MATH100 Practice

8/21/2019 MATH100 Practice

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T F There exists a real number tsuch that sin t= 1 and tan t= 1.

T F ln(x+y) = ln x ln y for all x,y >0.

T F tan(

x) =

tan xfor all x in the domain of tangent.

T F The function f(t) = 3 cos(400t) makes

200complete waves in the interval [0, 2].

It makes 400 waves in that interval, but its period is

200

T F The angle of 95

11 radians is coterminal with the angle of

7

11 radians.

T F sin1

sin

13

16

=

13

16

T F cos

cos1 .8947 =.8947

T F There exists no triangle with legsa = 3

3, b = 12, and angle A=

3.

sin B = 2 has no solutions

T F ln 1 = 0

T F sin(6t+ 2) is the graph of sin 6tshifted 13 units to the left.

T F sin1

4 =

2

2

This confuses sine with arcsine!

T F The range of sec x is (,1) (1,).

The correct range is (,1] [1,)

T F The equation cos 5x= 12 has 10 solutions for 0 x 2.

It makes 5 waves in that interval and for each wave there are two solutions.

T F The function cos(4t) is periodic with period 2.

The period is

1

2


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T F There exists a real number tsuch that sin t=1

3and cos t=

2

3

T F If the point (2, 3) is on the graph of a one-to-one function f then

1

2,1

3

is on

the graph off1

.

T F If f(1) =f(3) for a function f then f1 does not exist.

T F When dividing a polynomial fby a polynomial g , the remainder is a polynomialwith degree less than the degree ofg.

T F f(t) =t+ csc t is an odd function.

T F There exists a real number tsuch that sin t= 13

and cos t=2

3.

T F The function f(t) = 3 cos(400t) makes 200 complete waves in the interval [0, 2].

T F The angle of 95

11 radians is coterminal with the angle of

7

11 radians.

T F sin1

3 =

3

2

T F There exists no triangle with legsa = 3

3, b = 12, and angle A=

3.

T F The range of csc t is (

,

1]

[1,

).

T F sin(6t+ 2) is the graph of sin 6tshifted 13 units to the left.

T F The function f(x) =e2x is even.

T F ln(x+y) = ln x ln y

T F The function f(x) = x2 + 5 is always increasing.

T F The graph of a polynomial can have a horizontal asymptote.

T F A polynomial of degree 5 cannot have more than 5 roots.

T F An polynomial of odd degree must have at least one root of odd multiplicity.

T F The lines given by the equations 2x+y 2 = 0 and 4x 2y+ 18 = 0 are perpendicular.

T F f(x) =x2 + 3|x| is a polynomial function.

T F The function 2x is always decreasing.

T F The domain of f(x) =x2 7x+ 6 is (, 1) (1, 6) (6,).


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(When necessary, leave your answers in calculator ready form. )

1. Solve triangle ABCwhere a= 12.4, c= 6.2, A= 72.

C = sin11

2sin(72)

B = 180 72 sin1

1

2sin(72)

b = 12.4sin B

sin72

2. A 150-foot-long ramp connects a ground-level parking lot with the entrance of a building. Ifthe entrance is 8 feet above the ground, what angle does the ramp make with the ground?

sin1

8

150

3. Solve forxin the equation tan 4x= 8.

x=tan1 8

4 k

4. The population of a colony of fruit fliestdays from now is given by the functionp(t) = 1003t/10.(a) What will the population be in 15 days? In 25 days?

In 15 days there will be p(15) = 100 31.5 fruit flies.

In 25 days there will be p(25) = 100 32.5 fruit flies.

(b) When will the population be 2500? (Hint: Use properties of logarithms to answer thisquestion!)

10ln25

ln 3 days

5. If an angle in standard position oft radians has terminal ray lying on the line with slope 8 then

tan t= 8 .

6. Translate the given exponential statement into a logarithmic statement: e3.25 = 25.79.

ln(25.79) = 3.25


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7. Translate the given logarithmic statement into an exponential statement: log 750 = 2.88.

102.88 = 750

8. Fill in the following table:

sin x csc x tan1 x ln x ex

Domain (,) All reals exceptk (,) (0,) (,)

Range [1, 1] (,1] [1,)

2,

2 (,) (0,)

9. Compute

a. log28 = 3 b. eln931 = 931

c. sin1

3

2

=

3 d. sec

6 =

2

3

3

e. cos1

cos7

4

=

4 f. tan1(cos ) =

4

Note: 7

4 / [0, ]


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10. Compute sin

8.

Note:

8

is in the first quadrant so sin

8

> 0. Well pro-

ceed by using the half-angle formula for sine because

8 =

4

2

sin

4

2

=

1 cos 4

2

=

1

22

2

=

2

2

2

2

=

22

2

11. Evaluate sin

cos1

7

4

.

Suppose cos1

7

4 = u. Then, by definition u [0, ] and

cos u= 74

. Since weve let cos1 74

equalu, our job is now to find

sin u. Note that since u [0, ], sin u >0

By the Pythagorean property,

sin u =

1 cos2 u

=

1

7

4

2

=

1 7

16

= 3

4


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12. Find all solutions to the equation cos(2x) = sin x.

Note: cos(2x) = 1 2sin2 x. Making this substitution is critical.

1 2sin2 x = sin x

2sin2 x+ sin x 1 = 0

(2sin x 1)(sin x+ 1) = 0

This last equation implies that sin x = 1

2 or sin x = 1. This

yields three solutions sets:

x =

62k

x = 5

6 2k Note:

6 =

5

6

x = 2 2k Note: We dont get another

solution from this seed because

(2

) =3

2 is already

included in the set.

13. Consider the graph of the function below:

(a) What is the amplitude of the function?

4

(b) What is the period of the function?


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3 because it takes 3 units to make one complete wave.

(c) Write a rule for the function of the form A cos(bt+c).

First, we can use part (b) to find b. Period is 2

b = 3 so b =

2

3 .

For the rest, we observe that this looks like a cosine graph re-

flected vertically (so A= 4), and shifted right by 12

a unit. So the

phase shift is 1

2=c

b .

Usingb=2

3, we can solve for c to get c=

3.

So the rule is f(t) = 4cos

2

3 t

3

14. Prove the identity: sin x sin3xcos x cos3x = cot2x.

Dont worry about this problem. We didnt do enough with identitiesin our class to prove this.

15. Solve forx: log x+ log(x 3) = 1.

x= 5


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16. Given that tan x=4

3 and x 3

2, find sin2x and cos

3

+x

.

To get from tan x to sin x and cos x. We can either do this as in

the way demonstrated on HW 8 (because tan x = 43

implies that

the terminal ray lies on the line given by y = 4

3x.) Or we can use a

Pythagorean identity to find secant to find cosine.

Recall: 1 + tan2 x = sec2 x. And note that since x is in thethird quadrant, sine and cosine (and secant) will be negative.

Hence sec x =

1 +

4

3

2= 5

3 So cos x = 3

5 and thus

sin x = 4

5 . These are the values we need to find sin 2x andcos

3

+x

.

sin2x = 2 sin x cos x

= 2

3

5

4

5

= 24

25

cos

3+x

= cos

3cos x sin

3sin x

= 1

2

3

5

3

2

4

5

= 310

+4

3

10

= 4

3 310

17. Two ships leave Mathematica at 8 am. The first, traveling 8 knots (nautical miles per hour)makes a beeline south for South Pythagorolina. The second, at 4 knots, heads in a southeasterndirection to Trigonoville. Their courses differ by 30. How far apart are the two boats at 10:30am later that morning?

See solutions to HW 10.


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18. Give an example of polynomial function of degree 4 and leading coefficient -2:

2x4 + 17x3 31

19. Give an example of a function which is not a polynomial:

f(x) =

x

20. Forfto be a function we need that for every input there is a unique output.

21. Forfto be one-to-one we need that for any two distinct inputs we get distinct outputs

22. Solve forxin the equation: x1

2 + 3x1

4 10 = 0.

Let u= x1/4 So the equation can be rewritten as u2 + 3u 10 = 0.Thus u= 4

x=5 which has no solutions, and u= 4x= 2 which

implies that u = 16. So the only solution is 16.

23. Compute and simplify the difference quotient forf(x) =

2x+ 1. Recall the difference quotient

off is f(x+h) f(x)h

. (Hint: Rationalize the numerator to simplify!).

f(x+h) f(x)h

=

2(x+h) + 1 2x+ 1

h

=

2(x+h) + 1 2x+ 1

h

2(x+h) + 1 +

2x+ 1

2(x+h) + 1 +

2x+ 1

= 2(x+h) + 1 (2x+ 1)h

2(x+h) + 1 + 2x+ 1

= 2x+ 2h+ 1 2x 1h

2(x+h) + 1 +

2x+ 1

= 2h

h

2(x+h) + 1 +

2x+ 1


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24. Find the rule of a quadratic function which has vertex (4, 1) and passes through the point(2,11).

f(x) = 3(x 4)2 + 1

25. A group of friends are sharing the cost of a $210 cabin at the beach. At the last minute 2 peopledecide they cant afford it so they wont go. This increases the remaining friends share by $28each. How many people were in the group in the beginning? (Note: To receive full credit youmust set up an equation to solve, and solve it. A correct guess-and-check answer will receiveone point out of the total possible).

5 friends were originally going to share the cabin.

26. Find a numberk such that x + 2 is a factor ofx

3

+ 3x

2

+kx 2.

k= 1

27. How could we restrict the domain off(x) =3(x+ 1)2 4 to create a one-to-one function?What then would be the domain and range off1?

Restrict to [1,). The domain off1 would be (,4] and therange would be [

1,

)

28. Simplify and write the expression without radicals or negative exponents:3

6c4d14

3

48c2d2. Please

show all intermediary steps!

c2d4

2

29. Suppose you owe $700 on your credit card. Your credit card company charges 2% interest

monthly. Supposing you make no further purchases because your card is maxed out and youmake no payments because you are a poor college student (and you just paid for a cabin on thebeach!) write a rule for how much you will owe after t months. When will you owe $1400? Setup an equation (but dont solve it) to answer the last question.

f(t) = 700(1 +.02)t

The desired equation is 700(1.02)t = 1400. Ill owe $1400 af-

ter ln 2

ln 1.02 months. (Technically, it will be at the end of this

month.)


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30. Let f(x) = 2x

3x 1 . Find the inverse off.

f1(y) = y

3y

2

31. Find all real solutions to the given equation. Give both exact (calculator ready) answers.

(a) 213x = 3x+1

x= ln 2 ln 3ln 3 + 3 ln 2

(b) log 3

x2 + 21x=2

3

x= 4,25

Note: There are two solutions. We dont reject any. Why?

(c) ex +ex

ex ex = 2.

x=ln 3

2

(d) 2 3x2+3x 7 = 155

x= 4, 1

(e) log5(62.5x) = 4

x= 10

(f) e4y = 7

y= 4 ln(7 )

(g) log(x2 + 2) = 3.5

x=

103.5 2

Note: There are also two solutions to this equation.Why dont we reject any?


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32. Write as the logarithm of a single quantity.

(a) ln(e2y) + ln(ey) 3

ln(y2)

(b) log4(2x 3) 1

3log4(x) + 2 log4(2x 3)

log4

(2x 3)3

3

x

33. If an investment of $2000 grows to $2700 in three and a half years, with an annual interest ratethat is compounded monthly, what is the annual interest rate?

2000

1 + r

12

42= 2700 = r= 12

42

27

20 1

34. It takes 1000 years for a sample of 100mg of radium-226 to decay to 65mg. Find the half-life ofradium-226.

100 0.51000/h

= 65 = h= ln 113

1000 ln(0.5)

35. How old is a mummy that has lost 49% of its carbon-14?

Note: This requires knowing the half-life on Carbon-14: 5730 years.

If the amount of carbon-14 after t years is given by the equa-tion A(t) = m 0.5t/5730 then if a sample has lost 49% of itscarbon-14, then there is 51% remaining.

So we solve the equation m 0.5t/5730 = 0.51m The solutionis

t=5730 ln(0.51)

ln(0.5) years


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36. One hour after an experiment begins, the number of bacteria in a culture is 100. An hour later,there are 500.

(a) Find the number of bacteria at the beginning of the experiment and the number threehours later.

In one hour the population quintupled so assuming exponentialgrown, the population after t hours is modelled by the functionP(t) = A 5t. We can find the initial amount A, by solvingP(1) = 100 = A= 20. Finally, there will beP(4) = 2054 = 12, 500after four hours.

(b) How long does it take the number of bacteria at any given time to double?

ln 2

ln 5 hours

37. Find and simplify f(a 2) f(2)

a , forf(x) =ex + log2(2x). (You wont be able to simplify

much)

e(a2) + log2(2(a 2)) e2 log2(4)a

=ea2 + log(a 2) e2 1

a

38. Find the domain and range ofh1, where h(t) = 2 1012t.

We can either find h1 explicitly or just think about the graphof h to find the domain and range of h. The domain of h1(t) isthe range ofh(t). The function 10t has range (0,). None of thetransformations of 10t (horizontal refection, shifting, and stretchingand vertical stretching) will affect its range. So the domain ofh1(t) is also (0,).

The domain of h(t) is all real numbers, and thus is the rangeofh1(t).

39. Compute and simplify (f g)(2x+ 1) for f(x) = 3x1 and g(x) =x2 +x+ 1.

3(2x+1)2+(2x+1)+1 = 34x

2+6x+3


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40. Find the domain and range of the inverse function for g(x) = log2(3x 1).

Domain ofg1(x): All reals

Range ofg1(x):

1

3,

41. Find (f g)

1125

forf(t) = 21/t

2

and g(x) = log5(x).

21/

1125

2 log5

1

125

= 2125 +

3

2

42. An exponential function that decreases by 30% for every unit increase inx can be modeled by...

(a) f(x) = (0.30)x

(b) f(x) = (0.30)x(c) f(x) = (1.30)x

(d) f(x) = (0.70)x

(e) f(x) = (0.70)x


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43. Show that

62

4 =

23

2 by computing sin

11

12 with two different methods, the half

angle identities and the the addition identities.

First we compute sin1112

with the addition identity:

sin11

12 = sin

3

12+

8

12

= sin

4+

2

3

= sin

4cos

2

3 + sin

2

3 cos

4

=

2

2 (1

2 ) +

3

2

2

2

=

62

4

And now we compute sin11

12 with the half-angle identity

(note: sin11

12 >0)

sin11

12 = sin

116

2

=

1 cos 116

2

=

1

32

2

=

23

2

44. How many triangles are possible with angleA = 45, and sides a = 5, c =

5

6

2 . You mustjustify your answer for credit but you do not need to give the solution(s) if they exist.

sin45

5 =

sin C56

2

= sin C=

12

2 =

3> 1

This equation has no solutions. No triangles are possible.


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45. Consider the function 3 sin

2t+

2

.

(a) What is the amplitude of the function?

3

(b) What is the period of the function?

2

b =

(c) What is the phase shift?

cb

= 2

2 =

4

So the sine graph is shifted left

4

(d) Graph the function below. Remember to label your graph completely so that I can easilyread it.

(e) Write a rule for the function of the form A cos(bt+c).

3 cos(2t)

46. Prove the identity: cos x1 sin x = sec x+ tan x.

Multiply the left hand side by1 + sin x

1 + sin x ...

47. Given that tan x= 57

, find sin2x.


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See the solution to #16. But note that we dont need to know whatquadrant x is in because sin2x will be negative no matter what..(Why?!)

48. Evaluate: csc56

csc2 76 cot2 76+

sec43 csc6

cot 6076. Its really not as hard as it looks!

6 433

49. Sketch the complete graph off(x) = (x2 25)(4 3x)(x2 + 3x 10)(x 1) .

Your graph should be scaled nicely, and all holes, intercepts, and asymptotes clearly labeled.


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50. Let f(x) =

|3 11x| 5

2x 1 and g(x) = log2(x 3) + 1

(a) Find the domain offand write your answer in interval notation.

, 211

811

,

(b) Find the domain and range ofg , write your answer in interval notation.

(3,)

(c) Compute (f g)(11)

2

5

(d) Sketch the graph ofg(x)

(e) Sketch the graph ofg(x)


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51. Find the inverse ofg (x) = ln(x 4) + 5. Check your answer using the Round-Trip theorem.

g1 =ey5 + 4

Check that

g g1 (x) =x and that g1 g (x) =x52. Compute and simplify the difference quotient of f(x) = x2 2x+ 1. Recall the difference

quotient:f(x+h) f(x)

h

2x+h 2

53. Let f(x) =x2 2x 6(a) On what intervals is f increasing?

[1,)

(b) On what intervals is f(x) 0?, 1

7

1 +

7,

54. Write the rule of a function that is not a polynomial. Say why.

f(x) = 1

x=x1, which involves a negative exponent.

55. Sketch a graph of a function that could not possibly be the graph of a polynomial. Describewhy.

56. Write

4s6t4

3/2(8s6t3)2/3

as nta/sb for positive numbers n,a, andb.

2t8

s13

57. If logb2 =x and logb3 =y, evaluate logb281 in terms ofxand y.

x 4y


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58. Solve forx. 25x 8 5x = 12 (Hint: Letu = 5x.)

x= ln 6

ln 5 or x=

ln 2

ln 5

59. Solve forx. ln(3x+ 5) 1 = ln(2x 3)

x=3e+ 5

2e 3

60. If (2,3) is on the graph of an odd function, then so is the point .

(2, 3)

61. If (2,3) is on the graph off then the point is on the graph of 5f(3x) + 4.

(23 , 19)

62. If (2,3) is on the graph of a one-to-one functionf then is on the graph of f1.

(3, 2)

63. If (2,3) is on the graph of a parabola with axis of symmetry x = 1, then so is .

(0,3)

64. If the point (3, 4) is on the graph off and the point (3, 10) is on the graph ofg, then whatpoint is on the graph off g?

(a) (9, 40)

(b) (3, 40)

(c) (9, 14)

(d) Not enough information to determine

(e) None of the above


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65. If the discriminant of a quadratic function is negative then what can we say about its graph?

(a) It opens downward

(b) It has no xintercepts

(c) It has an absolute minimum

(d) It is always decreasing


66. Fifty wolves are placed into a newly acquired habitat. The wolf population over time is modeledby a rational function. Eventually, the population stabilizes at 225 wolves. What does this tellyou about the graph of the function modeling the population?

(a) It has a vertical asymptote atx= 225

(b) It has a horizontal asymptote aty= 225

(c) It has an absolute maximum value of 225

(d) It has a hole at x = 225

(e) The y-intercept is 225

(f) None of the above

67. The transformations necessary to changef(x) = 10x into 2 10x + 1 are . . .

(a) Reflect over y-axis, stretch vertically by a factor of 2, then shift up 1 unit.

(b) Reflect over y-axis, shift up 1 unit, then stretch vertically by a factor of 2.

(c) Reflect over y-axis, shift right 1 unit, then stretch vertically by a factor of 2.

(d) Reflect over x-axis, stretch vertically by a factor of 2, then shift up 1 unit.

(e) Reflect over x-axis, shift right 1 unit, then stretch vertically by a factor of 2.

68. If the point (a, b) is on the graph of an even function, then so is the point . . .

(a) (a,b)

(b) (a, b)(c) (a,b)(d) (b, a)



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69. A 12-ounce cocktail is made with 4 ounces of cranberry juice and the remainder, 7-UP. Then,jounces of cranberry juice are added to the cocktail. The new concentration of cranberry juicein the cocktail is . . .

(a) j

12(b)

4 +j

12

(c) 8 +j

12 +j

(d) 4 +j

12 +j

(e) 8

12 +j

70. The functionf(x) = 3x(x 3)(x+ 5)(x 6)2 . . .(a) . . . is not a one-to-one function.

(b) . . . does not have an inverse function.

(c) . . . is not an odd function.

(d) . . . does not have even degree.

(e) All of the above.

71. Use the points A= (3, 7) and B= (1,8).

(a) Find the distance betweenAand B .

241

(b) Find the midpoint of the line segmentAB.

1,12

(c) Write the equation of the circle with points on the diameterA andB.

(x 1)2 + (y+ 12)2 = 2414

(d) Find the slope of the line through the pointsAand B .

15

4


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(e) Find the equation of the perpendicular bisector toAB (the line that passes through themidpoint and is perpendicular to the line through A and B ). Hint: Parts (b) and (d) willbe helpful.

y+ 1

2

=

4

15

(x

1)

(f) At what x-value does the line from (e) have a y-value of 4?

x= 1278

72. Consider the polynomial functionR(x) = x5 + 4x4 + 2x3 12x2 9x.

(a) What must be a factor of the polynomial if 3 is a root of multiplicity 2?

(x 3)2

(b) Just from the function rule, find one additional root ofR. What does this fact imply mustbe a factor of the polynomial?

x= 0 is root x is a factor.

(c) If the last remaining root ofR is 1 and it has multiplicity 2, writeR(x) in its completelyfactored form.

R(x) = x(x 3)2(x+ 1)2

73. Find the value ofc such that the linear function containing the points (17, 38) and (1, 86)passes through the point (c, 0).

c= 893

74. Find values for the constantsa,b, andc so that the rational function f(x) =a(x 2)b(x+ 4)3

(x 2)2(x+ 4)chas a hole at x = 2, a vertical asymptote at x =

4, and a horizontal asymptote at y = 2.

Include brief notes about why you chose your values.

For the horizontal asymptote, we need a= 2 and that the degree ofthe numerator matches that of the denominator. They do, providedthat b + 3 = 2 + c, or c = b + 1. In order to produce a hole at x = 2,we needb 2. For the vertical asymptote at x = 4, we need c >3.The smallest solution is (a,b,c) = (2, 3, 4).


25/26

Part III: Graphs. Sketchthe graphs of the following functions on the given intervals. Make sureto label your axes!

a. sin x, [0, 2] b. csc x, [0, 2]

c. tan x, [32

,3

2 ] d. ln x, [1, 5]

e. 3sin5x, [0, 2] f. An angle of174

radians in standard po


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Extra credit

1. Complete the square to show that the vertex off(x) =ax2 +bx+cis the point

b

2a, c b

2

4a

.

2. Show thatx

c is always a factor ofxn

cn ifc >0. What can we say ifc

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