Calculus Placement Solution

I. FUNDAMENTALS

A. Real Numbers

(i) Graph the intervals (−5, 3] and (2, ∞) on the real number line.

Answer:

(ii) Express the inequalities x ≤ 3 and −1 ≤ x < 4 in interval notation.

Answer: (−∞, 3] and [−1, 4).

B. Exponents and Radicals

(i) Evaluate each expression.

(a)523

521(b)

(2

3

)−2

(c) 16−3/4

Solution: We have

(a)523

521= 523−21 = 52 = 25

(b)

(2

3

)−2

=

(3

2

)2

=22

32=

9

4

(c) 16−3/4 =1

163/4=

1

( 4√

16)3=

1

23=

1

8

(ii) Rationalize the denominator and simplify:

√10√

5 − 2

Solution: We have√

10√5 − 2

=

√10(

√5 + 2)

(√

5 − 2)(√

5 + 2)=

√10(

√5 + 2)

(√

5)2 − 22

=

√10(

√5 + 2)

5 − 4

=

√10(

√5 + 2)

1

=√

10(√

5 + 2)

=√

10√

5 + 2√

10

=√

2 · 5√

5 + 2√

10

=√

2√

5√

5 + 2√

10 = 5√

2 + 2√

10

1

C. Algebraic Expressions

(i) Simplify each expression. Write your final answer without negative exponents.

(a) (3a3b3)(4ab2)2 (b)

(3x3/2y3

x2y−1/2

)−2

(c)x2

x2 − 4− x + 1

x + 2(d)

y

x− x

y1

y− 1

x

Solution: We have

(a) (3a3b3)(4ab2)2 = (3a3b3)(42a2(b2)2) = (3a3b3)(16a2b4) = 48a3+2b3+4 = 48a5b7

(b)

(3x3/2y3

x2y−1/2

)−2

=

(x2y−1/2

3x3/2y3

)2

=

(x2−3/2

3y3+1/2

)2

=

(x1/2

3y7/2

)2

=(x1/2)2

32(y7/2)2=

x

9y7

(c)x2

x2 − 4− x + 1

x + 2=

x2

(x − 2)(x + 2)− x + 1

x + 2=

x2

(x − 2)(x + 2)− (x − 2)(x + 1)

(x − 2)(x + 2)

=x2 − (x − 2)(x + 1)

(x − 2)(x + 2)

=x2 − (x2 − x − 2)

(x − 2)(x + 2)

=x2 − x2 + x + 2

(x − 2)(x + 2)

=x + 2

(x − 2)(x + 2)

=1

x − 2

(d)

y

x− x

y1

y− 1

x

=

xy

(y

x− x

y

)

xy

(1

y− 1

x

) =

xy · y

x− xy · x

y

xy · 1

y− xy · 1

x

=

xy2

x− x2y

yxy

y− xy

x

=y2 − x2

x − y

=(y − x)(y + x)

x − y

= −(x − y)(y + x)

x − y

= −y + x

1

= −(y + x)

= −y − x

2

(ii) Perform the indicated operations and simplify.

(a) 3(x + 6) + 4(2x − 5) (b) (x + 3)(4x − 5) (c) (2x + 3)2 (d) (x + 2)3

Solution:

(a) We have

3(x + 6) + 4(2x − 5) = 3 · x + 3 · 6 + 4 · 2x − 4 · 5 = 3x + 18 + 8x − 20 = 11x − 2

(b) We have

(x + 3)(4x − 5) = x · 4x − x · 5 + 3 · 4x − 3 · 5 = 4x2 − 5x + 12x − 15 = 4x2 + 7x − 15

(c) We have(2x + 3)2 = (2x)2 + 2 · 2x · 3 + 32 = 4x2 + 12x + 9

since (a + b)2 = a2 + 2ab + b2.

(d) We have(x + 2)3 = x3 + 3x2 · 2 + 3x · 22 + 23 = x3 + 6x2 + 12x + 8

since (a + b)3 = a3 + 3a2b + 3ab2 + b3.

(iii) Factor each expression completely.

(a) 2x2 + 5x − 12 (b) x3 − 3x2 − 4x + 12 (c) x4 + 27x (d) x3y − 4xy

Solution:

(a) We first find two numbers a and b such that

a + b = 5 and ab = 2(−12) = −24

One can see that a = −3 and b = 8. We have

2x2 + 5x − 12 = 2x2 − 3x + 8x − 12 = x(2x − 3) + 4(2x − 3) = (2x − 3)(x + 4)

(b) We have

x3 − 3x2 − 4x + 12 = x2(x − 3) − 4(x − 3) = (x − 3)(x2 − 4) = (x − 3)(x2 − 22)

= (x − 3)(x − 2)(x + 2)

since a2 − b2 = (a − b)(a + b).

(c) We havex4 + 27x = x(x3 + 27) = x(x3 + 33) = x(x + 3)(x2 − 3x + 9)

since a3 + b3 = (a + b)(a2 − ab + b2).

(d) We havex3y − 4xy = xy(x2 − 4) = xy(x2 − 22) = xy(x − 2)(x + 2)

since a2 − b2 = (a − b)(a + b).

3

D. Equations

Find all real solutions.

(a) x+5 = 14− 1

2x (b)

2x

x + 1=

2x − 1

x(c) x2−x−12 = 0 (d) 2x2+4x+1 = 0

(e)√

3 −√

x + 5 = 2 (f) x4 − 3x2 + 2 = 0 (g) 3|x − 4| = 10

Solution:

(a) We have

x + 5 = 14 − 1

2x

2(x + 5) = 2

(

14 − 1

2x

)

2x + 10 = 28 − x

2x + 10 + x = 28 − x + x

3x + 10 = 28

3x + 10 − 10 = 28 − 10

3x = 18

3x

3=

18

3

x = 6

(b) We have2x

x + 1=

2x − 1

x

x(x + 1)2x

x + 1= x(x + 1)

2x − 1

x

2x2(x + 1)

x + 1=

(2x − 1)x(x + 1)

x

2x2 = (2x − 1)(x + 1)

2x2 = 2x2 + 2x − x − 1

2x2 = 2x2 + x − 1

2x2 − 2x2 = 2x2 + x − 1 − 2x2

0 = x − 1

x = 1

4

(c) Solution 1: We havex2 − x − 12 = 0

(x + 3)(x − 4) = 0

x = −3 or x = 4

Solution 2: We have

x1,2 =−(−1) ±

√

(−1)2 − 4 · 1 · (−12)

2 · 1 =1 ±

√49

2=

1 ± 7

2

therefore

x =1 − 7

2= −3 or x =

1 + 7

2= 4

(d) We have

x1,2 =−4 ±

√42 − 4 · 2 · 12 · 2 =

−4 ±√

8

4=

−4 ± 2√

2

4= −1 ±

√2

2

therefore

x = −1 −√

2

2or x = −1 +

√2

2

(e) We have√

3 −√

x + 5 = 2

(√

3 −√

x + 5)2 = 22

3 −√

x + 5 = 4√

x + 5 = −1

Since square root can not be negative, it follows that this equation has no solutions.

(f) If we set u = x2, then we get a quadratic equation in the new u:

u2 − 3u + 2 = 0

(u − 1)(u − 2) = 0

u = 1 or u = 2

x2 = 1 or x2 = 2

x = ±1 or x = ±√

2

(g) We have3|x − 4| = 10

|x − 4| =10

3

x − 4 = −10

3or x − 4 =

10

3

x =2

3or x =

22

3

5

E. Modeling with Equations

(i) Mary drove from Amity to Belleville at a speed of 50 mi/h. On the way back, she drove at

60 mi/h. The total trip took 42

5h of driving time. Find the distance between these two cities.

Solution: Let d represent the distance between the two cities. Let t represent the amount oftime she drove 50 mph. Then since 4 hours 24 minutes equals 4.4 hours, 4.4− t represents theamount of time driven at 60 mph.

Then d1 = 50t describes the outbound trip and d2 = 60(4.4 − t) describes the inbound trip,and, unless somebody picked Amity up and moved it while she was on the road or in Belleville,d1 = d2. Therefore:

50t = 60(4.4 − t) =⇒ 50t = 264 − 60t =⇒ 110t = 264 =⇒ t =264

110=

12

5

Therefore d = 50t = 50 · 12

5= 120 mi.

(ii) A rectangular parcel of land is 70 ft longer than it is wide. Each diagonal between oppositecomers is 130 ft. What are the dimensions of the parcel?

Solution: Let w = width, then w + 70 = length. Applying Pythagorean theorem:

w2 + (w + 70)2 = 1302

w2 + w2 + 2 · 70w + 702 = 16900

2w2 + 140w + 4900 = 16900

2w2 + 140w − 12000 = 0

w2 + 70w − 6000 = 0

(w − 50)(w + 120) = 0

w = 50 or w = −120

Toss out the negative solution: width is 50 feet, length is 50+70= 120 feet.

(iii) A bottle of medicine is to be stored at a temperature between 5◦C and 10◦C. Whatrange does this correspond to on the Fahrenheit scale? [Note: Fahrenheit (F ) and Celsius (C)

temperatures satisfy the relation C =5

9(F − 32).]

Solution: We have

5 ≤ 5

9F − 5

9· 32 ≤ 10

5 ≤ 5

9F − 160

9≤ 10

5 +160

9≤ 5

9F ≤ 10 +

160

9

41 ≤ F ≤ 50

6

F. Inequalities

Solve each inequality. Write the answer using interval notation, and sketch the solution on thereal number line.

(a) −4 < 5− 3x ≤ 17 (b) x(x− 1)(x + 2) > 0 (c) |x− 4| < 3 (d)2x − 3

x + 1≤ 1

Solution:

(a) We have−4 < 5 − 3x ≤ 17

−4 − 5 < −3x ≤ 17 − 5

−9 < −3x ≤ 12

12

−3≤ x <

−9

−3

−4 ≤ x < 3

(b) We know that the corresponding equation x(x−1)(x+2) = 0 has the solutions 0, 1, and −2.As shown in the Figure below, the numbers 0, 1, and −2 divide the real line into four intervals:(−∞,−2), (−2, 0), (0, 1), and (1,∞).

-• • •-2 0 1

On each of these intervals we determine the signs of the factors using test values. We choosea number inside each interval and check the sign of the factors x, x− 1, and x + 2 at the valueselected. For instance, if we use the test value x = −3 for the interval (−∞,−2) shown inFigure above, then substitution in the factors x, x − 1, and x + 2 gives

x = −3, x − 1 = −3 − 1 = −4, x + 2 = −3 + 2 = −1

All three factors are negative on this interval, therefore x(x−1)(x+2) is negative on (−∞,−2).Similarly, using the test values x = −1, x = 1/2 and x = 2 for the intervals (−2, 0), (0, 1) and(1,∞), respectively, we get:

-• • •-2 0 1– + – +

Thus, the solution of the inequality x(x − 1)(x + 2) > 0 is (−2, 0) ∪ (1,∞).

7

(c) We have

|x − 4| < 3 =⇒ −3 < x − 4 < 3 =⇒ −3 + 4 < x < 3 + 4 =⇒ 1 < x < 7

(d) We have2x − 3

x + 1≤ 1

2x − 3

x + 1− 1 ≤ 0

2x − 3

x + 1− x + 1

x + 1≤ 0

2x − 3 − (x + 1)

x + 1≤ 0

2x − 3 − x − 1

x + 1≤ 0

x − 4

x + 1≤ 0

As shown in the Figure below, the numbers 4 (at which the numerator ofx − 4

x + 1is 0) and −1 (at

which the denominator ofx − 4

x + 1is 0) divide the real line into three intervals: (−∞,−1), (−1, 4),

and (4,∞).

-• •-1 4

On each of these intervals we determine the sign ofx − 4

x + 1using test values. We choose a

number inside each interval and check the sign ofx − 4

x + 1at the value selected. For instance, if

we use the test value x = −2 for the interval (−∞,−1) shown in Figure above, then substitution

inx − 4

x + 1gives

−2 − 4

−2 + 1=

−6

−1= 6 > 0

Similarly, using the test values x = 0 and x = 5 for the intervals (−1, 4), and (4,∞), respectively,we get:

-• •-1 4+ – +

Thus, the solution of the inequality2x − 3

x + 1≤ 1 is (−1, 4].

8

G. Coordinate Geometry.

(i) (a) Plot the points P (0, 3), Q(3, 0), and R(6, 3) in the coordinate plane. Where must thepoint S be located so that PQRS is a square?

Answer: S(3, 6)

(b) Find the area of PQRS.

Solution: Let d = |PQ|. Since |SQ| = 6, we have

d2 + d2 = 62 =⇒ 2d2 = 36 =⇒ d2 = 18

It follows that the area of PQRS is d2 = 18.

(ii) (a) Sketch the graph of y = x2 − 4.(b) Find the x-and y-intercepts of the graph.(c) Is the graph symmetric about the x-axis, the y-axis, or the origin?

Solution:

(a)

(b) To find the x-intercepts, we solvex2 − 4 = 0

It follows that the x-intercepts are −2 and 2. To find the y-intercept, we set x = 0 in y = x2−4.It follows that the y-intercept is y = 02 − 4 = −4.(c) The graph is symmetric about the y-axis only.

9

(iii) Find the center and radius of each circle and sketch its graph.

(a) x2 + y2 = 25 (b) (x − 2)2 + (y + 1)2 = 9 (c) x2 + 6x + y2 − 2y + 6 = 0

Solution:

(a) The center is at (0, 0) and the radius is 5.

(b) The center is at (2,−1) and the radius is 3.

(c) We have

x2 + 6x + y2 − 2y + 6 = (x2 + 6x + 9) + (y2 − 2y + 1) − 4 = (x + 3)2 + (y − 1)2 − 22 = 0

Therefore the center is at (−3, 1) and the radius is 2.

10

H. Lines

(i) Let P (−3, 1) and Q(5, 6) be two points in the coordinate plane.(a) Plot P and Q in the coordinate plane.(b) Find the distance between P and Q.(c) Find the midpoint of the segment PQ.(d) Find the slope of the line that contains P and Q.(e) Find the perpendicular bisector of the line that contains P and Q.(f) Find an equation for the circle for which the segment PQ is a diameter.

Solution:

(a)

(b) The distance between P and Q is

d =√

(−3 − 5)2 + (1 − 6)2 =√

(−8)2 + (−5)2 =√

64 + 25 =√

89

(c) The midpoint of the segment PQ is(−3 + 5

2,1 + 6

2

)

=

(2

2,7

2

)

=

(

1,7

2

)

(d) The slope of the line that contains P and Q is

m =1 − 6

−3 − 5=

−5

−8=

5

8

(e) It follows from (d) that the slope of the perpendicular bisector of the line that contains P

and Q is −8

5, therefore an equation of the perpendicular bisector of the line that contains P

and Q is

y − 7

2= −8

5(x − 1)

which can be rewritten as

y = −8

5x +

51

10

(f) Since the midpoint of the segment PQ is

(

1,7

2

)

and the distance between P and Q is√

89,

it follows that an equation for the circle for which the segment PQ is a diameter is

(x − 1)2 +

(

y − 7

2

)2

=89

4

11

(ii) Write the linear equation 2x− 3y = 15 in slope-intercept form, and sketch its graph. Whatare the slope and y-intercept?Solution: We have

2x − 3y = 15 =⇒ −3y = −2x + 15 =⇒ y =2

3x − 5

Therefore the slope is2

3and y-intercept is −5.

(iii) Find an equation for the line with the given property.(a) It passes through the point (3,−6) and is parallel to the line 3x + y − 10 = 0.(b) It has x-intercept 6 and y-intercept 4.

Solution:

(a) We have3x + y − 10 = 0 =⇒ y = −3x + 10

Therefore the slope of the line is −3. It follows that an equation for the line that passes throughthe point (3,−6) and is parallel to the line 3x + y − 10 = 0 is

y − (−6) = −3(x − 3) =⇒ y = −3x + 3

(b) Since the y-intercept is 4, it follows that b = 4. Since the x-intercept is 6, it follows that

0 = 6m + b

Plugging in 4 into this equation, we get m = −2

3. Therefore an equation for the line that has

x-intercept 6 and y-intercept 4 is

y = −2

3x + 4

12

II. FUNCTIONS

A. Definition and Graphs of Functions

(i) Which of the following are graphs of functions? If the graph is that of a function, is itone-to-one?

Answer: (a) and (b) are graphs of functions, (a) is one-to-one.

(ii) Let f(x) =

√x + 1

x.

(a) Evaluate f(3), f(5), and f(a − 1).(b) Find the domain of f .

Solution:

(a) We have

f(3) =

√3 + 1

3=

√4

3=

2

3

f(5) =

√5 + 1

5=

√6

5

and

f(a − 1) =

√a − 1 + 1

a − 1=

√a

a − 1

(b) The domain of f is [−1, 0) ∪ (0,∞).

13

B. Average Rate of Change

Determine the average rate of change for the function f(t) = t2 − 2t between t = 2 and t = 5.

Solution: Since the average rate of change of a function f between two points t = a and t = bis

f(b) − f(a)

b − a

we have

Average rate =(52 − 2 · 5) − (22 − 2 · 2)

5 − 2=

15 − 0

3= 5

C. Transformations of Functions

(i) How is the graph of y = f(x − 3) + 2 obtained from the graph of f?

Answer: The graph shifts right 3 units, then shifts upward 2 units.

(ii) How is the graph of y = f(−x) obtained from the graph of f?

Answer: The graph reflects about the y-axis.

D. Quadratic Functions; Maxima and Minima

(i) (a) Write the quadratic function f(x) = 2x2 − 8x + 13 in standard form.(b) Sketch a graph of f .(c) What is the minimum value of f?

Solution:

(a) We have

f(x) = 2x2 − 8x + 13 = 2x2 − 8x + 8 + 5 = 2(x2 − 4x + 4) + 5 = 2(x − 2)2 + 5

(b)

(c) Since (x − 2)2 ≥ 0, it follows that

f(x) = 2(x − 2)2 + 5 ≥ 5

Therefore the minimum value of f is 5.

14

(ii) Let f(x) =

{

1 − x2 if x ≤ 0

2x + 1 if x > 0

(a) Evaluate f(−2) and f(1).(b) Sketch the graph of f .

Solution:

(a) We havef(−2) = 1 − (−2)2 = 1 − 4 = −3, f(1) = 2 · 1 + 1 = 3

(b)

E. Modeling with Functions

If 1800 ft of fencing is available to build five adjacent pens, as shown in the diagram below,express the total area of the pens as a function of x. What value of x will maximize the totalarea?

Solution: The function that models the area will be

A(x) = −3x2 + 900x

where x is the width of the pens. It follows that the area will attain its maximum when

x =900

2 · 3 = 150 ft

Using this we can find that the maximum area will be

A(x) = −3 · 1502 + 900 · 150 = 67500 ft

15

F. Combining Functions

If f(x) = x2 + 1 and g(x) = x − 3, find the following.

(a) f ◦ g (b) g ◦ f (c) f(g(2)) (d) g(f(2)) (e) g ◦ g ◦ g

Solution: We have

(a) f ◦ g = (x − 3)2 + 1

(b) g ◦ f = x2 + 1 − 3 = x2 − 2

(c) f(g(2)) = (2 − 3)2 + 1 = 2

(d) g(f(2)) = 22 − 2 = 2

(e) g ◦ g ◦ g = x − 3 − 3 − 3 = x − 9

G. One-to-One Functions and Their Inverses

(i) (a) If f(x) =√

3 − x, find the inverse function f−1.(b) Sketch the graphs of f and f−1 on the same coordinate axes.

Solution:

(a) We have:

Step 1: Replace f(x) by y:y =

√3 − x

Step 2: Solve for x:

y =√

3 − x =⇒ y2 = 3 − x =⇒ x = 3 − y2

Step 3: Replace x by f−1(x) and y by x:

f−1(x) = 3 − x2

Since the range of f(x) is all nonnegative numbers, it follows that the domain of f−1(x) isx ≥ 0. So,

f−1(x) = 3 − x2, x ≥ 0

(b)

16

(ii) The graph of a function f is given.(a) Find the domain and range of f .(b) Sketch the graph of f−1.(c) Find the average rate of change of f between x = 2 and x = 6.

Answer:

(a) Domain [0, 6], range [1, 7].

(b)

(c)5

4

17

III. POLYNOMIAL AND RATIONAL FUNCTIONS

A. Polynomial Functions and Their Graphs

Graph the polynomial P (x) = −(x + 2)3 + 27, showing clearly all x- and y-intercepts.

Answer:

B. Dividing Polynomials

Use long division to find the quotient and remainder when 2x5 + 4x4 − x3 − x2 + 7 is dividedby 2x2 − 1.

Answer: The quotient and remainder are x3 + 2x2 +1

2and

15

2, respectively.

C. Rational Functions

Consider the following rational functions:

r(x) =2x − 1

x2 − x − 2s(x) =

x3 + 27

x2 + 4t(x) =

x3 − 9x

x + 2u(x) =

x2 + x − 6

x2 − 25

(a) Which of these rational functions has a horizontal asymptote?(b) Which of these functions has a slant asymptote?(c) Which of these functions has no vertical asymptote?(d) Graph y = u(x), showing clearly any asymptotes and x-and y-intercepts the function

Answer:

(a) Only r and u have horizontal asymptotes.

(b) Only s has a slant asymptote.

(c) Only s has no vertical asymptote.

(d)

18

IV. EXPONENTIAL AND LOGARITHMICFUNCTIONS

A. Exponential Functions

Graph the function y = 2x−3.

Answer:

B. Logarithmic Functions

Sketch the graph of the function f(x) = log(x+1) and state the domain, range, and asymptote.

Answer: Domain (−1,∞), range (−∞,∞), asymptote x = −1.

C. Laws of Logarithms

(i) Evaluate each logarithmic expression.

(a) log3

√27 (b) log2 80 − log2 10 (c) log8 4

Solution: We have

(a) log3

√27 = log3

√33 = log3 33/2 =

3

2log3 3 =

3

2· 1 =

3

2

(b) log2 80 − log2 10 = log2

80

10= log2 8 = log2 23 = 3 log2 2 = 3 · 1 = 3

(c) log8 4 = log8 22 = log8(3√

8)2 = log8 82/3 =2

3log8 8 =

2

3· 1 =

2

3

19

(ii) Expand: log 3

√x + 2

x4(x2 + 4).

Solution: We have

log 3

√

x + 2

x4(x2 + 4)= log

(x + 2)1/3

x4/3(x2 + 4)1/3=

1

3log(x + 2) − 4

3log x − 1

3log(x2 + 4)

(iii) Combine into a single logarithm: ln x − 2 ln(x2 + 1) +1

2ln(3 − x4).

Solution: We have

ln x − 2 ln(x2 + 1) +1

2ln(3 − x4) = ln x − ln(x2 + 1)2 + ln

√3 − x4 = ln

x√

3 − x4

(x2 + 1)2

D. Exponential and Logarithmic Equations

Find the solution of the equation, correct to two decimal places.

(a) 10x+3 = 62x (b) 5 ln(3 − x) = 4 (c) log2(x + 2) + log2(x − 1) = 2

Solution:

(a) We have10x+3 = 62x

ln 10x+3 = ln 62x

(x + 3) ln 10 = 2x ln 6

x ln 10 + 3 ln 10 = 2x ln 6

x ln 10 − 2x ln 6 = −3 ln 10

x(ln 10 − 2 ln 6) = −3 ln 10

x =−3 ln 10

ln 10 − 2 ln 6≈ 5.39

(b) We have5 ln(3 − x) = 4

ln(3 − x) =4

5

3 − x = e4/5

x = 3 − e4/5 ≈ 0.77

(c) We havelog2(x + 2) + log2(x − 1) = 2

log2(x + 2)(x − 1) = log2 4

(x + 2)(x − 1) = 4

x = 2 or x = −3

Since x = −3 is not from the domain of log2(x + 2) + log2(x − 1), the only answer is x = 2.

20

E. Modeling with Exponential and Logarithmic Functions

(i) The initial size of a culture of bacteria is 1000. After one hour the bacteria count is 8000.(a) Find a function that models the population after t hours.(b) Find the population after 1.5 hours.(c) When will the population reach 15,000?(d) Sketch the graph of the population function.

Answer: (a) n(t) = 1000e2.07944t (b) 22, 627 (c) 1.3 h (d)

(ii) Suppose that $12,000 is invested in a savings account paying 5.6% interest per year.(a) Write the formula for the amount in the account after t years if interest is compounded

monthly.(b) Find the amount in the account after 3 years if interest is compounded daily.(c) How long will it take for the amount in the account to grow to $20,000 if interest is

compounded semiannually?

Answer: (a) A(t) = 12, 000

(

1 +0.056

12

)12t

(b) $14, 195.06 (c) 9.249 yr

21

V. TRIGONOMETRIC FUNCTIONS OF REALNUMBERS

A. The Unit Circle

The point P (x, y) is on the unit circle in quadrant IV. If x =√

11/6, find y.

Solution: We have

y = −√

1 − x2 = −

√√√√1 −

(√11

6

)2

= −√

1 − 11

36= −

√

25

36= −5

6

B. Trigonometric Functions of Real Numbers

(i) The point P in the figure at the right has y-coordinate4

5. Find:

(a) sin t (b) cos t (c) tan t (d) sec t

Answer: (a)4

5(b) −3

5(c) −4

3(d) −5

3

(ii) Find the exact value.

(a) sin7π

6(b) cos

13π

4(c) tan

(

−5π

3

)

(d) csc3π

2

Answer: (a) −1

2(b) −

√2

2(c)

√3 (d) −1

(iii) Express tan t in terms of sin t, if the terminal point determined by t is in quadrant II.

Solution: We have

tan t =sin t

cos t=

sin t

±√

1 − sin2 t= [since the terminal point is in quadrant II] = − sin t

√

1 − sin2 t

(iv) If cos t = − 8

17and if the terminal point determined by t is in quadrant III, find

tan t cot t + csc t

Solution: We have

tan t cot t + csc t = 1 +1

sin t= 1 +

1

±√

1 − cos2 t

Therefore if cos t = − 8

17and if the terminal point determined by t is in quadrant III, we get

tan t cot t + csc t = 1 − 1√1 − cos2 t

= 1 − 1√

1 −(

− 8

17

)2= 1 − 17

15= − 2

15

22

C. Trigonometric Graphs

(i) Find the amplitude, period, and phase shift of the function y = 2 sin

(1

2x − π

6

)

. Sketch

the graph.

Answer: Amplitude 2, period 4π, phase shift π/3.

(ii) The graph shown below is one period of a function of the form y = a sin k(x−b). Determinethe function.

Answer: y = 2 sin 2(x + π/3)

23

VI. TRIGONOMETRIC FUNCTIONS OF ANGLES

A. Angle Measure

(i) Find the radian measures that correspond to the degree measures 330◦ and -135◦.

Answer: 11π/6 and −3π/4

(ii) Find the degree measures that correspond to the radian measures4π

3and −1.3.

Answer: 240◦ and −74.5◦

B. Trigonometry of Right Triangles

Find tan θ + sin θ for the angle θ shown.

Solution: Since the hypotenuse of the triangle is√

22 + 32 =√

13, it follows that

tan θ + sin θ =2

3+

2√13

=2√

13 + 6

3√

13=

26 + 6√

13

39

C. Trigonometric Functions of Angles

Find the exact value of each of the following.

(a) sin 405◦ (b) tan(−150◦) (c) sec5π

3(d) csc

5π

2

Answer: (a)√

2/2 (b)√

3/3 (c) 2 (d) 1

24

D. The Law of Sines

Refer to the figure below. Find the side labeled x.

Solution: We havex

sin 69◦=

230

sin(180◦ − 52◦ − 69◦︸ ︷︷ ︸

59◦

)

therefore

x =230 sin 69◦

sin 59◦≈ 250.5

E. The Law of Cosines

Refer to the figure below.(a) Find the angle opposite the longest side.(b) Find the area of the triangle.

Solution:

(a) By the Law of Cosines we have

202 = 132 + 92 − 2 · 13 · 9 cosα

therefore

cos α =132 + 92 − 202

2 · 13 · 9 = −25

39It follows that

α = arccos

(

−25

39

)

≈ 129.9◦

(b) The semiperimeter of the triangle is

s =1

2(9 + 13 + 20) = 21

therefore by Heron’s formula the area of the triangle is

A =√

21(21 − 9)(21 − 13)(21 − 20) =√

21 · 12 · 8 · 1 =√

2016 = 12√

14 ≈ 44.9

25

VII. ANALYTIC TRIGONOMETRY

A. Trigonometric Identities

Verify each identity.

(a) tan θ sin θ + cos θ = sec θ (b)2 tanx

1 + tan2 x= sin 2x

Solution:

(a) We have

tan θ sin θ + cos θ =sin θ

cos θsin θ + cos θ =

sin2 θ

cos θ+ cos θ =

sin2 θ

cos θ+

cos2 θ

cos θ

=sin2 θ + cos2 θ

cos θ=

1

cos θ= sec θ

(b) We have

2 tanx

1 + tan2 x=

2sin x

cos x

1 +

(sin x

cos x

)2=

(

2sin x

cos x

)

cos2 x

(

1 +

(sin x

cos x

)2)

cos2 x

=2 sinx cos x

cos2 x + sin2 x=

2 sin x cos x

1= sin 2x

B. Addition and Subtraction Formulas

(i) Find the exact value of each expression.

(a) sin 8◦ cos 22◦ + cos 8◦ sin 22◦ (b) sin 75◦ (c) sinπ

12

Solution:

(a) We have

sin 8◦ cos 22◦ + cos 8◦ sin 22◦ = sin(8◦ + 22◦) = sin 30◦ =1

2

(b) We have

sin 75◦ = sin(45◦ + 30◦) = sin 45◦ cos 30◦ + cos 45◦ sin 30◦

=

√2

2·√

3

2+

√2

2· 1

2=

√6

4+

√2

4=

√6 +

√2

4

(c) We have

sinπ

12= sin

π/6

2=

√

1 − cos(π/6)

2=

√

1 −√

3/2

2=

√

2 −√

3

4=

1

2

√

2 −√

3

26

(ii) For the angles α and β in the figures, find cos(α + β).

Solution: We have

cos(α + β) = cos α cos β − sin α sin β =2√5·√

5

3− 1√

5· 2

3=

2√

5 − 2

3√

5=

10 − 2√

5

15

C. Double-Angle, Half-Angle, and Sum-Product Identities

(i) (a) Write sin 3x cos 5x as a sum of trigonometric functions.(b) Write sin 2x − sin 5x as a product of trigonometric functions.

Solution:

(a) We have

sin 3x cos 5x =1

2[sin(3x + 5x) + sin(3x − 5x)] =

1

2[sin(8x) + sin(−2x)] =

1

2[sin(8x) − sin(2x)]

(b) We have

sin 2x − sin 5x = 2 cos2x + 5x

2sin

2x − 5x

2= 2 cos

7x

2sin

−3x

2= −2 cos

7x

2sin

3x

2

(ii) If sin θ = −4

5and θ is in quadrant III, find tan(θ/2).

Solution: We have

tan(θ/2) =1 − cos θ

sin θ=

1 +√

1 − sin2 θ

sin θ=

1 +

√

1 −(

−4

5

)2

−4

5

=1 +

3

5

−4

5

= −2

27

D. Inverse Trigonometric Functions

(i) Graph y = sin x and y = sin−1 x, and specify the domain of each function.

Answer:

(ii) Find the exact value of cos

(

tan−19

40

)

.

Solution 1: We have

cos x = ± 1√1 + tan2 x

Since −π/2 < tan−1 θ < π/2, it follows that cos (tan−1 θ) > 0. Therefore

cos x =1√

1 + tan2 x

hence

cos

(

tan−19

40

)

=1

√

1 + tan2

(

tan−19

40

) =1

√

1 +

(9

40

)2

=40√

402 + 92

=40

41

Solution 2: Put θ = tan−19

40, so tan θ =

9

40. Then

cos

(

tan−19

40

)

= cos θ =40

41

����������

θ

9

40

41

28

E. Trigonometric Equations

Solve each trigonometric equation in the interval [0, 2π).

(a) 2 cos2 x + 5 cosx + 2 = 0 (b) sin 2x − cos x = 0

Solution:

(a) We have2 cos2 x + 5 cos x + 2 = (2 cosx + 1)(cosx + 2) = 0

therefore2 cos x + 1 = 0 or cosx + 2 = 0

Since the second equation is impossible, it follows that

cos x = −1

2

which gives x =2π

3,

4π

3.

(b) We havesin 2x − cos x = 2 sinx cos x − cos x = cos x(2 sin x − 1) = 0

therefore

cos x = 0 or sin x =1

2

which gives x =π

6,

π

2,

5π

6,

3π

2.

29

VIII. POLAR COORDINATES AND VECTORS

A. Polar Coordinates

(i) Convert the point whose polar coordinates are (8, 5π/4) to rectangular coordinates.

Solution: We have

x = 8 cos5π

4= 8

(

−√

2

2

)

= −4√

2, y = 8 sin5π

4= 8

(

−√

2

2

)

= −4√

2

(ii) Find two polar coordinate representations for the rectangular coordinate point (−6, 2√

3),one with r > 0 and one with r < 0, and both with 0 ≤ θ < 2π.

Solution: We have

r = ±√

(−6)2 + (2√

3)2 = ±√

48 = ±4√

3

and

tan θ =2√

3

−6= − 1√

3

It follows that two polar coordinate representations are (4√

3, 5π/6) and (−4√

3, 11π/6).

B. Graphs of Polar Equations

Graph the polar equation r = 8 cos θ. What type of curve is this? Convert the equation torectangular coordinates.

Solution: We multiply both sides of the equation by r, because then we can use the formulasr2 = x2 + y2 and r cos θ = x. We have

r = 8 cos θ

r2 = 8r cos θ

x2 + y2 = 8x

x2 − 8x + 16 + y2 = 16

(x − 4)2 + y2 = 42

which is a circle of radius 4 and center (4, 0).

30

C. Vectors

Let u be the vector with initial point P (3,−1) and terminal point Q(−3, 9).(a) Express u in terms of i and j.(b) Find the length of u.

Solution:

(a) u = (−3 − 3)i + (9 − (−1))j = −6i + 10j

(b) |u| =√

(−6)2 + 102 =√

136 = 2√

34

D. The Dot Product

(i) Let u = 〈1, 3〉 and v = 〈−6, 2〉.(a) Find u− 3v.(b) Find |u + v|.(c) Find u · v.(d) Are u and v perpendicular?

Solution:

(a) We haveu− 3v = 〈1 − 3(−6), 3 − 3 · 2〉 = 〈19,−3〉

(b) Since u + v = 〈1 + (−6), 3 + 2〉 = 〈−5, 5〉, it follows that

|u + v| =√

(−5)2 + 52 = 5√

2

(c) We haveu · v = 1 · (−6) + 3 · 2 = 0

(d) Since u · v = 0, it follows that u and v are perpendicular.

(ii) Let u = 3i + 2j and v = 5i − j.(a) Find the angle between u and v.(b) Find the component of u along v.(c) Find proj

vu.

Solution: We first find |u|, |v| and u · v:

|u| =√

32 + 22 =√

13, |v| =√

52 + (−1)2 =√

26, u · v = 3 · 5 + 2 · (−1) = 13

(a) We have

cos θ =u · v|u||v| =

13√13√

26=

1√2

therefore the angle between u and v is π/4.

(b) The component of u along v is |u| cos θ =√

13 · 1√2

=

√26

2.

(c) We have

projvu =

(u · v|v|2

)

v =

(13

(√

26)2

)

(5i − j) =5

2i − 1

2j

31

IX. SYSTEMS OF EQUATIONS AND INEQUALITIES

A. Systems of Linear Equations in Two Variables

Find all solutions of the system.

(a)

{x + 3y = 7

5x + 2y = −4(b)

{6x + y2 = 10

3x − y = 5

Answer: (a) x = −2, y = 3. (b) x = 1, y = −2 or x = 5/3, y = 0.

B. Systems of Linear Equations in Several Variables

Find all solutions of the system or show that no solution exists.

(a)

x − y + 2z = 0

2x − 4y + 5z = −5

2y − 3z = 5 = −2

(b)

2x − 3y + z = 3

x + 2y + 2z = −1

4x + y + 5z = 4

Answer: (a) x = 5/2, y = 5/2, z = 0. (b) No solution.

C. Systems of Inequalities

Graph the solution set of the system of inequalities. Label the vertices with their coordinates.

(a)

2x + y ≤ 8

x − y ≥ −2

x + 2y ≥ 4

(b)

{x2 + y ≤ 5

y ≤ 2x + 5

Answer:

32