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 5 + 5 = x 5 = x + 5 5 = 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
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 5 3 + 3) 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 4 4 4 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).
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 http://www.wolframalpha.com) 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
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
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
c) y = b x 1 d) y = b x 3 e) y = b x+1 + 3 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
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
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 = 11. 13.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 3 0.03114014 0.03 The rounded answer for c) is x = 1 ln 4 0.693147 0.693. The rounded answers to e) are ±1.174. 14. a) i) According to the compound interest formula B = 18, 000 1 +.09 ) 365 1 365 = 18, 0001.0004657534) 4380 = $5, 997.18. 8
ii) Solve 500, 000 = 18, 0001.0004657534) 365t. Hence 500000 18000 = 1.0004657534)365t and so 365t log1.0004657534) = log ) 500000 18000 so t = 1 log ) 500000 18000 36.94 years. 365 log1.0004657534) b) The amount of time it would take investment amount P at 7% compounded monthly is obtained by solving P = P 1 +.07 ) 1t 1 for t. Thus ln) = 1t ln 1 +.07 ), or 1 or approximately 9 years and 11 months. ln) t = 1 ln 1 +.07 ) 9.930955715 years. 1 15. 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 = 10 7.6 I 0 = 39, 810, 717I 0. c) M 1 = 107.6 I 0 M 10.9 = 10 7.6.9) = 10 4.7 50, 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) 500/5730.739N 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