Solutions for Problem Set 2

MATH 131, Fall 2007 September 17, 2007

Section 1.5

4. The graphs y = e^{x} and y = 8^{x} are increasing exponentials; y = e^{−x} is the reflection of y = e^{x} across the y-axis and y = 8^{−x} is the reflection of y = 8^{x} across the y-axis. Those two are decreasing exponentials.

8. Shift y = 4^{x} 3 units to the right (y-axis intercept becomes (0, 4^{−3}) = (0, \frac{1}{64})).

10. Take y = e^{x}, stretch vertically by a factor of 2, then shift up one unit. 13.

a. y = e^{x} − 2

b. y = e^{x−2}

c. y = −e^{x} d. y = e^{−x} e. y = −e^{−x}

14.

a. y = 8 − e^{x} (same as reflecting across the x-axis, then shifting up by 8

units). b. y = e^{4−x} = e^{−(x−4)} (same as shifting right by 4 units, then reflect

across the y-axis).

17. From the given points, 6 = Ca and 24 = Ca^{3}. Hence 4 = \frac{Ca^{3}}{Ca} = a^{2}, so a = 2 and C = 3. f(x) = 3 · 2^{x}.

18. Same method as 17: 2 = Ca^{0} = C and 2/9 = Ca^{2} = 2a^{2}. So a^{2} = 1/9, and a = 1/3. f(x) = 2 · (\frac{1}{3})^{x}.

20. Definitely take option II. The reason is that since each month has at least 28 days, on the last day you will get at least 2^{27} pennies. This is $(.01)134, 217, 728 = $1, 342, 177.28, which is more than a million dollars, just on the last day.

26.

a. There are 6 doublings in the first three hours, so 500 · 2^{6} = 32000.

b. After t hours, there are 500 · 2^{2t} bacteria.

c. 40 minutes = \frac{2}{3} hour, so 500 · 2^{4/3} ≐ 1260 bacteria.

d. The graph is a standard increasing exponential, with vertical axis intercept at 500. The population reaches 10000 when 100000 = 500 · 2^{2t}, so 200 = 2^{2t}, so t = \frac{1}{2} log_{2}(200) hours (about 3.82 hours).

Section 1.6

4. Apparently one-to-one, since no duplicate f(x)-values appear in table.

6. Not one-to-one. Note that f(x) = 0 for three different x-values in the picture. The graph does not pass the horizontal line test.

10. The graph y = 10 − 3x is a line with slope −3, so it crosses horizontal lines only once. This is a one-to-one function.

12. Not one-to-one, since, for instance cos(0) = 1 = cos(2π)

14. Not one-to-one, since peoples’ heights typically stay roughly constant for extended periods (e.g. between growth spurts as children, or once they reach maturity).

15. f^{−1}(9) = 2.

16 b. For any function f that is one-to-one, if 5 is in the range of f, f(f^{−1}(5)) = 5. (This has nothing to do with the given formula for f(x).)

21. y = \sqrt{10 − 3x} implies x = \frac{1}{−3}(y^{2} − 10) = \frac{10}{3} − \frac{y^{2}}{3} . So f^{−1}(x) = \frac{10}{3} − \frac{x^{2}}{3} .

24. y = 2x^{3} + 3 implies x = (\frac{y−3}{2})^{1/3} = ^{3} \sqrt{\frac{y−3}{2}}, so f^{−1}(x) = ^{3} \sqrt{\frac{x−3}{2}}.

26. y = \frac{1+e^{x}}{1−e^{x}} implies y − ye^{x} = 1 + e^{x}, so e^{x} = \frac{y−1}{y+1}, and x = ln (\frac{y−1}{y+1}). The

inverse function is f^{−1}(x) = ln (\frac{x−1}{x+1}).

30. Reflect the graph across the line y = x. The inverse function graph should be decreasing from about (−1, 4), through (0, 2), to (1, 0).

34. a. log_{8}(2) = 1/3, since 8^{1/3} = 2. b. ln(e^{\sqrt{2}}) = \sqrt{2}.

36. a) Using properties of logs and exponentials, 2^{log_{2}(3)+log_{2}(5)} = 2^{log_{2}(15)} = 15.

48.

a. Take ln of both sides in e^{2x+3} = 7. We get 2x + 3 = ln(7), so x = \frac{ln(7)−3}{2}.

b. Take exponential of both sides in ln(5 − 2x) = −3 to get 5 − 2x = e^{−3}. Then x = \frac{5−e^{−3}}{2} .

50 a. Take exponentials twice, ln(ln(x)) = 1 implies ln(x) = e, so x = e^{e}.

54. Let f(x) = ln(2 + ln(x)). The domain of this composition is the set of x in the domain of ln(x) such that 2+ln(x) is in the domain of the “outside” ln. Recall that ln(x) has domain the set of all positive reals. But 2 + ln(x) > 0 only for x > e^{−2}. So the domain is the set of all real x with x > e^{−2}. The inverse function is found by the usual process: y = ln(2 + ln(x)) when e^{y} = 2 + ln(x), so e^{e^{y}−2} = x. Thus f^{−1}(x) = e^{e^{x}−2}. The domain of this function is the same as the range of f(x), which is the set of all real x.

Appendix C

2. a. −315^{∘} = −\frac{7π}{4} radians. (The negative angle means measure clockwise, not counterclockwise as usual.) b. 36^{∘} = \frac{π}{5} radians.

4. a. −\frac{7π}{2} = −630^{∘}. b. \frac{8π}{3} = 480^{∘}.

16. x = 25 cos(40^{∘}) ≐ 19.15111 (using Maple).

24. Square out the left:

(sin(x) + cos(x))^{2} = sin^{2}(x) + 2 sin(x) cos(x) + cos^{2}(x).

Regrouping the terms, we have

= (sin^{2}(x) + cos^{2}(x)) + 2 sin(x) cos(x) = 1 + sin(2x),

by the basic identity 7, and the double angle formula 15a.

30. 2 sin^{2}(x) = 1 when sin(x) = ±\frac{1}{\sqrt{2}}, so x = \frac{π}{4} , \frac{3π}{4} , \frac{5π}{4} , \frac{7π}{4} (the first two give +\frac{1}{\sqrt{2}} the other two give −\frac{1}{\sqrt{2}}).

32. | tan(x)| = 1 when tan(x) = ±1, so sin(x) = ± cos(x). This happens at the same x-values given in problem 30: x = \frac{π}{4} , \frac{3π}{4} , \frac{5π}{4} , \frac{7π}{4} .

37. Shift the graph y = cos(x) by \frac{π}{3} units to the right.

38. Compress the graph y = tan(x) by a factor of 2 horizontally. This puts vertical asymptotes at x = ±\frac{π}{4} , ±\frac{3π}{4} , etc.

41. a. This is the unique angle θ in the interval [−\frac{π}{2} , \frac{π}{2} ] with sin(θ) = .5, so θ = \frac{π}{6} . b. arctan(−1) = tan^{−1}(−1) = −\frac{π}{4} .

42. a. tan^{−1} \sqrt{3} = \frac{π}{3} . b. arcsin(1) = sin^{−1}(1) = \frac{π}{2} .

43. a. sin(sin^{−1}(0.7)) = 0.7 (see problem 16 b in Section 1.6 above). b. sin (\frac{5π}{4} ) = −\frac{1}{\sqrt{2}}, so arcsin (−\frac{1}{\sqrt{2}}) = −\frac{π}{4} . (Values of arcsin = sin^{−1} are all in the interval [−\frac{π}{2} , \frac{π}{2} ].)

44. a) If θ = arctan(2), then tan(θ) = 2, so we can get θ in a right triangle with opposite side 2 and adjacent side 1 (draw the picture!). Then the hypotenuse is \sqrt{5}, so sec(θ) = \sqrt{5} (hypotenuse/adjacent).

b) First use the double angle formula:

sin(2 sin^{−1}(3/5)) = 2 sin(sin^{−1}(3/5)) cos(sin^{−1}(3/5)) = 2·\frac{3}{5}·\sqrt{1 − (\frac{3}{5})^{2}} = \frac{24}{25}.

Note: the value \frac{4}{5} = cos(sin^{−1}(3/5)) is obtained by a calculation similar to what we did in part a. If θ = sin^{−1}(3/5), then θ is an angle in a right triangle with opposite side 3 and hypotenuse 5. The adjacent side is \sqrt{25 − 9} = 4, so cos(θ) = 4/5 (adjacent/hypotenuse).