Solutions to Final Review Sheet. The Math 5a final exam will be Tuesday, May 1 from 9:15 am 12:15 p.m.

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1 Math 5a Solutions to Final Review Sheet The Math 5a final exam will be Tuesday, May 1 from 9:15 am 1:15 p.m. Location: Gerstenzang 1 The final exam is cumulative (i.e., it will cover all the material we covered in class this semester). However, the material covered since the last exam (Section.7, Modeling with Functions, Sections and Trigonometry) will make up at least 60% of the exam: Section.7: One-to-one Functions and their Inverses Modeling with Functions Section 4.1: Exponential Functions Section 4.: The Natural Exponential Function Section 4.: Logarithmic Functions Section 4.4: Laws of Logarithms Section 4.5: Exponential and Logarithmic Equations (and Inequalities) Trigonometry: Section 6.1: Angle Measure Section 6.: Trigonometry of Right Triangles Trig Ratios in a Circle, Families of Angles, Unit Circle, Trig Equations and Trig Graphs You must bring your student ID to the final exam. On the exam, you will be asked to show all your work. We are grading your ability to clearly communicate your understanding of how to solve each problem much more than your ability to get the correct answer. In fact, a correct answer with little or no work might not receive any points at all. Showing your work clearly also gives us justification for assigning partial credit. No calculators will be allowed on the exam. Our exams are closed book: no notes, books, calculators, cell phones or internetconnected devices will be allowed. The solutions to the review sheet are available on LATTE. OVER for Practice Exam & Problems

2 Practice Exam (Problems 1-19) 1. Simplify: (a b 1 c) abc (a b 1 c) abc a 6 b c abc a 7 b c 4 or b a 7 c 4. Solve the following equations: (a) x x 1 0 (b) 1 x 1 ± ( 1) 4(1)( 1) (1) 1 ± ± 5 (c) 1 + x + 1 x x + 1 x ( 1 (1 + x)(x) 1 + x + 1 ) (1 + x)(x) x x + (1 + x) (x + x ) x + 1 x + x 1 x 1 x ± 1 x x x 4 0 This fraction is equal to 0 only if the numerator is 0, i.e., only if x 0, but x is never 0, so there are no solutions.. Evaluate the following. If an expression doesn t exist, just say so. (a) ln e - e e1/ (b) ln ln e e (c) log 1 0 ln e /

3 (d) log 0 undefined 4. Suppose Janet bought an iphone last year for $650. She found out that if she wanted to sell it now, she could sell it for $60. To determine how much it might be worth in the future, you decide to model the current sale value for her with a linear function so that the value, V, of the iphone is a function of time, with time t 0 corresponding to when she bought the phone. (a) Find a linear equation expressing the value of the iphone as a function of time (in years). We have two points: (0, 650) and (1, 60), so the slope is m The equation is then V (t 0) or V 90t (b) Sketch a graph of your linear function from part 4a. (c) What does the slope of your line represent? (Explain in words.) The slope of the line represents how much the value of the iphone decreases each year. 5. Find the inverse function of f(x) x +7. Swap variables and then solve for y: x y +7 log x log ( y +7 ) log x y + 7 log x 7 y log x 7 y So f 1 (x) log x 7.

4 6. Let h(x) cos(x x). Find two functions f(x) and g(x) such that h(x) (f g)(x). Note: You may not use the functions f(x) x or g(x) x. Take g(x) x x and f(x) cos x. 7. Solve the following equations: (a) e 4k (b) ln(x + 1) + ln(x ) ln e 4k 00 e 4k ln 00 ln e 4k ln 00 4k ln 00 k 4 ln(x + 1) + ln(x ) ln 4 ln((x + 1)(x )) ln 4 ln(x x ) ln 4 e ln(x x ) e ln 4 x x 4 x x 6 0 (x )(x + ) 0 This gives x or x, but x is not in the domain of ln(x + 1) (or of ln(x )), so we reject it as a solution. The only solution is x. 8. Let f(x) x f(x + h) f(x). Find x 1 h and simplify as much as possible.

5 f(x + h) f(x) h x+h x x+h 1 x 1 h 1 [ x + h h x + h 1 x ] x 1 1 [ ] (x + h)(x 1) h (x + h 1)(x 1) x(x + h 1) (x + h 1)(x 1) 1 [ ] x x + hx h h (x + h 1)(x 1) x + xh x (x + h 1)(x 1) 1 [ ] x x + hx h (x + xh x) h (x + h 1)(x 1) 1 [ ] x x + hx h x xh + x h (x + h 1)(x 1) 1 [ ] h h (x + h 1)(x 1) 1 (x + h 1)(x 1) 9. Evaluate the following. If an expression doesn t exist, just say so. All angle measures are in radians. (a) sin π (b) tan π 1 0 undefined (c) csc 11π 6 (d) cos 15π In each of the following, find the values θ in [0, π] that make the equation true. (a) tan θ tan θ tan θ 1 tan θ 1 Family is π 6, quadrants are II and IV, so θ 5π 6 (b) 4 cos θ 0 11π and θ 6

6 4 cos θ 0 4 cos θ cos θ 4 cos θ ± Family is π; for +, the quadrants are I and IV, so θ π 6 6 quadrants are II and III, so θ 5π and θ 7π 6 6 (c) θ sin θ sin θ 0 11π and θ ; for 6 θ sin θ sin θ 0 sin θ(θ 1) 0 sin θ(θ 1)(θ + 1) 0 So sin θ 0 or θ 1 0 or θ From sin θ 0, we get θ 0, θ π and θ π. We also have θ 1. We would have θ 1, but 1 is not in the interval [0, π], so we reject this solution. 11. Make an accurate sketch of the function 1 x < x f(x) (x + 1) 1 x < 0. cos x + x 0 on the axes below. Note: Your graph must extend to both ends of the x-axis π π π π 1

7 1. Find all angles in the interval [0, π] that are in the θ π 7 family. (Of course, π 7 is not one of our special angles.) We want the angles that are π 7 away from the x-axis, i.e., θ π 7, 6π 7, 8π 7, and 1π 7 1. Suppose you are on a city committee tasked with designing a new park. The area of the rectangular park will be 1, 000m. Your committee decides to have a brick wall on one side of the park that will cost $90/m. The other three sides will have a low stone wall that costs $0/m. Express the total cost of building the surrounding walls as a function of the length of the brick wall. Let l be the length of the brick wall, and w be the length of the other side of the rectangular park. Then the cost of building the walls is Cost 90l + 0l + 0w + 0w 10l + 60w We want this to be a function of l so we use the other information in the problem to find w in terms of l. In particular the area is 1000 l w so w Plugging this l into the cost function gives ( ) 1000 C(l) 10l + 60 l 14. Find the domain of the function g(x) notation. We need x 16 x analysis: x 16 x. Write your answer in interval 0. This is undefined at x and has zeros at x ±4. Do sign So the domain is [ 4, ) [4, ).

8 15. Let f(x) be the function graphed below: 5 y (a) Find the domain of f(x). [ 4, 1) ( 1, ] [, 4] (b) Find the range of f(x). ( 1, 4] (c) For what values, if any, is f(x) 1? x, 1, (d) Find (f f)(). f(f()) f(4) (e) On what intervals is f(x) increasing? (0, ) (, 4) (f) On what intervals is f(x) positive? [ 4, ) ( 1, 0) (0, ] [, 4] (g) Is f(x) one-to-one? Why or why not? No. It does not pass the horizontal line test, so there are some y values (e.g., y 1) which have more than one associated x-value. x 16. Solve the following inequality. Write your answer in interval notation. (ln x + 1)(ln x ) > 0 We find the zeros: ln x ln x 1 x e 1 and ln x 0 ln x x e. We do sign analysis:

9 So the solution set is (0, e 1 ) (e, ). 17. Consider the equation sin θ cos θ 0. Solve this equation (i.e., find all values of θ in the interval [0, π] that make the equation true) by using one of the two methods below: (you can choose whichever method you prefer) Factor the left hand side and then solve, OR sin θ cos θ 0 (sin θ cos θ)(sin θ + cos θ) 0 sin θ cos θ or sin θ cos θ From sin θ cos θ we get θ π 4 and θ 5π 4 and from sin θ cos θ we get θ π 4 and θ 7π 4. Solve by using a trig identity to write the above equation in terms of cos θ. From the identity sin θ + cos θ 1, we have sin θ 1 cos θ, so: sin θ cos θ 0 1 cos θ cos θ 0 1 cos θ 0 1 cos θ 1 cos θ ± 1 cos θ From + 1 cos θ we get θ π and θ 7π 4 4 and θ 5π. 4 and from 1 cos θ we get θ π Make an accurate sketch of the function { ln(x 1) + 1 x > 1 f(x) e x x 1.

10 Note: You must label (with both x- and y-coordinates) at least points on each piece of the graph. (e+1,) 1 (,1) (0,-1) (-1,-e) 19. Determine whether each of the following two statements is true or false. You must justify your answer. Note: In mathematics, true means that the statement must always be true. False means that the statement may sometime be false. (a) True or False: The domain of f(x) ln(ln x)) is (1, ). True: The domain of ln x is (0, ), so we need the input (ln x) to be greater than 0, i.e., we need ln x > 0. This is true for x > 1, i.e., for x in (1, ). (b) True or False: Let p(x) 0 be a quadratic equation. If the discriminant of p(x) is equal to 1, then there is exactly one solution to the equation above. False: Since the discriminant is positive, the equation p(x) 0 will have two solutions. OVER for Practice Exam

11 Practice Exam (Problems 0-6) 0. (6 points) Convert to exponential form and simplify: x y z 1 5 x y z x y z 1 5 x 1 y z 1 5 x y z x 1 y 1 z x 1 1 y 1 1 z 5 1 x 1 6 y z (11 points) Solve the following equations: (a) 1 x + 1 x [ 1 x x + 1 ] (x ) x 1 + x x (b) e x e x e x 7 0 x x 1 x 1 ± 1 4()( 1) ( 1) x 1 ± 1 No solutions, since e a is positive for any a. (c) (ln x )(ln x + 7) 0 ln x 0 or ln x ln x or ln x 7 x e or x e 7. (8 points) Evaluate the following. If an expression doesn t exist, just say so. (a) ln 0 Does not exist. (The domain of ln x is (0, + ).) (b) ln e e ln e e ln e1 (c) log 5 1 log e 1 ln e 1 1 ln e 1 1

12 (d) log 10 log (8 points) Sketch the graph of the following function on the axes below. Make sure you clearly label at least points (with x- and y- coordinates) on each piece of the graph (so 4 points total). { x if x 0 f(x) ln x if 0 < x 4 (-1,) (0,1) 1 (1,0) (e,-1) (5 points) Verify that f(x) x + x 1 Check f(g(x)) x: and g(x) x + x f(g(x)) ( x+ x ) + ( x+ x ) 1 ( ( x+ x ) + ( x+ x ) 1 ) are inverses. (Show all work.) x x (x + ) + (x ) (x + ) 1(x ) x x 6 x + 6 x + 9x 9 x

13 Check g(f(x)) x: x+ x 1 g(f(x)) + ( x+ x 1) ( x+ x 1 + ) ( x 1 ) x+ x 1 x 1 x + + (x 1) (x + ) (x 1) x + + 6x 9x + 6 9x + 9x 9 x Since f(g(x)) x and g(f(x)) x, we know f and g are inverses of each other. 5. (6 points) A model for the number of people N in a college community who have heard a certain rumour is given by N P (1 e 0.15d ) where P is the total population of the community and d is the number of days elapsed since the rumour began. In a community of 1000 students, how long will it take until half of the students have heard the rumour? Note: You won t be able to find a number you should solve until you get an expression that you would plug into your calculator to get a number (1 e 0.15d ) e 0.15d 0.5 e 0.15d 0.5 e 0.15d ln 0.5 ln(e 0.15d ) ln d ln 0.5 d (8 points) Let f(x) f(x + h) f(x). Find x h and simplify as much as possible.

14 f(x + h) f(x) h (x+h) x h x x (x+h) (x+h) x (x+h) h x (x +xh+h ) x (x+h) x (x+h) h x x +6xh+h ) x (x+h) x (x+h) h x x 6xh h ) x (x+h) h 6xh h x (x+h) ( h h( 6x h) x (x + h) 6x h x (x + h) ) 1 h 7. (8 points) Evaluate the following expressions. If the expression does not exist, just say so. All angle measures are in radians. (a) cos(5π/6) cos(5π/6) (b) tan(π/) tan(π/) 1 0 undefined (c) sin(7π) sin(7π) 0 (d) csc( 4π/) csc( 4π/) 1 sin( 4π/) 1 8. (10 points) Solve the following equations. (a) x 16x x 16x x 16x 0 x(x 16) 0 x(x 4)(x + 4) 0 x 0, x 4, x 4

15 (b) log x + log(x ) log(x + 4) log x + log(x ) log(x + 4) log(x(x )) log(x + 4) 10 log(x(x )) 10 log(x+4) x(x ) x + 4 x x x + 4 x x 4 0 (x + 1)(x 4) 0 x 1, x 4 But x 1 is not in the domain of the log functions in our original equation, so we throw this out. The only answer is x (5 points) Find the domain of the function f(x) 1. Write your answer in interval ln(x) notation. There are two things to consider: the domain of ln x and the denominator. The domain of ln x is (0, + ). We also cannot have the denominator equal to 0, so we solve ln x 0 x 1, which means x 1 is not in the domain of f. So the domain of f(x) 1 is (0, 1) (1, + ). ln(x) 0. (15 points) Solve the following inequalities. Write your answers in interval notation. (a) x 9 x + x 6 0 x 9 (x + )(x ) First factor: x + x 6 (x + )(x ). Set numerator and denominator equal to 0 to find the places where the expression might change sign. We get x, x and x. Sign Analysis:

16 x 9 So 0 on (, ) (, ) [, + ). x + x 6 (b) (ln x + 1)(ln x 1) < 0 This one is already factored for us, so we set each factor equal to 0: ln x ln x 1 x e 1 1 e ln x 1 0 ln x 1 x e Sign Analysis: (Note that the domain of ln x is (0, + ), so we don t need to include negative numbers in our sign analysis.)

17 So (ln x + 1)(ln x 1) < 0 on ( 1 e, e). 1. (8 points) The secant line to the function f(θ) cos(θ) between 0 and π is the line going through the points (0, 1) and (π, 1). (a) Find an equation for this line. Find the slope: m 1 1 π 0 π Equation of the line: y 1 (θ 0) π y π θ + 1 (b) Find an equation for the line perpendicular to the line you found in part 1a and going through the point (π/, 0) A perpendicular line will have slope m π. Equation of the line: y 0 π (θ π ) y π θ π 4. (11 points) Consider the given function h(x) sin (x). (a) Find functions f(x) and g(x) such that h(x) (f g)(x). Note: you may not use f(x) x or g(x) x in your answer. One pair is f(x) x and g(x) sin x. A different correct answer is f(x) x and g(x) sin x. (b) Solve h(x) 1. h(x) 1 sin (x) 1 sin (x) 1 sin(x) ± 1 The family of angles is the π 4 and II, so this gives answers of θ π and θ π 4 4 Quadrants III and IV, so this gives answers of θ 5π and θ 7π. 4 4 family. From sin(x) 1, we have Quadrants I. From sin(x) 1, we have

18 (c) Is h(x) a 1-1 function? why or why not? No. From b above, we know that h(π/4) 1 and h(π/4) 1. Since two values of θ are sent to the same y value, h is not 1-1. (It would not pass the Horizontal Line Test.). (8 points) A wire of length 1 feet is cut into two pieces and each piece is bent into a square. Let x be the length of the side of one square and y be the length of the side of the other square. Write the sum of the areas of the two squares as a function of x. The sum of the areas will be: Total Area x + y. We want this as a function of x, so we go back to the problem for more information. The perimeter of one square is 4x and the perimeter of the other is 4y. Together, they need to add to 1 (the total length of the wire) so we know that 4x + 4y 1. Solving for y gives y x. Plug this back into the formula for total area above to get: which can be simplified to: Total Area x + ( x), Total Area x 6x + 9. OVER

19 4. (1 points) On the axes below, sketch the graph of a function f(x) that satisfies all the following properties: f(x) 1 x on (1, ) f(x) is increasing on ( 4, ) (, 0) f(x) is decreasing on (0, 1) (1, ) f(x) is negative on [ 4, ) (, 1] f(x) is positive on (, ) (1, ) the domain of f(x) is [ 4, ) (, ) Note: There are many possible correct answers. After you finish, go back and make sure your function satisfies all the listed properties. If you make any changes, go back and check the other conditions again. OVER

20 5. (15 points) Use transformations to graph the function f(θ) cos(θ) + 1, on the grid below. Make sure your graph extends from π to π. Then answer the following questions. 4 1 π π π π π π π π (a) Find the range of f. The range of f is [ 1, ]. (b) On what interval(s) in (0, π) is the function increasing? f is increasing on (π, π). (c) On what interval(s) in [0, π] is the function negative? Hint: It may help to solve the equation f(θ) 0. Solve: f(θ) 0 cos(θ) cos(θ) 1 So, on the interval [0, π], we have θ π and θ 4π. We can then see from the graph that f(θ) is negative on the interval ( π, 4π). OVER

21 6. (6 points) Determine whether each of the following two statements is true or false. You must justify your answer. Note: In mathematics, true means that the statement must always be true. False means that the statement may sometime be false. (a) True or False: If θ and φ are coterminal 1 angles then cos(θ) cos(φ). TRUE. By definition cos(θ) is equal to the x value of the point where the terminal side of θ intersects the unit circle. Likewise, cos(φ) is equal to the x value of the point where the terminal side of φ intersects the unit circle. Since θ and φ are coterminal, their terminal sides are the same, so the point where these intersect the unit circle are the same, so cos(θ) cos(φ). (b) True or False: Given two angles θ and φ, if cos(θ) cos(φ), then θ and φ must be coterminal. FALSE. For example, take θ π and φ 4π. Obviously, θ φ, but we do have cos(θ) 1 and cos(φ) 1, so cos(θ) cos(φ). OVER for Practice Problems 1 Recall: Two angles in standard position are coterminal if they have the same terminal (ending) side.

22 7. (a) Practice Problems (Problems 7-80) (a b c ) (a 1 b c 5 ) a 6 b 6 c 6 a b 9 c 15 a b 15 c 9 (b) (a) x y 9xy 4 1 x y x 1 y 4 (7) 1 x y 5 xy 5 4 (b) x6 x 6 4 ( ) x x 6 1 x 1 9. ( 8) 1 ( 8) 1 ( ) There are many correct answers. Here is one: x + x 5. Another is 4x + πx ( x y ) ( x + y ) [ (x y) ( x y ) ( x + y )] (x y) (x y)(x y) x xy + y 4. (a) 6y 4 96x 4 6(y 4 16x 4 ) 6(y + 4x )(y 4x ) 6(y + 4x )(y + x)(y x) (b) 1x + x 4 64 x 4 + 1x 64 (x + 16)(x 4) (x + 16)(x + )(x ) 4. (a) y 9 y 49 y 8y + 15 y + 5y 14 which y + y 6 y 1y + 5 (b) 1+ x x x x 9 1+ x 9 + x + x x (x + )(x ) (y )(y + ) (y 7)(y + 7) (y + 7)(y ) (y 5)(y ) x x x (x + )(x ) x 9 (x + )(x ) (x ) (x + )(x ) x + (y + )(y ) (y 7)(y 5), (x + )(x ) + x(x + ) x (x + )(x ) 44. (a) x x 1 x x 1 0. To solve this, we use the quadratic formula, getting x 1 ± 5. (b) x 4 6x x 4 6x 0 x (x 6) 0 x 0, ± 6. (c) x x 1 0 x 0 x ±. (d) x + x 8x < 0. Factoring gives us x(x 4)(x + 7) 0, so the zeros are x 0, 4, 7. We do a number line analysis: x x x x(x 4)(x + 7)

23 (e) So the solution set is (, 7) (0, 4). 1 x 1 0. First we must write the LHS as a single fraction. We get x + 0. The numerator has no zeros. The zeros of the denominator are x(x + ) x 0 and x. A number line analysis gives us the following: x x x(x + ) (f) So the solution set is (, ) (0, + ). x e x 5e x 0 ex (x 5) 0. The zeros of the numerator are x ± 5. x + x + The zero of the denominator is x. We do a sign analysis: e x x x e x (x 5) x So the solution set is (, ) [ 5, 5] (a) The slope is m The equation is y 1 1 (x 6) or y 5 6 (x + ). For the graph, see the top line in the picture below. 1 (b) Since the lines are parallel, the slope of the new line is m 1. The line goes through the point (1, 1), so its equation is y 1 1 (x 1) or y 1 x +. For the graph, see the bottom line in the picture below.

24 y! 0.5 ( x! 6) + 1 y (! 0.5 ( x! 1) + 1)

25 46. Note that the line y 7 is a horizontal line, so any line perpendicular to it must be vertical. In this case, the vertical line goes through the point (, ), so its equation is x. 47. We know that p kd. When d feet, p 118 lb/in, so k 59. So p 59d. When d 5, p lb/in. 48. No, the day of the week is not a function of the forecast since there is an element in the set of forecasts (the domain) that corresponds to two different elements in the set of days (the range). More specifically, the element sunny in the set of forecasts corresponds to two different days: Tuesday and Thursday. 49. (a) (b) f(x + h) f(x) x + h + 1 x + 1 h h x + x h f(x + h) f(x) h (x + 1) (x + h + 1) (x + h + 1)(x + 1) h h (x + h + 1)(x + 1) (x + h + 1)(x + 1) h h h (x + h + 1)(x + 1) 1 h (x + h + 1)(x + 1). ( ) (x + h) 5(x + h) + 1 x + 6xh + h 5x 5h + 1 x + 5x 1 h h ( ) x 5x + 1 6xh + h 5h h 6x+h (a) f(x) 1 x e x e x. We solve x e x e x 0, getting e x (x ) 0. Recall that e anything > 0, so the only solutions are the solutions to x 0, namely x ±. So the domain of f(x) is { x : x ± }. (b) f(x) ln(x 5x). The domain of ln x is (0, + ), so the domain of f(x) is all x such that x 5x > 0. We solve this inequality in the usual way. The zeros of x 5x x(x 5) are x 0 and x 5, and a sign analysis gives us the following: x (x 5) x(x 5) So the domain of f(x) is (, 0) 5, + ). OVER

26 51. (a) The graph of f(x) is shown below: (b) The graph of g(x) is shown below: (i) f(x) is positive on (, 1] (0, + ) and negative on ( 1, 0). (ii) f(x) is increasing on ( 1, 1) and decreasing on (, 1) (1, + ). OVER

27 5. The answer to problem #5 is given in the back of the book. For problem #1: the average value equals f(0) f( ) 1 ( 11) ( ) For problem #64 on page 10: the graphs for (a) and (b) are shown in the first row below, the graphs for (c) and (d) are shown in the second row, the graphs for (e) and (f) are shown in the third row.

28 16. The graphs for (a) and (b) are shown in the first row below, the graphs for (c) and (d) are shown in the second row, and the graphs for (e) and (f) are shown in the third row.

29 5. Let x and y be the dimensions of the rectangular pen. Let a be the area of the pen. We want to write a as a function of x. Clearly, a x y, but this is not a function of x, since it also contains the variable y. However, we know that the rancher has 0 feet of fencing, so x + y 0, so we can write y in terms of x: y 0 x. So a(x) x(0 x) 0x x square miles. 54. We ll call the cost of fencing in the pen c. The heavy-duty fencing for the sides perpendicular to the river costs $10 per mile. So it will cost $10x to fence in each of the two sides perpendicular to the river. The cheaper fencing costs $80 per mile. So it will cost $80(0 x) to fence in the side parallel to the river. So c(x) 10x + 10x + 80(0 x) x dollars. 55. Let s call the radius of the cylinder r and the height h, and let s call its surface area s. We know that s πr + πrh, but this is not a function of r since it also contains the variable h. However, we know that the volume of the can is 100 cubic centimeters. Remember that volume of a cylinder is πr h, so here πr h 100. Hence we can write h in terms of r: h 100 πr. So ( 100 ) s(r) πr + πr πr πr + 00 square centimeters. r 56. Let s be the length of the side of the base of the box and let h be its height. Let C be the cost of manufacturing the box. The base of the box will cost 6s dollars. The top of the box will cost 4.50s dollars. The sides of the box will cost 4.50(4sh) dollars. So the cost of the box is C 6s s (4sh) 10.50s + 18sh. We need to write C as a function of s alone. We know that the volume of the box is 0, so s h 0. Solving this for h gives h 0. So the final answer is s ( 0 ) C 10.50s + 18s s. 57. Since the boat travels at a speed of 45 mph, it travels 45t miles in t hours. So t hours after the boat weighs anchor, it is t miles due east of Rockport Harbor. Thus the function giving the distance of the boat from Rockport Harbor at time t is d(t) t. Remember that t represents the number of hours that have passed since 8:00 am. Clearly t 0. Since the boats stops moving at pm, t 7. So the domain of d(t) is [0, 7]. 58. We know that p is a linear function of d, so the graph of the equation relating p and d is a line. Moreover, d is the independent variable and p is the dependent variable. Since p 15 when d 0, the point (0, 15) is on the line. Similarly, since p 45 when

30 d, the point (, 45) is on the line. So the slope of the line is 0 10, and the 11 equation of the line is p d (a) i. (gh)(0) g(0) h(0) ( f ) ii. (1) f(1) k k(1) ln iii. (f h)(0) f(h(0)) f( ) 1 iv. (k g)( ) k(g( )) k ( e ( ) k(e ) ln(e ) + + ) 1 v. (h h)() h(h()) h(0) (b) The y-intercept of f(x) is y The x-intercept of f(x) is the solution to the equation x + 1 0, which is x 1. e0 The y-intercept of g(x) is y 0 1. The x-intercepts of g(x) are the 1 e x solutions to the equation x 0. This equation has no solutions, so g 1 has no x-intercepts. The function k(x) ln x+ has no y-intercept, since x 0 is not in the domain of k(x). The x-intercept of k(x) is the solution to the equation ln x + 0. We get ln x x e. (c) (k f)(x) ln(x+1)+. To find (k f) 1 (x), we write this as y ln(x+1)+ and solve for x. We get y ln(x + 1) + y ln(x + 1) e y x + 1 x e y 1. So (k f) 1 (x) e x (a) We start with y 4 + x 1 and solve for x. We get y 4 x 1 (y 4) x 1 (y 4) + 1 x. So x (y 4) + 1 and therefore f 1 (x) (x 4) + 1. (b) We start with y e x+1 and solve for x. We get ln y ln e x+1 ln y x + 1 ln y 1 x x ln y 1. So f 1 (x) ln x The graph of f(x) is shown on the left below. f(x) is one-to-one since the graph passes the horizontal line test. The graph of f 1 (x) is shown on the right below. Note that the points (0, ), (1, 0) and (e, ) are marked on the graph of f(x) and the points (, 0), (0, 1) and (, e) are marked on the graph of f 1 (x).

31 4. The graph of f(x) is shown on the left below. f(x) is one-to-one since the graph passes the horizontal line test. The graph of f 1 (x) is shown on the right below. Note that the points (0, ), (1, 0) and (e, ) are marked on the graph of f(x) and the points (, 0), (0, 1) and (, e) are marked on the graph of f 1 (x). 6. f(x) x + 5x. Note that we can t find the formula for f 1 (x). But we do know the following property of inverse functions: (f f 1 )(x) x for all x in the domain of f 1 (x) and (f 1 f)(x) x for all x in the domain of f(x). So (f f 1 )() and (f 1 f)( 4) 4. ( ) ( ) 9 6. (a) log, since (b) log 8 6, since ( ) 6 8. (c) ln e e ln 1 4 ln(e 1 ) e ln ( ) ln x 1 ln x + ln y ln(x4 ) ln(x 1 ) + ln(y x 4 y ) ln ln(x 7 y ). 65. Problem #9 on page 1: The solution is in the back of the text. Ignore the part about the horizontal asymptote. Problem #48 on page : f(x) log a x, so we must find a. The point (9, ) is on the graph of f(x), so f(9) log a 9 a 9 a. So f(x) log x. Problem #5 on page : The solution is in the back of the text. Problem #60 on page : The graph is shown below. The domain of f(x) is (, 0), the range is all reals, and the function has a vertical asymptote at x 0. x 1

32 ( e, ) ( 1, 1) ( 1 e, 0) Problem #70 on page 0: We have the equation log S log c + k log A. (a) Solving for S: log S log c + k log A log S log c + log A k log S log(ca k ) S ca k. (b) If k, we have S ca. If we double the area, then the area is A, so we get S c(a) S c(8a ) S 8 ca. So S is eight times greater, meaning that the number of species is eight times greater. Problem #85 on page 9: ( ) P (a) ln h Exponentiating both sides gives P 100 e h 7 P 100e h 7. (b) When h 4, the pressure is 100e 4 7 kpa. 66. The function has the form f(x) Ca x ; we must find C and a. The point (0, ) is on the graph of f(x), so f(0) Ca 0 C(1) C. So f(x) a x. The point (, 1) is also on the graph of f(x), so f() 1 a 1 a 4 a. So the function is f(x) x. 67. (a) x x 7 0 x 0. This equation has no solutions, so the solution set of x 0 is. x 7

33 (b) Factor the equation e x e x 15 0 like a quadratic: (e x + )(e x 5) 0 e x or e x 5. The equation e x has no solutions, since e anything > 0. The equation e x 5 has solution x ln 5. So the solution to e x e x 15 0 is x ln 5. ( ) ( x ) (c) ln x ln(x + 1) ln ln ln e ln x x+1 e ln x x + 1 x + 1 x x + x. This is not a valid solution, so the solution set is the empty set. (d) x ln x x 0. We find the zeros of x ln x x: x( ln x 1) 0 x 0 or x e 1 e. We throw out x 0 since it is not in the domain of ln x. Using the number line technique, we get the following. x ln x x( ln x 1) e So the solution set is (0, e ]. (e) log 7 x x 49 x ±7. Both solutions are valid. (f) ln + ln x ln( x) ln + ln x ln( x) ln(x ) ln( x) e ln(x) e ln( x) x x x + x 0 (x )(x + 1) 0 x or x 1. The second solution is not valid, so x. (g) (ln x) ln x < 0. First we solve (ln x) ln x 0. We get (ln x)(ln x ) 0 x 1 or x e. We then use a number line analysis, getting the following: ln x ln x (ln x)(ln x ) e Note that good test points for the three intervals, from left to right, are e 1, e, and e 4. The solution to the inequality is (1, e ). (h) ln x x e 4 0. First find the zeros of the numerator and the denominator. Numerator: ln x 0 ln x x e. Denominator: x e 4. We get the number line analysis shown below.

34 ln x x e ln x x e e e 4 Note that good test points for the three intervals, from left to right, are e, e, and e 5. The solution to the inequality is (0, e ] (e 4, ). 68. (a) For h(x) sin(x x + ): let f(x) sin x and let g(x) x x +. (b) For h(x) 4 ln x + + 5: let f(x) 4 x + 5 and let g(x) ln x +. Another possibility: let f(x) 4 x and let g(x) ln x. (c) For h(x) e cos x+x : let f(x) e x and let g(x) cos x + x. 69. We know that cot θ AB BC. We re given that cot θ and that AB 8, so 8 BC BC 4. Then we use the Pythagorean theorem to get AC: AC AC An angle with positive measure that is coterminal with θ π radians is α π +π 9 9 0π radians. An angle with negative measure that is coterminal with θ π radians is 9 9 β π 16π π radians (a) 00 π 00 10π radians (b) π radians π π 18 (c) 4π radians 4π π (a) sin 70 1 (b) cos 10 1 (c) tan( 40 ) (d) csc (e) cot 0 is indefined (f) sec 7π 6 (g) sin 5π (h) cot 1π 6 (i) tan 5π 4 1 (j) cos( π) 1 7. Since sec θ 5, cos θ > 0. We re told that tan θ < 0. Since cos θ > 0 and tan θ < 0, sin θ < 0, so θ is in Quadrant IV.

35 (a) Since sec θ 5, cos θ 1 5 (b) First we need to find sin θ. We can draw a triangle or we can use the identity sin θ + cos θ 1. Here we ll use the identity. We get sin θ + ( 1 5) 1 sin θ 4 5 sin θ ± 4 5 sin θ ± 6 5. Since sin θ < 0, sin θ 6 Then cot θ cos θ sin θ 1/5 6/ (c) csc θ 1 sin θ 1 6/ (a) Note that cos θ and that sin θ 7. There are several ways to find k, but k k ( 7 ) ( ) the simplest is to use the identity sin θ + cos θ 1. We get k 9. So k ±. Since θ is in the third k k k quadrant, k. (b) sin θ 7 (c) Since cos θ sin θ, tan θ 7 cos θ. 75. (a) cos θ θ 5π, θ 7π 6 6 (b) tan θ 1. Recall that tan θ sin θ. So if tan θ 1, then sin θ cos θ, cos θ i.e., sin θ and cos θ are opposites of each other. So θ must be in the π family; 4 moreover, θ must be in the second or fourth quadrant. So θ π or θ 7π. 4 4 (c) cos θ 1 cos θ ±1 cos θ 1 or cos θ 1. So θ 0, θ π or θ π. (d) csc θ 4 sin θ 1 sin θ ± 1 sin θ 1 or sin θ 1. Then 4 θ π, 5π, 7π 11π or It is impossible, since 1 cos θ 1 for all values of θ. 77. We have f(x) sin(x) and let g(x) cos(x + π ). (a) f( π 4 ) sin( π 4 ) sin( π ) 1 (b) g( π 4 ) cos( π 4 + π ) cos( π 4 ) 1 (c) f( π ) sin( π ) sin( 4π ) (d) (f g)( π) f(g( π)) f ( cos( π + π )) f ( cos( π )) f(0) sin 0 0 (e) (g f)(x) cos(sin x + π ) (f) (g f)(0) cos(sin 0 + π ) cos(0 + π ) cos( π ) The graph of f(x) is shown below. Note: multiples of π are marked on the x-axis. k k 5.

36 The graph of f(x) is shown below. Note: multiples of π are marked on the x-axis. 80. f(t) e t sin t (a) At t π 6 seconds, the position is f(π 6 ) e π 6 sin π 6 e π 6 1 e π 6. This is a positive number, so it is below its equilibrium position. (b) The average velocity of the spring between t π 0 e π (1) π π 4e π units/second. and t π is f(π) f(π/) π π/ (c) To find when the weight passes through its equilibrium position, we solve f(t) 0 for t. We get e t sin t 0 e t 0 or sin t 0. The first equation has no solutions. The solutions to the second equation are t 0, t π, t π, etc. So the weight passes through its equilibrium position for the second time at time t π seconds.