MATH 121 SAMPLE FINAL PARTIAL SOLUTIONS 12/4/1996 D. Radford

1. Set f(x) = x^{4} + x^{3} − 2x^{2} − 4x − 8. The rational solutions to f(x) = 0 are among ±1, ±2, ±4, ±8. That there are 2 rational solutions is given. Graphing y = f(x) yields that ±2 are the two. Thus x − 2, x + 2 divide f(x); hence x^{2} − 4 divides f(x). Long division gives f(x) = (x^{2} − 4)(x^{2} + x + 2). The roots of x^{2} +x+2 are −\frac{1}{2} ± \frac{\sqrt{7}}{2}i. Thus 2, −2, −\frac{1}{2} + \frac{\sqrt{7}}{2}i, −\frac{1}{2} − \frac{\sqrt{7}}{2}i are the roots. Followup question: What is the factorization of f(x) into linear factors?

2. P (x) = a(x − (−4))(x − 3)^{2} = a(x + 4)(x − 3)^{2} since P (x) has degree 3, x = −4 and x = 3 are roots of multiplicity 1 and 2 respectively. Implicit is that the coefficients of P (x) are real numbers. Since P (x) ⟶ +∞ as x ⟶ −∞ we conclude that a < 0. (a) .... (e) P (x) has roots x = −4, x = 3. Since P (−100) > 0, P (0) < 0, P (4) < 0, it follows that P (x) < 0 on (−4, 3) ∪ (3, ∞). The other parts are straightforward.

3. (a) ln(x + 1) + ln(x + 7) = ln(x + 2) + ln(x + 3) means ln(x + 1)(x + 7) = ln(x+2)(x+3). Since ln is one-one it follows that (x+1)(x+7) = (x+2)(x+3). Expanding both sides we have x^{2} + 8x + 7 = x^{2} + 5x + 6 and thus 3x = −1. Therefore x = −\frac{1}{3}. Check to see that this is in fact a solution by substituting into the original equation.

(b) log_{a}c = log_{b}c/log_{b}a, so log_{7}13 = \frac{ln 13}{ln 7}. Work out the exact value to 4 decimal places.

4. Let z = e^{x} Then \frac{1}{z} = e^{−x} ≠ 0 since the range of an exponential function consists of positive numbers. The equation e^{x} − e^{−x} = 6 is thus z − \frac{1}{z} = 6. Therefore, z^{2} − 1 = 6z so z^{2} − 6z − 1 = 0. By the quadratic formula e^{x} = z = \frac{6±\sqrt{40}}{2}. Therefore x = ln(\frac{6±\sqrt{40}}{2}). Note \frac{6−\sqrt{40}}{2} < 0. Thus there is one answer, x = ln(\frac{6+\sqrt{40}}{2}). Work out the answers to 4 decimal places.

Since 2^{3x+1} = 3^{2x+1}, (3x + 1) ln 2 = (2x + 1) ln 3. Therefore (3 ln 2 − 2 ln 3)x = ln 3 − ln 2 or (ln 8 − ln 9)x = ln 3 − ln 2. Thus (ln \frac{8}{9})x = ln \frac{3}{2} and

x = ln \frac{3}{2} ln \frac{8}{9} . (Work out the answer to 4 decimal places.)

5. Q(t) = Q(0)e^{kt}, so 5Q(0) = Q(5.83) = Q(0)e^{5.83k} means that 5 = e^{5.83k}. Therefore ln 5 = 5.83k and hence k = \frac{ln 5}{5.83}. At the time to triple in size Q(t) = 3Q(0). Thus 3Q(0) = Q(t) = Q(0)e^{kt} means 3 = e^{kt} and therefore t = \frac{ln 3}{k} = \frac{ln 3}{( \frac{ln 5}{5.83} )} = \frac{5.83 ln 3}{ln 5}. Q(1) = Q(0)e^{k} so \frac{Q(1)}{Q(0)} = e^{k}. (Work out the explicit

answers. Note that if P (t) = P a^{kt} then P (0) = P.)

6. The slope of the line passing L thru points (a_{1}, b_{1}) and (a_{2}, b_{2}), when a_{1} ≠ a_{2}, is m_{L} = \frac{b_{2}−b_{1}}{a_{2}−a_{1}} . For points (7, 8) and (8, 7) we compute m_{L} = \frac{7−8}{8−7} = −1. Thus y − 7 = (−1)(x − 8) is an equation of the line (point slope form). The slope-intercept form is y = −x + 15.

Let L_{perp} be the line perpendicular to the line L determined by 2x − 5y = 10, and suppose L_{perp} passes thru (2, 2). The slope-intercept form of L is y = \frac{2}{5}x−2. Thus m_{L} = \frac{2}{5}. Since m_{L}m_{L_{perp}} = −1 it follows that m_{L_{perp}} = −\frac{5}{2}. Thus L_{perp} has equation y − 2 = −\frac{5}{2}(x − 2). The slope-intercept form of the equation of L_{perp} is thus y = −\frac{5}{2}x + 7. 7. f(x) = 3 − \sqrt{x^{3} − 8}. Thus the domain of f(x) consists of all numbers x such that x^{3} − 8 ≥ 0, or x^{3} ≥ 8, or x ≥ 2 since y = x^{3} is an increasing function. The domain of f(x) is [2, ∞). To find the range of y = f(x) note that \sqrt{x^{3} − 8} ≥ 0, so y = 3 − \sqrt{x^{3} − 8} ≤ 3 − 0 = 3. If y ≤ 3 then y = 3 − \sqrt{x^{3} − 8} has solution x = ((3 − y)^{2} + 8)^{\frac{1}{3}} . Therefore the Range of f(x) is (−∞, 3]. Note that y = f(x) is one-one. Thus f(x) has an inverse, and we have shown that f^{−1}(x) = ((3−x)^{2}+8)^{\frac{1}{3}} . Domain f(x) = Range f^{−1}(x) and thus Range f(x) = Domain f^{−1}(x) hold generally for a function with an inverse. Thus [2, ∞) = Domainf(x) = Rangef^{−1}(x) and Domainf^{−1}(x) = Rangef(x) = (−∞, 3].

Let y = g(x) = \frac{2x−1}{3x+2}. To find g^{−1}(x) we set x = \frac{2y−1}{3y+2} and solve for y in terms of x: (3y + 2)x = 2y − 1, (3x − 2)y = −1 − 2x and hence y = \frac{−1−2x}{3x−2} = \frac{1+2x}{2−3x}. Therefore g^{−1}(x) = \frac{1+2x}{2−3x}. \{x | x ≠ −\frac{2}{3}\} = Domain g(x) = Range g^{−1}(x) and \{x | x ≠ \frac{2}{3}\} = Domain g^{−1}(x) = Range g(x).

8. (f ∘ g)(3) = f(g(3)) = f(\frac{3+2}{3−2}) = f(5) = \sqrt{19 + 5^{3}} = \sqrt{144} = 12. (g ∘ f)(5) = g(f(5)) = g(12) = \frac{12+2}{12−2} = \frac{14}{10}.

9. Solve 1 − 4x − x^{2} = |2x − 3|. Case 1: 2x − 3 ≥ 0. The equation becomes 1 − 4x − x^{2} = 2x − 3, or equivalently x^{2} + 6x − 4 = 0. Thus x = \frac{−6±\sqrt{52}}{2} = −3 ± \sqrt{13} by the quadratic formula. Since 2x − 3 ≥ 0, necessarily x = −3+\sqrt{13}. In this case 2x−3 = −6+2\sqrt{13}−3 = 2\sqrt{13}−9 < 0. No solution from this case. Case 2: 2x − 3 < 0. The equation in this case is 1 − 4x − x^{2} = −(2x − 3) or equivalently x^{2} + 2x + 2 = 0. By the quadratic formula there are no real roots. Conclusion: There are no real solutions to 1 − 4x − x^{2} = |2x − 3|.

Consider |2x + 1| = |x|. Then ±(2x + 1) = ±x as |x| = ±x, so we solve 2x + 1 = ±x. If 2x + 1 = x then x = −1. If 2x + 1 = −x then x = −\frac{1}{3}. Substituting these values into the original equation, we see they are both solutions.

10. y = 20e^{Kx}. If (1, 30) is a point of the graph then 30 = 20e^{K1} = 20e^{k}, so \frac{30}{20} = e^{K} and ln \frac{3}{2} = K. note y(0) = 20 is the initial value of y. Thus when y has doubled, 40 = y = 20e^{Kx} so 2 = e^{Kx} and ln 2 = Kx. Therefore x = \frac{ln 2}{K} = \frac{ln 2}{ln \frac{3}{2}} .

If the graph of y = 20e^{Kx} has the point (30, 1) then 1 = 20e^{K30} or \frac{1}{20} = e^{K30}. Therefore ln \frac{1}{20} = 30K or \frac{− ln 20}{30} = K.

11. P (t) = P (1 + \frac{r}{k})^{kt} is the return on an initial investment of P after t years with an annual interest rate of 100r% compounded k times annually. Comparing investment schemes we need to solve P (1 + \frac{r}{12})^{12} = P (1 + .0575) as t = 1. Thus (1+ \frac{r}{12})^{12} = 1+.0575 and 12 ln(1+ \frac{r}{12}) = ln 1.0575. Therefore ln(1 + \frac{r}{12}) = \frac{ln 1.0575}{12} so 1 + \frac{r}{12} = e^{\frac{ln 1.0575}{12}} and r = 12(e^{\frac{ln 1.0575}{12}} − 1). The annual percentage rate is 100r%.

Compounding continuously is given by P (t) = P e^{rt}. In this comparison we have P e^{r} = P (1 + .0575) and hence e^{r} = 1.0575. Therefore r = ln 1.0575.

Assume that we compound continuously at an annual rate of 100%, starting with $1. After 1 year we have P (1) = 1e^{r1} = e dollars as r = 1 in this case.

12. f(x) = \frac{x^{2}−5x+4}{(x+1)(x^{2}−3)} = \frac{(x−1)(x−4)}{(x+1)(x+\sqrt{3})(x−\sqrt{3})}. To solve f(x) < 0 we compute the real roots of the denominator and numerator of f(x) and use these as endpoints of intervals on the real number line. These roots are (in order) −\sqrt{3}, −1, 1, \sqrt{3}, 4. f(x) will be positive or negative on each of these intervals. The calculations f(−10) < 0, f(−1.5) > 0, f(0) < 0, f(1.5) > 0, f(3) < 0, f(10) > 0 show that f(x) < 0 on (−∞, −\sqrt{3}) ∪ (−1, 1) ∪ (\sqrt{3}, 4).

Solve g(x) = x + \frac{1}{x} ≥ 0. Write g(x) = \frac{x^{2}+1}{x}. The real roots of the numerator and denominator of g(x) are 0. As g(−1) < 0 and g(1) > 0 we conclude that g(x) > 0 on (0, ∞). Since x = 0 is not in the domain of g(x) the solutions to g(x) ≥ 0 consist of (0, ∞).

13. The most general form of the function is f(x) = \frac{ax^{2}+bx+c}{(x−1)(x+1)(x−5)} + x^{3} − x. (Generally if f(x) = \frac{p(x)}{q(x)} and p(x) = Q(x)q(x) + R(x), where R(x) = 0 or DegR(x) < Degq(x), then f(x) = \frac{p(x)}{q(x)} = Q(x) + \frac{R(x)}{q(x)} has end behaviour asymptote y = Q(x).) Taking a = b = 0 and c = 1 we have f(x) = \frac{1}{x^{3}−5x^{2}−x+5} +x^{3} −x = \frac{1+(x^{3}−5x^{2}−x+5)(x^{3}−x)}{x^{3}−5x^{2}−x+5} = \frac{x^{6}−5x^{5}−2x^{4}+10x^{3}+x^{2}−5x+1}{x^{3}−5x^{2}−x+5}. Replace x^{3} − x by x − x^{2} and redo the previous calculation.

14. Recall that sin(x + \frac{π}{2}) = cos x and cos(x − \frac{π}{2}) = sin x. Also function (\frac{π}{2} −x) = cofunctionx; thus sin(\frac{π}{2} −x) = cos x, cos(\frac{π}{2} −x) = sin x, cot(\frac{π}{2} −x) = tan x, etc. Recall that sin(−x) = − sin x and cos(−x) = cos x; that is sin is an odd function and cos is an even function. Thus sin(x−π) = sin((x−π)+2π) = sin(x+π) = sin((x+ \frac{π}{2})+ \frac{π}{2}) = cos(x+ \frac{π}{2}) = cos(−x− \frac{π}{2}) = sin(−x) = − sin x. Also, sin(x + \frac{3π}{2}) = sin((x + \frac{3π}{2}) − 2π) = sin(x − \frac{π}{2}) = − sin(\frac{π}{2} − x) = − cos x. Try to derive these identities, and the others, by looking at the unit circle.

15. cos(\frac{π}{8}) = cos(\frac{1}{2}(\frac{π}{4})) = \sqrt{\frac{1+cos \frac{π}{4}}{2}} = \sqrt{\frac{1+ \frac{\sqrt{2}}{2}}{2}} = \sqrt{\frac{2+\sqrt{2}}{4}} = ^{\frac{\sqrt{2+\sqrt{2}}}{2}}. sin(\frac{π}{12}) =

sin(\frac{π}{3} − \frac{π}{4}) = sin(\frac{π}{3}) cos(\frac{π}{4})−cos(\frac{π}{3}) sin(\frac{π}{4}) = (\frac{\sqrt{3}}{2})(\frac{\sqrt{2}}{2})−(\frac{1}{2})(\frac{\sqrt{2}}{2}) = \frac{\sqrt{2}}{4}(\sqrt{3}−1). Note that cos 75^{o} = cos(45^{o} + 30^{o}) = cos 45^{o} cos 30^{o} − sin 45^{o} sin 30^{o}..... , and cos 165^{o} = cos(180^{o} − 15^{o}) = − cos 15^{o} = − cos(\frac{1}{2}(30^{o})).

16. y = cot x = tan(\frac{π}{2} − x) = − tan(x − \frac{π}{2}). See problem 14 above. The last equation follows since y = tan x is an odd function. Since cot = − tan(x − \frac{π}{2}) we can describe the graph of y = cot x as derived from the graph of y = tan x by (1) a horizontal shift right of \frac{π}{2} and (2) reflection through the x-axis. Note that a is a zero of y = tan x if and only if x = a is a vertical asymptote of y = cot x, and a is zero of y = cot x if and only if y = a is a vertical asymptote of y = tan x.

y = cos x = sin(x + \frac{π}{2}), so the graph of y = cos x is obtained from the graph of y = sin x by a horizontal shift to the left \frac{π}{2}.

17. tan(cos^{−1} \frac{x}{8}) = \frac{sin(cos^{−1} \frac{x}{8} )}{cos(cos^{−1} \frac{x}{8} )} = ±\frac{\sqrt{1−cos^{2}(cos^{−1} \frac{x}{8} )}}{\frac{x}{8}} = ±^{\frac{\sqrt{1− \frac{x^{2}}{64}}}{\frac{x}{8}}} . Show that the domain of this composite is [−8, 0) ∪ (0, 8]. Show that if x ∈ [−8, 0) ∪ (0, 8] then tan(cos^{−1} \frac{x}{8}) = \frac{\sqrt{64−x^{2}}}{x}. Also solve this problem by constructing a right triangle with hypoteneus of length 8 and adjacent side of length x.

18. sin(3x) = sin(2x+x) = sin 2x cos x+cos 2x sin x = (2 sin x cos x)(cos x)+ (1 − 2 sin^{2} x) sin x = 2 sin x(1 − sin^{2} x) + (1 − 2 sin^{2} x) sin x = 3 sin x − 4 sin^{3} x. Thus sin 30^{o} = 3 sin 10^{o} − 4 sin^{3} 10^{o}, so as sin 30^{o} = \frac{1}{2} we deduce that 4 sin^{3} 10^{o} − 3 sin 10^{o} + \frac{1}{2} = 0.

19. By the Law of Sines \frac{sin 34^{o}}{6} = \frac{sin α}{7}. Therefore sin α = \frac{7}{6} sin 34^{o}. If −\frac{π}{2} ≤ α ≤ \frac{π}{2} then α = sin^{−1}(\frac{7}{6} sin 34^{o}) ≈ 40.72^{o}. In any case the (approximate) possibilities for α are 40.72^{o} and 180^{o} − 40.72^{o} = 139.28^{o}. Since h = 6 sin 34^{o} ≈ 3.55 both are realized.

Case 1: α ≈ 40.72^{o}. Then γ ≈ 180^{o} − (34 + 40.72)^{o} = 105.28^{o}. Therefore \frac{sin 105.28^{o}}{c} ≈ \frac{sin 34^{o}}{6} and consequently c ≈ 6(\frac{sin 105.28^{o}}{sin 34^{o}}) ≈ 10.35. Check by comparing the values \frac{sin 34^{o}}{6} , \frac{sin 40.72^{o}}{7} , \frac{sin 105.28^{o}}{10.35} which should be about the same by the law of sines. In this case the area is given by Area = \frac{1}{2}ac sin β = \frac{1}{2}(7)(10.35) sin 34^{o} ≈ 20.26.

Case 2: α ≈ 139.28^{o}. Check that γ ≈ 6.72^{o}, c ≈ 1.26.

(Note: In neither case are all of the angles acute.)

20. By the Law of Cosines we have 5^{2} = 7^{2} + 8^{2} − 2(7)(8) cos α. Thus 25 = 47 + 64 − 112 cos α, or 112 cos α = 113 − 25 = 88. Therefore cos α = \frac{88}{112} = \frac{11}{14}. The range of cos^{−1} is [0, π]. Thus α = cos^{−1} \frac{11}{14} ≈ 38.21^{o}, or about .67 radians.

To find the area we first compute the semiperimeter s = \frac{1}{2}(a + b + c) = \frac{1}{2}(5 + 7 + 8) = 10 and use the formula Area = \sqrt{s(s − a)(s − b)(s − c)} =

\sqrt{10(5)(3)(2)} = 10\sqrt{3}.

21. linear velocity = radius × angular velocity, or v = rω. 40 miles per hour = (40)(5280) feet per hour = \frac{(40)(5280)}{(60)(60)} feet per second. Thus \frac{(40)(5280)}{(60)(60)} = \frac{2}{3}ω as 8 inches = \frac{2}{3} feet. Thus ω = \frac{(3)(40)(5280)}{(2)(60)(60)} = 88 radians per second. This translates to \frac{ω}{2π} ≈ 14.01 revolutions per second, or (\frac{ω}{2π})(60) ≈ 840.34 revolutions per minute.

When the angular speed is ω = 2π radians per second, the linear speed is rω = (\frac{2}{3})(2π) ≈ 4.1888 feet per second, or (\frac{2}{3})(2π)(\frac{(60)(60)}{5280}) ≈ 2.8560 miles per hour.

60 miles per hour = (60)(\frac{5280}{60}) = 5280 feet per minute. So 5280 = \frac{2}{3}ω means that ω = (\frac{3}{2})(5280) = 7920 radians per minute which gives \frac{ω}{2π} = \frac{7920}{2π} ≈ 1260.51 revolutions per minute (rpm).

22. Write y = f(x) = a sin(bx + c) + d. We may assume that a, b > 0. Note that y_{max} = a + d and y_{min} = −a + d. Therefore d = \frac{1}{2}(y_{max} + y_{min}) and a = \frac{1}{2}(y_{max} − y_{min}). Since y_{max} = 4.8 and y_{min} = 2.0 we have d = \frac{1}{2}(4.8 + 2.0) = \frac{1}{2}(6.8) = 3.4 and a = \frac{1}{2}(4.8 − 2.0) = \frac{1}{2}(2.8) = 1.4. Thus y = 1.4 sin(bx + c) + 3.4. Now if a maximum occurs at x = x_{1} and the next minimum occurs at x = x_{2} then x_{2} −x_{1} is half a period. Therefore the period is 2(8.8) = 17.6. Since |b|(period) = 2π, we conclude that |b| = \frac{2π}{17.6}. Thus y = 1.4 sin((\frac{2π}{17.6})x + c) + 3.4 = 1.4 cos((\frac{2π}{17.6})x + c − \frac{π}{2}) + 3.4. A specific function which is consistent with the given requirements is y = f(x) = 1.4 sin((\frac{2π}{17.6})x)+3.4 = 1.4 cos((\frac{2π}{17.6})x− \frac{π}{2})+3.4 = 1.4 cos((\frac{2π}{17.6})(x− \frac{17.6}{4}))+3.4.

23. z = −1 + i\sqrt{3} = a + ib where a = −1 and b = \sqrt{3}. Thus r = |z| = \sqrt{a^{2} + b^{2}} = \sqrt{(−1)^{2} + (\sqrt{3})^{2}} = 2. Writing z = r(cos θ+i sin θ) = 2(−\frac{1}{2} +i\frac{\sqrt{3}}{2}) we have that cos θ = −\frac{1}{2} and sin θ = \frac{\sqrt{3}}{2}. The reference angle is \frac{π}{3} and we may take θ = π − \frac{π}{3} = \frac{2π}{3}. Thus z^{6} = r^{6}(cos 6θ + i sin 6θ) = 2^{6}(cos 6(\frac{2π}{3}) + i sin 6(\frac{2π}{3}) = 64(cos 4π + i sin 4π) = 64 + i0 = 64.

Let ω = −2−2i. Then |ω| = 2\sqrt{2} so ω = 2\sqrt{2}(−\frac{1}{\sqrt{2}} −i\frac{1}{\sqrt{2}}) = 2\sqrt{2}(cos \frac{5π}{4} + i sin \frac{5π}{4}). Therefore ω^{5} = (2\sqrt{2})^{5}(cos 5(\frac{5π}{4}) + i sin 5(\frac{5π}{4})) = 2^{5}(2)^{\frac{5}{2}} (cos \frac{π}{4} + i sin \frac{π}{4}) = 2^{7}\sqrt{2}(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i) = 2^{7}(1 + i) since \frac{25π}{4} = 6π + \frac{π}{4}. Note that z^{10} = 2^{10}(cos 10(\frac{2π}{3})+i sin 10(\frac{2π}{3})) = 2^{10}(cos \frac{20π}{3} +i sin \frac{20π}{3}) = 2^{10}(cos \frac{2π}{3} +i sin \frac{2π}{3}) = 2^{10}(−\frac{1}{2} + i\frac{\sqrt{3}}{2}) = 2^{9}(−1 + i\sqrt{3}). Thus ω^{5}/z^{10} = \frac{2^{7}\sqrt{2}}{2^{10}}(cos(\frac{π}{4} − \frac{2π}{3}) + i sin(\frac{π}{4} − \frac{2π}{3})) = \frac{\sqrt{2}}{8}(cos(−\frac{5π}{12}) + i sin(−\frac{5π}{12})) = \frac{\sqrt{2}}{8}(cos \frac{5π}{12} − i sin \frac{5π}{12}). Noting \frac{5π}{12} = \frac{π}{6} + \frac{π}{4} complete this calculation. We will compute ω^{5}/z^{10} another way. Note that ω^{5}/z^{10} = \frac{2^{7}(1+i)}{2^{9}(−1+\sqrt{3})} = \frac{1}{4}(\frac{1+i}{−1+\sqrt{3}}) = \frac{1}{4}(\frac{1+i}{−1+i\sqrt{3}})(\frac{−1−i\sqrt{3}}{−1−i\sqrt{3})}) = \frac{1}{4}(\frac{−1+\sqrt{3}}{4} +

\frac{−1−\sqrt{3}}{4}i) = \frac{−1+\sqrt{3}}{16} − \frac{1+\sqrt{3}}{16}i.

24. −32i = 32(0 − i) = 32(cos \frac{3π}{2} + i sin \frac{3π}{2}). Thus the 5^{th} roots of z are

ω = 2(cos( \frac{\frac{3π}{2} +2πk}{5}) + i sin( \frac{\frac{3π}{2} +2πk}{5})) = 2(cos(\frac{3π+4πk}{10}) + i sin(\frac{3π+4πk}{10})) for k = 0, 1, 2, 3, 4. For k = 2 for example, ω = 2(cos \frac{11π}{10} + i sin \frac{11π}{10}) which needs to be worked out with a calculator.

1 = 1 + 0i = 1(cos 0 + i sin 0). Therefore the 8^{th} roots of unity are ω = 1(cos \frac{0+2πk}{8} +i sin \frac{0+2πk}{8}) = cos \frac{πk}{4} +i sin \frac{πk}{4} for k = 0, 1, 2, 3, 4, 5, 6, 7. See the text for the rest of the details of the calculation.