0 or x = 1 are not in domain increase: (1, ) decrease: ( , 1) y / e. // 2 1 2 2 y 2(x 1) (x 1) (2x)(2x) // 2 2 2 2 2 // 2 2 y 2(x 1) [x 1 2x ] 2(x 1) y (x 1) y // always concave down ( , 1) (1, ) f. asymptotes 2 x 1 lim ln(x 1) 2 x 1 lim ln(x 1) so x = 1 are vertical asymptotes 4 -3 -2 -1 0 1 2 3 4 + + -5 -4 -3 -2 -1 1 2 3 4 5 6 -5 -4 -3 -2 -1 1 2 3 4 0 4.b 4 -3 -2 -1 0 1 2 3 4 + 4 -3 -2 -1 0 1 2 3 4
Image of page 13
14 eg Sketch y = x + ln x given the x intercept at (0.57, 0) x + lnx is only defined for x > 0 y / = 1 1 x c.p. at 1 1 1 0, 1 x 1, but x 0 so no c.p. x x fox x > 0, 1 1 x > 0 so function is continuously increasing y / = 1 // 2 2 1 1 1 1 x , y ( 1)x x x y // < 0 so graph is concave down. Formula Math 30 Math 31 t d o y y (2) Write t d o y y (2) as t tln2 ln2 d d o o y y (e ) y (e ) y total amount Make k = ln 2 d , so kt o y y e y o initial amount y = y o e kt t total time y total d doubling time y o initial amount k constant t total time Note that if P = P o e kt then kt kt o o dP d d P (e ) e P dt dt dt kt kt o o d P e (kt) k(P e ) kP dt This means that in a population growing exponentially the rate of population growth is directly proportional to the population. eg 1. You start with 75 bacteria. In 12 hours, you have 150 bacteria. Find: a. k y = y o e kt 150 = 75e k(12) 2 = e 12k ln 2 = ln(e 12k ) ln 2 = 12k k = ln 2 12 so ln2 t 0.0577622t 12 y 75e or y 75e 4.c 5 5 0
Image of page 14

Subscribe to view the full document.

15 b. number of bacteria after 6 hours ln2 ln2 t (6) 12 12 y 75e 75e 106 c. time to get 3000 bacteria ln 2 t 12 ln 2 t 12 3000 75e 40 e ln 2 ln 40 t 12 12ln 40 t 63.9 ln 2 d. rate of increase at 80 hours ln 2 t 12 ln 2 t / 12 ln 2 (80) / 12 / y 75e ln 2 y 75e 12 75ln 2 y e 12 y 440 bacteria / hour e. the time before the rate of growth becomes 750/hr ln 2 t / 12 ln 2 t 12 ln 2 t 12 ln 2 y 75e 12 ln 2 750 75e 12 750x12 120 e 75ln 2 ln 2 120 ln 2 ln t ln 2 12 120 ln 2 ln t ln 2 12 120 ln ln 2 t 89.2 hr. ln 2 12 4.d
Image of page 15
16 2. A population increases by 15%/a. a. Find k. Start at 1, after one year, population will be 1.15. 1.15 = 1e k(1) ln 1.15 = k (0.139762) b. If the population was 280 000 in 1995, find the: i. population in 2003 y = y o e ln 1.15t y = 280 000e ln 1.15(8) y = 856 526 ii. population in 1990 y = 280 000e ln 1.15( 5) y = 139 209 iii. year that population will be 1 400 000 1 400 000 = 280 000e ln 1.15t 5 = e ln 1.15t ln 5 = ln 1.15t t = 11.5 (around 2007) iv. rate of increase in 1999 y = 280 000e ln 1.15t y / = 280 000e ln 1.15t (ln 1.15) y / = 280 000e ln 1.15(4) (ln 1.15) y / = 68 444 people/year v. year the rate of increase would be 100 000 people y / = 280 000e ln 1.15t (ln 1.15) 100 000 = 280 000e ln 1.15t (ln 1.15) 2.555 = e ln 1.15t ln 2.555 = ln 1.15t t = 6.7 (around 2002) Assignment Read p 385 390, skip p 393 396, Sketch 2 x y e pg. 384 # 5 a, c, 6, 7 a, b, 8 a, b, 10 b
Image of page 16
  • Fall '14
  • Berry
  • Math, Exponents, Natural logarithm, Logarithm, Binary logarithm, e. lim e

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern