Math 121, Practice Questions for Final (Hints/Answers)

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1 Math 11, Practice Questions for Final Hints/Answers) 1. The graphs of the inverse functions are obtained by reflecting the graph of the original function over the line y = x. In each graph, the original function is the dashed line, and the inverse function is the solid line, the line y = x is dotted. a) b) c) d).a) F Gx)) = Gx)) 5 = x + 5 GF x)) = F x) + 5 = x = x 5 = x = x. ) = x. ) Because both GF x)) = x and F Gx)) = x, F and G are inverse functions. The graph below shows F dashed line), G solid line) and y = x on the same coordinate axes. 1

2 b) pqx)) = qx) 5 = x inverse functions. x x 5 5 x = x 5x 5) xx 5) = 3x + 5 x 10x x. Therefore, p and q are not 3.a) Let x = y 5 3 and solve for y. Then y 5 = x + 3 and so y = x + 3) 1 5. Thus f 1 x) = x + 3) 1 5. Verification: ff 1 x)) = f 1 x)) 5 3 = x + 3) 3 = x ), and f 1 fx)) = fx) + 3) 1 5 = x ) 1 5 = x 5 ) 1 5 = x. ) b) Let x = y y implies yx 1) = x, and then y = Verification: gg 1 x)) = and solve for y. Then y )x = y and so xy x = y or xy y = x which g 1 x) g 1 x) = x x 1. Thus g 1 x) = x x 1 x x 1 = g 1 gx)) = gx) gx) 1 = x x x x 1 = x x x ) = x = x ) ) 1 x c) The inverse function is h 1 x) =. 4 x x 1, x 1. x x x ) = x = x ) Verification follows by using the inverse property of logs and exponentials: h 1 hx)) = 1 hx) 4) = 1 log1/4 x 4) = x, and ) 1 x hh 1 x)) = log 1 h 1 x) = log 1 = x The graph of h dashed line) and h 1 solid line), along with y = x dotted line) are given below. 4. a) Yes because f has an inverse function. b) Yes because g has an inverse function. c) The domain of g is [, 0] and the range of g is [3, 10]. d) We know g ) = 3 because f3) = and g0) = 10 because f10) = 0. There is not enough information to find g3) and g10).

3 5. a) 3 x = 81, or 3 x = 3 4 and so x = 4. b) 3x 1 = 5 and so 3x 1 = 5 and so x =. c) x = ln e π = π ln e = π, thus x = π. d) 3x + = 4x + 1) and so 3x + = 4x + 4, and so x =. 6. a) The value of b is 3. This is because the graph of y = log b x passes through the point 3, 1). The graphs are given below, they and the graphs for the answers to Question 7,9 and 10 were generated by Wolfram Alpha which is available for on-line use free of charge. The following are hints for sketching the graphs by hand, and then computer generated graphs are given. b) The graph of y = b x is the reflection of the graph of y = log b x across the line y = x. c) The graph of y = log 1/b x) is the reflection of the graph of y = log b x over the x-axis. d) The graph of y = log b x is symmetric about the y-axis it is the graph of y = log b x and its reflection over the y-axis). e) Shift the graph of y = log b x -units to the right, and 3-units up. f) Reflect the graph of y = log b x over the y-axis. b) y = b x c) y = log 1 x) b d) y = log b x e) y = log b x ) + 3 3

4 f) y = log b x). 7.a) The graph of y = log b x passes through the point, 1). Thus 1 = log b, or in exponential form, b 1 = which means b = 1 = 1/. The hints for the rest of this problem are similar to Question 6, except in e) shift the graph of y = log b x 3-units to the left and -units down. Also,note the graphs of y = log 1/ x) and y = ) 1 x are given in the answer to Question 1 since in that problem gx) = 1 x ) and g 1 x) = log 1 x). b) y = b x c) y = log 1 x) b d) y = log b x e) y = log b x + 3) 4

5 f) y = log b x). 8. a) The graph of y = b x passes through the point 1, 4) and so b = 4. The graph of y = log b x is obtain by reflecting the graph of y = b x across the line y = x. See the answer to question 3 for a graph of y = log 1 x)). 4 b) The functions g and h in Question 1b) and c) are exponential functions. In fact gx) = 1 and hx) = x. ) x 9. The graphs are given below. Hints on drawing the graphs by hand are as follows. a) The graph of y = log b x is obtained by reflecting the given graph of y = b x over the line y = x. b) The graph of y = b x is obtained by reflecting the given graph of y = b x over the y-axis. c) Shift the graph of y = b x from b) one unit down. d) Shift the given graph of y = b x three units to the right and units down. e) Shift the given graph of y = b x one unit to the left and 3 units up. f) This graph is symmetric about the y-axis. For x 0, it is the same as the graph of y = b x, the other half is obtained by reflecting that part over the y-axis to make the graph symmetric. a) y = log b x b) y = b x 5

6 c) y = b x 1 d) y = b x 3 e) y = b x f) y = b x 10. The graphs are given below. The following provide suggestions on how to construct the graphs by hand. a) Note that if x 0, the graph looks like the graph of y = 3 x and the function is even, so the rest of the graph is a reflection about the y-axis. b) x = 3 y. c) The graph of y = log 3 x is obtained by reflecting the graph of y = 3 x over the line y = x or plot x = 3 y by choosing the y-values first). d) The graph of y = log 3 x is a reflection of the graph from c) of y = log 3 x over the x-axis. e) The graph of y = log 3 x )) is the graph from d) of y = log 3 x shifted units to the right. f) The graph of y = 4 log 3 x ) is the graph from e) of y = log 3 x ) shifted 4 units up. g) Note that y = log 3 x 4 is an even function so its graph is symmetric about the y-axis, and for x > 0, log 3 x 4 ) = 4 log 3 x vertical stretch of factor of 4 of the graph in c)). 6

7 a) y = 3 x c) y = log 3 x) d) y = log 3 x e) y = log 3 x ) f) kx) = 4 log 3 x ) g) y = log 3 x 4 ) 11. a) ) x log 3 z 5 b y 4 = log b x 3 z 5 ) log b y 4 ) = 1 log bx 3 z 5 ) 4 log b y 4 ) = 1 [log bx 3 ) + log b z 5 )] 4 log b y 4 ) = 3 log b x + 5 log b z 4 log b y. b) 4 log b x 3 z 3 ) log b z y) + 3 log b xy = log b x 3 z 3 ) 4 log b z y) + log b xy/) 3 = log b x 1 z 1 ) log b z y) + log b x 3 y 3 /8) x 15 z 1 y 3 ) x 15 y z 10 ) = log b 8z = log y b. 8 7

8 c) The domain of f is {x x > 4}, the range of f is {y < y < }. d) The domain of g is {x < x < } and the range of g is {y y > 3}. e) The domain of h is {x x 0}. f) The range of k is {y y < 3}. 1. a) 1 ) 3 = 8. b) log 048 = a) 53 x ) 49 x ) = 0 implies 53 x ) = 43 ) x, so 53 x ) = 43 x ). Divide both sides by 43 x ) to obtain 5/4 = 3 x. Thus log5/4) = x log 3. Thus x = log5/4) log 3. b) The equation implies log[x+4)x+1)] = log4x+10) note all terms x+4, x+1 and 4x+10 must be positive, so x > 1. Thus x + 4)x + 1) = 4x + 10, or x + 5x + 4 = 4x + 10 and so x + x 6 = 0, or x + 3)x ) = 0. Thus x = 3 or x =, but we know x > 1, so the only solution is x =. c) Multiply both sides of the equation by e x + e x ) to obtain e x e x = e x + e x. This simplifies to e x = 4e x. Multiply both sides of this by e x to obtain e x = 4. Thus x = ln 4. and so x = 1 ln 4 ) x + 1 d) First logx + 1) log3x 1) =, and so log =. Rewriting this in exponential form 3x 1 we have x + 1 3x 1 = 10. Therefore x + 1 = 300x 100, or 98x = 101 and so x = 101. You should plug this in and verify 98 that it works because sometimes extraneous solutions are introduced when combining the logs. e) Multiply both sides by 3 to obtain 10 x + 10 x = 15. Then multiply both sides by 10 x to obtain 10 x ) + 1 = 1510 x ). Let u = 10 x and then u 15u + 1 = 0. Thus u = 15 ± 1, and so 15 + ) 1 15 ) 1 x = log and x = log. f) The rounded answer for a) is x = log5/4) log The rounded answer for c) is x = 1 ln The rounded answers to e) are ± a) i) According to the compound interest formula B = 18, ) = 18, ) 4380 = $5,

9 ii) Solve 500, 000 = 18, ) 365t. Hence = )365t and so 365t log ) = log ) so t = 1 log ) years. 365 log ) b) The amount of time it would take investment amount P at 7% compounded monthly is obtained by solving P = P ) 1t 1 for t. Thus ln) = 1t ln ), or 1 or approximately 9 years and 11 months. ln) t = 1 ln ) years Before solving any of these problems, notice that 10 M = I, and so I = 10 M I 0 is an another I 0 form of the equation. a) The magnitude is M = log1, 589, 54) 7.1. b) From the scratch work formula, the intensity is I = I 0 = 39, 810, 717I 0. c) M 1 = I 0 M 10.9 = ) = , 118. Therefore, the magnitude 7.6 earthquake was I 0 approximately 50,118 times as intense as a magnitude.9 earthquake. 16. Let t be measured in hours and let t 0 = 1 noon. Now P 0 = 1000 is the population when t = 0. Thus P t) = 1000e kt and so we need to find k. Also, P.5) = 1300, therefore 1300 = e.5k. Therefore k = ln1.3).547. Thus: a) At 11:30am, t = 1/ and so there were approximatley 1000e.5).547) 769 bacteria present in the vat. b) At 1:30pm, t = 1.5 and so there were approximately 1000e 1.5).547) 197 bacteria in the vat. 17. a) Solve.68N 0 = N 0.5) t/5730 and so.68 =.5) 5/5730 and so ln.68) = t ln.5) 5730 or t = 5730 ln.68) ln.5) 3188 years Thus the scroll is approximately 3188 years old. b) Nt) = N 0.5) 500/ N 0. This implies that we would expect to find approximatly 73.9% of the original amount of the Carbon-14 remaining in the bone. 9