Chapter 3: Exponentials and Logarithms

Transcription

1 Chapter 3: Eponentials and Logarithms Lesson See graph at right. kf () is a vertical stretch to the graph of f () with factor k. y 5 5 f () = 3! 4 + f () = 3( 3! 4 + ) f () = 3 (3! 4 + ) f () =!( 3! 4 + ) f () =! (3! 4 + ) 3-. See graph below. f (k) is a horizontal stretch to the graph of f () with factor k. f () = (! ) 3! 4 (! ) + =! y 5 f () = 3! 4 + f () = ( 3 ) 3! 4 ( 3 ) + = 3 7! f () = (!3) 3! 4(!3) + f () = (!) 3! 4(!) + =! =! See graph at right.! f () reflects f () across the -ais. f (!) reflects f () across the y-ais. y 5 5 f () = 3! 4 + f () = (!) 3! 4(!) + f () =!( 3! 4 + ) 3-4. The results agree with 3-. y f () = 4! f () = (3) 4! 4(3) + 3 f () = (!) 4! 4(!) + 3 = 8 4! = 6 4! f () = ( 3 ) 4! 4 ( 3 ) + 3 = 4 8! f () = (! ) 4! 4 (! ) + 3 = 4 6! + 3 CPM Educational Program 0 Chapter 3: Page Pre-Calculus with Trigonometry

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3 3-5. f () f ()! f () f (!) Review and Preview See graph at right. a. Vertical stretch f () = 3 3 f () = (3) 3 5 y f () = 3 f () = ( 3 ) 3 b. Horizontal compression 5 c. Horizontal stretch 3-7. g() = (3) 3 = = 7 3, which is a vertical stretch Original f () : f() a. New function: f() 0.5 Each y-value is halved, thus this is a vertical compress and the new epression is f () b. New function: f() We can see this new function is horizontally compressed with factor epression is f (). thus the new c. New function: f() We can see all of the y-values are reversed from the original. Thus, the new epression is f (!). CPM Educational Program 0 Chapter 3: Page 3 Pre-Calculus with Trigonometry

4 3-9. a. 0.7 = 7 0 so 00.7 = b = = 0 0!7 = 0 0 ( ) 7 0 = ( 0 ) 7 c. 0 0 ( ) 7 Thus c = d. When not using a calculator, taking the root first makes the number you are raising to a certain power that much smaller, so the work is simplified. ( ) 7 0 = ( 0 ) 7 = 5.0 e. 0 0 f = 5.0. The answer is the same. g = 5.9, too large = 4.898, too small = is eact up to h. Since we can rewrite 0 = 5 as log 0 5 =, calculating log 5 = gives several more decimal places instantly = ( 3 ) = 3!!and Thus 3 = 3 " and!! = ".!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!3 = 3 3 = 3! 3" = 3 ". Also,!5 = (5 ) = 5 5!!and! 5 Thus 5 = 5 /5 and = 5. = 5 /5. The only that satisfies this equation is = 0.!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! = The slope of the line connecting A(!, 3) and B(4,!5) can be found by: m =!5!3 4+ =!8 6 =! 4 3. The midpoint of the line can be found by: M =!+4 ( ) = (,!), 3!5 Point-slope form: y! y = m(! ) The slope of the perpendicular bisector is the negative reciprocal of the line connecting the two points, or m = 3. Using the midpoint (,!) in point-slope form we get: 4 y + = 3 (! ) a. Clockwise angle is negative. Angle is! "! " 6 =! " radians away from zero. 3 b. Counter-clockwise angle is positive. Angle is! +! +! 6 = 7! radians away from zero. 6 CPM Educational Program 0 Chapter 3: Page 4 Pre-Calculus with Trigonometry

5 3-3. a.!pat is isosceles with PA = TA. Thus, PA b. sin P = TA PT = PA PT = 4 8 c. cos P = PA PT = 4 8 = = 3-4. a. See graph at right. b. From the graph, we can see the zeros are = 0, and the range is! < y " 4. c. See graph at right. d. # h() = f (! ) = (! ) for! " (! ) < $ %&! (! ) for " (! ) < 4 ( ) + ( TA) = ( PT ) = 8 = 64 ( PA) + ( PA) = ( PT ) = 64 ( PA) = 64 ( PA) = 3 PA = 3 = (! )!(! )! = 4 #!!!!!!!!!!!!!!!!!!!!!!!! = (! ) for! " < $ %& 3! for " < 5 e. From part (c) we see the zeros are =, 3 and the range is! " y " 4. Lesson a. y = a!b with (, y) = (,8) and (, y) = (4,6) we get: 8 = a! b 6 = a! b 4 b. In the first equation, solve for a : a = 8. Substitute this value for a into the second b equation: 6 = 8 b!b4 = 8!b "! b 4 = 8!b or 6 = 8b c. Solve for b : a = 8 b 8!b = 6 b = 6 8 = 9 b = 3 d. Use b = 3 to solve for a : a = 8 b = 8 3 = 8 9 = CPM Educational Program 0 Chapter 3: Page 5 Pre-Calculus with Trigonometry

6 e. The final equation y = a! b can be written as: y =! t S( t) 36g 8g 9g 4.5g.5g.5g 0.563g 3-7. Using S(t) = A!b t and the points (t, S(t)) = (0, 36) and (t, S(t)) = (,8) (or any two points) we get: 36 = A!b 0 " 36 = A 8 = A!b " 8 = 36!b " b = Thus, S(t) = A! b t may be written as: S(t) = 36!( ) t t (min) Number of half-lives B(t) 60g 30g 5g 7.5g 3.75g.875g 0.938g 3-9. Using B(t) = A!b t and the points (t, B(t)) = (0, 60) and (t, B(t)) = (3, 30) we get: 60 = A!b 0 " A = 60!!!!!!!!!!!!!!!!!!!!!!!!b 3 = " b = 30 = A!b 3 " 30 = 60!b 3 Thus B(t) = A!b t may be written as: B(t) = 60!(0.794) t. ( ) /3 " b = a. There will be t 3 units in t minutes because t = number of half-lives or because each half 3 life is 3 minutes. b. B(t) = 60! (0.5) t 3 because C = 60 is the original amount and because t is the number of 3 half-lives in t minutes. During every half-life, the amount of Bromine-85 gets multiplied by = Using Total = C! (percent) kt where the original amount saved is C = $400, and the percent of savings left after each year (i.e. k = ) is! 0. = 0.9 % we get: Total = 400!(0.9) t. After t = 5 years, the total amount left is: Total = 400! (0.9) 5 = $36.0. CPM Educational Program 0 Chapter 3: Page 6 Pre-Calculus with Trigonometry

7 Review and Preview a. We have two points: (t, d) = (0,0) and (t, d) = (5, 90). First, write a system of two equations, then solve for k and m. 0 = km 0! 0 = k!!!!!!!!!!!!!!!!!!!!m = = km 5! 90 = 0m The particular equation is 9 d = 0! 5 b. d = 0! = 9 5 " 0.96 ( ) t ( ) 5 = The temperature of the coffee is: = 30.48! F 3-3. a. f () = 3! reflected over the -ais is f () =!(3! ) =!3 +. b. f () = 3! reflected over the y-ais is f () = 3(!)! (!) = a. 4 0 =.587 b. ( 0 ) 4 =.5 c. 0 0, 000 = a. Circumference equation: C =!r =!0cm = 0!cm Area equation: A =!r =!(0cm) = 00!cm b. of the circle is shaded so the area and arc length of the NON-shaded portion are: 4 A = 3 4 (00!cm ) = 75!cm!!!!!!!!C = 3 (0!cm) = 5!cm a.!y b. 5 +!4 c. +! 3-7. a !!! + = " or k !!! + = k=3 b !!! + 30 = ()+ + ()+ + (3)+ +!!! (4)+ = " k+ k= 6 " c !!! + 6 = () 3 + () 3 + (3) 3 +!!! + (6) 3 = k 3 k= 3+ +!!! = " k+ k= CPM Educational Program 0 Chapter 3: Page 7 Pre-Calculus with Trigonometry

8 3-8. a. A = (, 4) and B = (5,!). d = (5! ) + (!! 4) = 5 b. midpoint = +5 ( ) = (3,), 4! c. The slope of the line is: m =!!4 5! =! 6 4 =! 3 Point-slope form is: y! y = m(! ). Use point A = (, 4) to get: y! 4 =! 3 (! )!!or!!y =! 3 (! ) + 4 Use point B = (5,!) to get: y + =! 3 (! 5)!!or!!y =! 3 (! 5)! 3-9. a. At time t = 0 Jenny s heart rate was 85 beats per minute (bpm). b. Jenny s heart rate reached 40 bpm around 7-0 minutes in to her run. Her heart was at least 40 bpm for 5-8 minutes. c. The area under the curve represents the total number of heartbeats in the 5 minutes she was on the treadmill a. Let a = 3, b = IN, c = 6. Using the Pythagorean theorem: 3 + b = 6 b. Using SOH CAH TOA we find cos P = PI = 3 PN 6 = c. sin P = IN = 3 PN 36 = 3 b = 6! 3 = 3 3 = IN Lesson a. f () = () transforms f () by a horizontal stretch by by. b. f () = () =! = 4 = a, thus a = 4. c. g() = a = 4 gives a vertical stretch by 4. or a horizontal compression d. f () = () =! = 4,!!!g() = a = 4,!!!4 f () = 4( ) = 4 Thus f () = g() = 4 f () because they have the same equation. e. Yes, two different transformations may give the same result. This is not true for any function f (). Eample: Eponent Laws: a+b = a! b 3-3. a. Letting y = and g() = 3! we see that g() is a vertical stretch of y by a factor of 3. b. See graph at right. c. The y-intercept of g() occurs when = 0 i.e. g(0) = 3! 0 = 3. This gives the point (0,3). CPM Educational Program 0 Chapter 3: Page 8 Pre-Calculus with Trigonometry

9 3-33. a. h() is a vertical stretch times 3 and a shift up by. b. See graph at right. c. From the graph of h() we can see that the horizontal asymptote occurs at y =. d. The y-intercept occurs when = 0. i.e. h(0) = 3! 0 + = 4. This gives the point (0, 4). e. The y-intercept occurs when = 0. i.e. y = A! + B when = 0 gives y = A! 0 + B = A + B. This gives the point (0, A + B) ! 3 + =! 3! 3 =! 3! 3 = 6! a. y = is shifted left units and stretched vertically by 3 to get y = k(). b. The y-intercept of y = k() = 3! (+) occurs when = 0. y = k(0) = 3! (0+) = 3! = giving the point (0,). c. To ensure m() = A! has the same y-intercept, use the point (, m()) = (0,) to get: = A! 0 = A " A = Thus, m() =!. d. k() = 3! (+) = 3!! = 3!! =! = m() Yes, k() = m() a. f ( + ) = 3! 4 (+) = 3! 4! 4 = 6!(3! 4 ) = 6 f () b. f (! ) = 3" 4 (!) = 3" 4! " 4 = 4 "(3" 4 ) = 4 f () ! 4 (+) = 6! 4! 4 = 6!6! 4 = 96! 4 = A! 4 " A = a. y = 7 (+3) = 7 3! 7 = 343(7 ) b. y = (5! ) + 7 = (5! " 5 ) + 7 = 5 (5 ) + 7 = 0.48(5 ) a. 5 = (5 ) = 5 b. 3!3 = 3!3 " 3 = 3!3 " (3 ) = 7 " (9) 3..3 Review and Preview 3-4. a. 5! 3 (+) = 5! 3! 3 = 5! 3! 3 = 45! 3 = A! 3!!"!!A = 45 b. 5! 5(+4) = 5! 5! 5 4 = 5 "! 5 4! 5 = 5 "+4! 5 = 5! 5 = 5! 5 = A! 5!!#!!A = 5 c. 6! (+4) = 4!! 4 = 4! 4! = 4+4! = 8! = 56! = A!!!"!!A = 56 d. 3! 3(") = 3 "! 3 "! 3 = 3 ""! 3 = 3 "3! 3 = 3 3! 3 = 7! 3 = A! 3!#!A = 7 CPM Educational Program 0 Chapter 3: Page 9 Pre-Calculus with Trigonometry

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11 3-4. a. 6! (4+3) = 6! 4! 3 = 6! 3! 4 = 48!( 4 ) = 48!(6) b. 8! 3 (3"4) = 8! 3 3! 3 "4 = 8! 3 "4! 3 3 = 8! 3 4! (33 ) = 8 8! (7) = 9! (7) a. 8 (+3) = 3 ( 3 ) (+3) = = = 5 3 =!4 =! 4 3 b. 7 = ( 9 ) (!) (3 3 ) = 3 ( )(!) 3 6 = (3! ) (!) 3 6 = 3!+ 6 =! + 8 = = 4 f (!5) =, f (!) =, f (!) =, f (0) =, f () = 0, f () =!, f (3) = 0, f (4) =, f (5) = g(!5) = 6, g(!) = 6, g(!) = 4, g(0) =, g() = 0, g() =, g(3) = 4, g(4) = 6, g(5) = 6 From inspection we see the minimum is shifted to the left by, there is a vertical stretch by and a vertical shift by. Thus, g() = f ( + ) a. c. t + t t 3 = t + t t = t + t! t = t + t b. + y = 3! = 3 d. = +y =! +y = +y =! = =! = = a. " 4i 3! = (4 # 3 3! ) + (4 # 4 3! ) + (4 # 5 3! ) + (4 # 6 3! ) = = 74 i=3 b ( ) = 0.4 ( ) = ! ! ! ! ! i i= a. b. c. 5 " CPM Educational Program 0 Chapter 3: Page Pre-Calculus with Trigonometry

12 3-48. a. From Pythagorean theorem we know: CA + BA = CB!!!! = CB b. sin C = 5 5 = c. =! = = CB 50 = CB 5 = CB a. " 4n! 7 = (4 # 3! 7) + (4 # 4! 7) + (4 # 5! 7) + (4 # 6! 7) + (4 # 7! 7) = n=3 8 b. " (!) j = (!) + (!) 3 + (!) 4 + (!) 5 + (!) 6 + (!) 7 + (!) 8 =! +! +! + j= Lesson = 3, because two of the points have an -value of a. Yes. {(, 6), (!5,), (!7, 4) }is a function because every input () has only one output (y). b. Answers vary. Eample: (, 6), (!5,), (!7, 4), (, 7) corresponds to two outputs: 6 and 7. { } is not a function because the input 3-5. a. = y b. = y is not a function because it does not pass the vertical line test. c. y =, y = 4, y = 3! 4 all have inverses that are not functions. y a. See graph above right. If y = (! ), then the inverse relation is = (y! ). Solving for y yields y = ± +. A range that will make this a function is y! or y!. b. See graph below right. If y =! 4, then the inverse relation is = y! 4. Solving for y yields y = ± + 4. A range that will make this a function is y! 0 or y! 0. y CPM Educational Program 0 Chapter 3: Page Pre-Calculus with Trigonometry

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14 3-55. g() =. Replace g() with y: y = y. Switch and y: = + + y+. Solve for y: (y + ) = y y + = y y! y =! y(! ) =! y =!! =! = g! () Yes. f and f! undo each other. The situation is symmetric f! () = Replace f! () with y: y = Switch and y: = y Solve for y:! 6 = y 3 (! 6) 3 = y (! 6) 3 3 = f ()!!!!!or!!! f () =! 6 Review and Preview a. Yes. Switch and y. b. = y does not pass the vertical line test. c. The graph of y = is the reflection of = y over the line y =. # d. y = = $ for! 0 %" for < f () = + Replace f () with y: y = Solve for y: (y + ) = y y + = y y! y =! y(! ) =! y. Switch and y: = + y+ y =!! "! = f! (y) CPM Educational Program 0 Chapter 3: Page 4 Pre-Calculus with Trigonometry

15 3-60. ( 5 ) (!3) =!!!!!!!!!!!3(! 3) =! 5 ( )(!3) =!!!!!!!!!!!!!!!6 + 9 =! !3(!3) = 5!!!!!!!!!!!!!!!!!!!!6 =!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! = 6 y 3-6. a. + y = y =! y =!! y =!y =! y =! + b. See graph at right. Points of intersections are: (, 0) and ( 5, ) 3-6. a. + y + y = +y +y y = +y! y +y = y b. +y!!y =!(!y)(+y) =!(!y ) = y (+y) (+y) (+ y) a. = 8 (!) = ( 3 ) (!) = 3!3 = 3! 3 3 = = 3 b. 4! = 8 (+) ( )! = ( 3 ) (+)! = 3+6! = 3 + 6!6 = 5 =! 6 5 c. 7 ( +3) = 7 ( +3) = = 0 =!3 =! Let!A = 35!. Let B be the point of the first measurement with angle 4!. Let D be the point at the peak of the mountain. Draw a vertical line from B to create a smaller, right triangle. Let E be the intersection of the vertical line emerging from B. Find the height of the line BE by: tan(35) = BE, BE = 00 tan(35) = We also know the following angles:!abe = 90!,!BAE = 35!!BEA = 80! " 90! " 35! = 55!!BED = 80! " 55! = 5!!EBD = 80! " 90! " 4! = 48!!EDB = 80! " 5! " 48! = 7! CPM Educational Program 0 Chapter 3: Page 5 Pre-Calculus with Trigonometry

16 3-64. Solution continued from previous page. Knowing this, we can find BD : BD sin(5) = sin(7) Now we can find h: sin(4) = h h = sin(4) BD = sin(7)! sin(5) = 5, Solution continues on net page. h = 3, 779. ft a. For the left piece of f () the slope of the line is m = and the y-intercept is b =. Using the equation for a line ( y = m + b ) we get: f () = + for < 0. The right piece of f () is a parabola shifted to the right by and down by 4. Using the equation for a parabola y = ( + h) + k and letting h =! and k =!4 to get the appropriate shifts, we get: f () = (! )! 4 for! 0. # Combine these to get: f () = + for < 0 $ %& (! )! 4 for " 0 b. See graph at right. g() = f ( + ) + #( + ) + + for + < 0 = $ %& (( + )! )! 4 + for + " 0 # + 5 for <! = $ %&! 3 for "! To find b so that there is only one intersection point, the discriminant must equal 0. The discriminant is the part of the quadratic formula that is inside of the Square root. + b =! 6 + 6! 8 + (6! b) = 0 Using the Quadratic Formula:, = 8± = 8!4(6!b) 8± 40+4b There is only one intersection point if: b = 0 40 =!4b b =!0 g() CPM Educational Program 0 Chapter 3: Page 6 Pre-Calculus with Trigonometry

17 Lesson a. b. = 5 y y 65 4 / Undefined 0 Undefined 5 /5.36 / a. log 5 (65) = 4 because 5 4 = 65. b. y = log 5 () a. log 5 (65) = 4 b. log 5 (5) = 3 because 5 3 = 5. c. log 5 (5) = because 5 = a. log 8 = 3 because 3 = 8. b. log 6 (96) = 4 because 6 4 = 96. c. log 7 49 ( ) =! because 7! = 49. d. log 9 3 = because 9/ = a. y = 7 can be rewritten as = log 7 y. b. log 4 = y can be rewritten as 4 y =. c. y = can be rewritten as y = log. d. W k = B can be rewritten as log W B = K. e. K = log W B can be rewritten as B = W K. f. log 3 P = Q can be rewritten as P = 3 ( ) Q log00 =, log(3.6) = 0.5, etc indicates that the calculator uses base 0 because: 0 = 00,0 0.5 = 3.6, etc. CPM Educational Program 0 Chapter 3: Page 7 Pre-Calculus with Trigonometry

18 Review and Preview log b = a means that b a =. If b 0, then will alternate between positive and negative values or be undefined a. From Pythagorean theorem: (RA) + n = (n) (RA) = (n)! n (RA) = 4n! n (RA) = 3n RA = n 3 b. i. sin R = n n = c. sin R cos R = 3 =! 3 = 3 = tan R ii. cos R = n 3 n = 3 iii. tan R = n = = 3 n a. log 5 5 = because 5 = 5 b. log 5 = 0 because 5 0 = c. log 5 5 ( ) =! because 5! = a. log 3 8 = 4 because 3 4 = 8 b. 3 = 9. 9 goes into the log 3 machine. log 3 9 = because 3 = 9. c. 3 = y. log 3 y = log 3 3 = because 3 = 3. d. log 3 = y e. = 3 y f. Yes. 3 > 0, so y = log 3 means = 3 y Area of a triangle is A = b! h where b = 5cm. Let!A = 36!,!B = 5! then, if we call this triangle!abc then!c = 80! " 36! " 5! = 9!. Using the Law of Sines: CB sin(36º ) = 5 sin(9º ) so CB = 5 sin(36º )! 0.08cm sin(9º ) From C, draw a line perpendicular to AB. Let D be the point where this line intersects AB. Because!BCD is a right triangle we know: sin(5º ) = CD = sin(5º )!0.08 = 4.60cm. Thus, the height of the triangle is h = 4.60 cm and the area is: A = (5cm)(4.60cm)=3.95cm CD 0.08 CPM Educational Program 0 Chapter 3: Page 8 Pre-Calculus with Trigonometry

19 3-78. a. To break the interval (, 5) into si pieces, the width,, of the rectangles will be (5! ) = 6; =. The area of each rectangle is A = b! h where b = and h = 3. The area in sigma notation using left-endpoint rectangles is: 3(0.5k + ) b. To use right-endpoint rectangles: 3(0.5k + ) 6! k= c. To use midpoint rectangles: 3(0.5k +.5) 5! k=0 d. To estimate the area using trapezoids, average the left- and right-endpoint results f (!4) =!6, f (!3) =!, f (!) =,! f (!) = 3, f (0) =, f () =!,! f () =, f (3) = 3, f (4) = 5 g(!) = 7, g(0) =, g() =!, g() =!, g(3) =!, g(4) =, g(5) = 0, g(6) =!, g(7) =!4 It is clear from the graph that g() is a vertically flipped version of f (). The function is then shifted to the right by 3 and up by to get: g() =! f (! 3) +. 5! k= a. 8(4) = 3 ( 3 )( ) = (! ) 3 3+ =! =! =! 7 =!9 =! 9 7 b. ( 5 ) (+) = 5 (5! ) (+) = ((5 3 ) ) 5!! = 5 3/!! = 3!4! 4 = 3!4 = 7 =! a. f () = g() when =! + + 4!!"!!!! 4 = 0. = ± (!4)!4()(!4) ± 36 = = ±6 =!, () 4 4 When =!, f (!) = g(!) =. When =, f () = g() = 4. Thus, the two intersection points are (!,) and (, 4). b. " (! + + 4! )d = 9 units! CPM Educational Program 0 Chapter 3: Page 9 Pre-Calculus with Trigonometry

20 Lesson a. Let y = log. This can be rewritten as y =. The inverse of this function is: = y. b. The domain of = y is all real numbers, the range is y > 0. c. See graph at right. d. See graph at right. e. See graph at right. f. Domain: > 0, Range: all real numbers. g. Domain: > 0, Range: all real numbers. h. The graphs have the same general shape. The graph of y = log is vertically stretched. They have the point (, 0) in common. y a. To get y = 5 log, stretch y = log vertically by a factor of 5. The vertical asymptote is = 0. b. To get y = log! 3, shift y = log down 3 units. The vertical asymptote is = 0. c. To get y = log (! 3), shift y = log 3 units to the right. The vertical asymptote is = 3. d. To get y =! log, reflect y = log across the -ais. The vertical asymptote is = 0. e. To get y =! log ( + 4) +, shift y = log 4 units to the left then reflect y = log across the -ais, and shift if up by unit. The vertical asymptote is =! a. Domain: > 0. Range: all real numbers. b. Domain: > 0. Range: all real numbers. c. Domain: > 3. Range: all real numbers. d. Domain: > 0. Range: all real numbers. e. Domain: >!4. Range: all real numbers. y =! log y =! log ( + 4) + y = log! 3 y = 5 log y = log y = log (! 3) f () = log = 0 when = 0 = 04. f () = log = 0 when = 0 =, 048, CPM Educational Program 0 Chapter 3: Page 0 Pre-Calculus with Trigonometry

21 To get a y-value of 0 you would have to go, 048, 576 units or 0.5 inches!, 048, 576 units = 6,44 inches or = foot! 6,44 inches =, feet unit inches or = mile =, feet = 4.37 miles 580 feet Review and Preview See graph at right. f () : Domain: all real numbers. Range: y > 0. There are no zeros. g() : Domain: > 0. Range: all real numbers. Zeroes: = f () = 3 and f! () = log 3 a. f (4) = 3 4 = 8 b. f! (8) = log 3 8 = 4 c. f (!) = 3! = 9 d. f! 9 ( ) = log 3 9 =! e. f See graph at right. f () = log 5 ( + ) Domain: >! Range:!" < y < " ( ) = 3 = 3 f. f! ( 3 ) = log 3 3 = 3-9. a. 4! y = ( + y)(! y) b. 9z! y = (3z + y)(3z! y) 3-9. f () = +3. Let y = +3 y+3. Switch and y: = y. Solve for y: y = y y! y = 3 y(! ) = 3 y = 3! f! () = 3! a. Vertical Stretch by a factor of. b. Horizontal compression by a factor of. f () f () CPM Educational Program 0 Chapter 3: Page Pre-Calculus with Trigonometry

22 Solution continues on net page. CPM Educational Program 0 Chapter 3: Page Pre-Calculus with Trigonometry

23 3-93. Solution continued from previous page. c. Flipped over the -ais. d. Flipped over the y-ais.! f () f (!) a.! " + " 6 ( ) =! 7" 6 b.! +! +! 4 = 7! a. As increases from to 4, y increases from 36 to 8. Thus, the function is increasing. b. Use the form y = A!b and the points (, 36) and (4, 8). First, substitute in (, 36): 36 = A!b. Solve for A : 36!b " = A. Use this value of A when substituting in the point (4, 8) : 8 = 36! b "!b 4. Solve for b: 8 = 36!b.5 = b.5 = b Use this value to solve for A : 36!b " = A 36! (.5) " = A 6 = A.5 = b Thus, the eponential function that passes through the two points is: y = f () = 6(.5). c. To have a horizontal asymptote of y = 0, the function in part (b) must be shifted up by 0: f () = 6(.5) + 0 CPM Educational Program 0 Chapter 3: Page 3 Pre-Calculus with Trigonometry

24 a. (! y ) ( + y ) + y! y! y! y y! y b. y! y y + y y! y y + y + + y! +y + ( y! )+ +y ( ) +y y +y CPM Educational Program 0 Chapter 3: Page 4 Pre-Calculus with Trigonometry

25 Lesson a. Yes the entries are correct. b. No, this relation does not hold. c. log + log 3 = = d. log + log 4 = = log 6 = log + log 3 = log 6 e. log 3 + log 4 = =.079 log =.079 log 3 + log 4 = log log 8 = log + log 4 = log 8 f. log + log y = log y a. log log 3 7 = 5 and log 3 (9! 7) = log 3 43 = 5, thus, log log 3 7 = log 3 (9! 7). b. log log 4 4 = 5 and log 4 (8! 4) = log 3 3 = 5, thus, log log 4 4 = log 4 (8! 4). c. Both sides =! Problem 3-98 does not prove that log b + log b y = log b y. Showing a relation holds for three cases does not show it holds for all cases. This pattern may break down with more eamples. We have not shown why this pattern is always true, yet a. log 6! log = 0.778! 0.30 = and log 3 = 0.477, thus, log 6! log = log 3. b. log 8! log 4 = 0.903! 0.60 = 0.30 and log = 0.30, thus, log 8! log 4 = log. c. log! log y = log y 3-0. Let M = 8 and N =. Then log 8 ( ) = log 4 = 0.60 log 8 log = = 3 ( )! log 8 log 8 log 3-0. a. log = (0.30) = 0.60 = log 4 b. 3 log = 3(0.30) = = log 8 c. 4 log = 4(0.30) =.04 = log 6 d. n log = log n e.! log 5 = log 5! = log 5 f. 0.5 log 64 = log = log Let log = N so that 0 N =. Let log y = M so that 0 M = y. log y ( ) ( y ) = 0N 0 M = 0N!M ( ) = log 0 N!M ( ) = log! log y log y ( ) = N! M CPM Educational Program 0 Chapter 3: Page 5 Pre-Calculus with Trigonometry

26 3-05. Let log = M so that 0 M =. ( ) n = 0 Mn ( ) = Mn n = 0 M log n = log 0 Mn log n = n log Review and Preview a. ln(3! ) =.79 and ln 3 + ln =.79. Thus, ln(3! ) = ln 3 + ln. ln 3 = and ln 3! ln = Thus, ln 3 = ln 3! ln. ln 3 =.97 and ln 3 =.97. Thus, ln 3 = ln 3. The three log laws work in these and all other cases. b. log b b = because b = b. c. Eample, ln 3 = = 3 when =.78. The base of the natural logarithm is = a. log 3 8 = 4 because 3 4 = 8. b. log 5 5 = because 5 = 5. c. log 4 6 =! because 4! = a. log 4 + log! log 5 = log(4 " )! log 5 = log 8! log 5 = log 8 5 b. log M + log N = log M + log N = log M! N ( ) = log MN log 4 = y can be rewritten as 4 y =. 4 y = y = can be rewritten as log = y. Solving for y gives: log = y a.! +! +! 6 = 8! 3 b.! "! " 6 =! 5" a. f () = y = 3! 5. Switch and y: = 3 3y! 5. Solve for y: 3 = 3y! = 3y = y = f! () CPM Educational Program 0 Chapter 3: Page 6 Pre-Calculus with Trigonometry

27 3-. Solution continued from previous page. b. g() = y =! y!. Switch and y: = 3! 3!y c. h() = y = log 3 (! ). Switch and y: = log 3 (y! ). Solve for y: 3 = y! y = 3 + = h! () Solution continues on net page.. Solve for y: (3! y) = y! 3! y + = y 3 + = y + y 3 + = y( + ) y = 3+ + = g! () 3-. a. Since this is a right triangle we know: AC + BC = AB b. Using the law of sines: c. sin A = d. sin 45! = always. n sin A = n sin 90! n sin A = n sin A = n n = no matter the value of n. e. cos 45! = cos A = n n = always. (n ) + (n ) = AB n + n = AB 4n = AB n = AB or 3-3. f () will be continuous at = 0 if a + b = a + 5 when = 0 or b = 5. f () will be continuous at = if a + 5 = 3! b when = or a + 5 = 3! b = 3! 5 =! a =!! ( ) + = 6 ( ) ( ) = 8( ( ) ) = 8 ( 4 ) a. y = 6 b. y = 00(5)! = 00(5) (5)! = 4(5 ) = 4(5) a =! 7 CPM Educational Program 0 Chapter 3: Page 7 Pre-Calculus with Trigonometry

28 Lesson Given equation.05 = Take logs of both sides log.05 = log Power Law for logs log.05 = log Divide both sides by log.05 = log log.05 Use calculator to evaluate = Let = # of years and a = prices. Then prices have doubled at a. The inflation function is a(+ 0.05). Prices have doubled when a = a(.05)! = (.05). log = log(.05) log = log(.05) = log log(.05) = 4.07 = 4.07 years = 40 and 40 = 600, so maybe 40.5 = 400? Actual answer: =.64 log 8 = 3 because 3 = 8. log 3 7 = 3 because 3 3 = a. Yes he is right. If log 6 60 = then we know that 6 = 60. b. We currently do not know how to find log c. 6 = 60 log 6 = log 60 log 6 = log 60 = log 60 log 6 = 3.03 d. Yes, because the -value that satisfies 6 = 60 also satisfies = log a. = log 3 can be rewritten as 3 =. Using the previous method: b. To graph y = log 3, we can use y = log log 3. c. The range of y = log 3 is all real numbers. The calculator plots this function by plotting discrete points and then connecting them with segments. The net -value to the left of = 0. is = 0 and log 3 0 is undefined, so the graph stops at = = log 3 = log log 3 = log = log log 3 =.83 CPM Educational Program 0 Chapter 3: Page 8 Pre-Calculus with Trigonometry

29 3-0. a. log 3 8!.9w because 3 = 9 so 3.9! 8. b. Let log 3 8 =. This can be rewritten as 3 = 8. Taking the log of both sides as before: c. log 3 = log 8 log 3 = log 8 = log 8 log 3 Now set the two values of equal to each other: = log 3 8 = log 8 log 3 w log 8 log 3 =.893 d. log 3 8 =.893 can be rewritten as: = 8. It checks. 3-. a. The initial amount is 0mg, hence 0(0.9) 0 = 0. Also, each hour, 90% of what was there the hour before remains. b. 0(0.9) t = when (0.9) t = = 0.6. Taking the log of both sides: 0 c. t = log(0.6) = or t = 4 hours 5 mintues. log(0.9) Review and Preview log(0.9) t = log(0.6) t log(0.9) = log(0.6) t = log(0.6) log(0.9) 3-. a. Estimate 0 = 36 has a solution!.9 because 0 = 400. Actual solution: 0 = 36 log 0 = log 36 = log 36 log 0 =.9 b. Estimate (7.3) = 4.8 has a solution! 0.6 because (7.3) = 7.3. Actual solution: (7.3) = 4.8 log(7.3) = log(4.8) = log(4.8) log(7.3) = c. Estimate 60(0.5) = 8 has a solution! 4. because when = 4 : Actual Solution: 60(0.5) = 8 (0.5) = 8 60 = 0.05 log(0.5) = log(0.05) = log(0.05) log(0.5) = (0.5) 4 = 60 ( ) 4 = 60 4 = 60 6 = 0 CPM Educational Program 0 Chapter 3: Page 9 Pre-Calculus with Trigonometry

30 3-3. a. log 3 = 6 6 = 3 = ( 6 ) 3 = = a. 00(.05) = 000 (.05) = = 5 log(.05) = log 5 = log 5 log.05 = a. log m! 3 log n + log p 3-6. = log m! log n 3 + log p = log m p n 3.05 = (.05 ).05 =.05 =.935 b. log 4 + log 4 3 = log 4 3 = 4 = 3 = 4 3 = 6 3 b. 0(.5)! 400 = 600 0(.5) = = 50 log.5 = log50 = log 50 log.5 = 4.69 b. (log a + log b! 3 log c) = (log a + log b! log c 3 ) = ( ) log ab c 3 = log ab c 3 ( ) = log ab c b. log b = can be rewritten as b =. Solving for b: b ( ) = , b = a. Using Pythagorean theorem we get: PQ + n = (n) = (n)! n = 4n! n = 3n b. sin P = n n = c. cos P = n 3 n = 3 d. From the Law of Sines we know: n sin Q = PQ = 3n = n 3 n sin P ( ) = 30! We also know sinq = sin(90) = so: n = n sin P!!!!!sin P =!!!!!P = sin" e. The ratios do not depend on the value of n, so they will always be the same as those found in (b) and (c). CPM Educational Program 0 Chapter 3: Page 30 Pre-Calculus with Trigonometry

31 3-9. a.! +! = 3! b.!"! "! "! " 4 =! 5" a. The equation of a circle with center (h, k) and radius r is (! h) + (y! k) = r. Here, the center is (0, 0) and the radius is r = so the equation of the circle is: + y =. b. Let = then ( ) + y = y =! 4 = 3 4 y = ± 3 4 = ± f () =! can be approimated with: #(! 3) + 4 for " 3 # f () = $ or f () = $ + for! 3 %!(! 3) + 4 for < 3 %" + 7 for < 3 Closure Merge Problem temp. 300º 3-3. a. See graph at right. 75º b. f (t) = am t + k where k is where equilibrium value of the pie, i.e. the room temperature. k = 75! time F c. Using the form f (t) = y = am t + 75 and the points (t, y) = (, 33) and (t, y) = (5, 88) we can find values for a and m. 88 = am = am + 75! 33 " 75 = am 48 = am 48m " = a! 88 = 48m " m " 75 = 48m 5" 3 = 48m = m3 ( ) 3 m = 3 48 m = and a = 74.5, thus: f (t) = 74.5( ) t + 75 d. When t = 0, f (0) = 74.5( ) = = e. When the internal temperature of the pie reaches 0!. 48m! = a " ( ) # $ %! 48 ( 3 48 )! 3 = a = a CPM Educational Program 0 Chapter 3: Page 3 Pre-Calculus with Trigonometry

32 0 = 74.5( ) t = 74.5( ) t = ( )t Closure Problems 0.64 = ( ) t log 0.64 = t log( ) log 0.64 log( ) = t t = 35.7 minutes CL a. b. c. d. CL a. As increases from 4 to 6, y decreases from 80 to 0, thus the function is decreasing. b. The function is quartered as increases units, thus it is halved as increases by unit. c. The base or multiplier for the eponential function is. Using the form of an eponential function y = m ( ) and either point (, y) = (4, 80) or (, y) = (6, 0) we can find m. 80 = m ( ) 4 80 = 6! m " 80!6 = m 80 = m y = 80 ( ) CL a. f () = 3() + = 3!! = 3!! =! b. f () = 90(.5)! = 90 ".5 ".5! = 90 ".5! ".5 = 40 ".5 CPM Educational Program 0 Chapter 3: Page 3 Pre-Calculus with Trigonometry

33 c. f () = 4 3 (3)+ = 4 3! 3! 3 = 4 3!(3 )! 3 = 4! 9 d. f () = 64(4)! = 64 " 4 " 4! = 64 "(4 ) " 6 = 4 " CPM Educational Program 0 Chapter 3: Page 33 Pre-Calculus with Trigonometry

34 CL a. f () = (! 3) 3 +. Rewrite as y = (! 3) 3 +. Switch and y and solve for y: 3 3 = (y! 3) 3 +! = (y! 3) 3! = y! 3! + 3 = y = f! () b. g() =! + = y! y+ (y + ) = y! y + = y! + = y! y + = y(! ) can be rewritten as y =!. Switch and y and solve for y: + +! = y = g! () c. h() = 5! can be rewritten as y = 5!. Switch and y and solve for y: = 5! y 5 = y log = log y 5 log 5 = y log log 5 log = y = h" () d. k() = log 3 (! 5) can be rewritten as y = log 3 (! 5). Switch and y and solve for y: = log 3 (y! 5) 3 = y! = y = k! () CL a. log 6 = 4 because 4 = 6. b. log 3 9 ( ) =! because 3! = 9 c. log 5 5 = because 5 = 5 CL a. Horizontal shift of units to the right, vertical stretch by a factor of, reflection over -ais. b. Domain: (,!) c. See graph at right. CPM Educational Program 0 Chapter 3: Page 34 Pre-Calculus with Trigonometry

35 CL a. log 3! log 3 (! ) = b. 3 log () + log (7) = log 5 (5) log First notice that log 5 5 = 3. 3! = 3 log + log 7 = 3 3 =! log 3 + log 7 = 3 9(! ) = log 7 3 = 3 9! 8 = 3 = = 8 = 8 8 = = 3 ( ) 3 = ( 3 ) 3 = CL a. 0(.05)! 50 = 50 0(.05) = 300 (.05) = 5 c. 3 log 4 (! ) = 6 log 4 (! ) = = log.05 5 = 4 =! 6 + = = 8 log 5 log.05 = = 3 b. 5(! ) 3.7 = 50 (! ) 3.7 = 50 5 = 0 3 ( ) 3.7! =.385 ((! ) 3.7 ) 3.7 = 0 3 = CL 3-4. a. Using the Law of Sines, sin 90 = base sin 45 b. sin 90 = base sin 60 base = sin 60 sin 90 = 3 base = sin 45 sin 90 = The process repeats for the height of the triangle. sin 90 = height sin 30 sin 30 height = sin 90 = CL 3-4. a.! 6 b.! 4 c.! 3 CL CPM Educational Program 0 Chapter 3: Page 35 Pre-Calculus with Trigonometry

36 See graph at right. CPM Educational Program 0 Chapter 3: Page 36 Pre-Calculus with Trigonometry