Name: Key

Math 004 Final Exam SolutionsMonday, August 4, 2008

8:00–10:00 AM

This exam contains 10 pages. Each page is worth 10 points. Two blank pages are included atthe end to be used as scrap paper for computations and checking answers. You will not need acalculator for this test. You may begin when all students have received the exam.

Instructions

• Stay calm.

• Read each question carefully.

• Write clearly and legibly.

• Show me what you know. If you get stuck on one prob-lem, move on and come back to it. Show the steps in yourcalculations and the reasons for what you’re doing.

• Please box in or circle your answers, and make sure theanswer to a question is on the page where the question isasked.

• If you want scratch work graded for possible partialcredit, mark clearly to which problem it belongs.

Academic Integrity is expected of all students of Cornell Uni-versity at all times, whether in the presence or absence of mem-bers of the faculty.

Understanding this, I declare I shall not give, use, or receiveunauthorized aid in this examination.

Signature of the Student

Page 1: /10

Page 2: /10

Page 3: /10

Page 4: /10

Page 5: /10

Page 6: /10

Page 7: /10

Page 8: /10

Page 9: /10

Page 10: /10

TOTAL: /100

Formulas

Distance formula: d(A,B) =√

(xA − xB)2 + (yA − yB)2

Midpoint formula: mid(A,B) =(xA + xB

2,yA + yB

2

)Slope of a line:

yB − yAxB − xA

General equation of a line: ax+ by = c

Slope-intercept equation of a line: y = mx+ b

Point-slope equation of a line: y − y0 = m(x− x0)

Equation of a circle: (x− a)2 + (y − b)2 = r2

Equation of a parabola: y = ax2 + bx+ c or x = ay2 + by + c

Equation of an ellipse centered at the origin:x2

a2+y2

b2= 1

Equation of a hyperbola centered at the origin:x2

a2− y2

b2= 1 or

y2

b2− x2

a2= 1

Pythagorean identity: cos2 θ + sin2 θ = 1

Angle sum formulas:

{cos(α+ β) = cosα cosβ − sinα sinβsin(α+ β) = sinα cosβ + cosα sinβ

Compound interest formula: A = P(1 +

r

n

)ntContinuously compounded interest formula: A = Pert

1. Answer the following questions simply and completely:

(a) What is the effect on the graph of a function f(x) when x is replaced by x− 5?

Answer: The graph is moved 5 units to the right.

(b) Is −3x2 + x+√x a polynomial in x? Explain why or why not.

Answer: No; it contains a square root of x, while a polynomial can only have positiveinteger powers of x.

(c) Name the four kinds of conic sections.

Answer: circle, parabola, ellipse, hyperbola

(d) What is the base of the natural exponential and natural logarithm functions? Give itsname and an approximation with at least two digits.

Answer: e, which is approximately 2.718281828

(e) Draw a picture to explain why (x+ y)2 = x2 + 2xy + y2.

x

y

x y

x2 xy

yx y2

The large square has an area of (x+ y)2.Because multiplication is commutative,yx = xy, and so the four small rectanglestogether total x2 + 2xy + y2.

1

2. (a) Simplify the expression2(x3)−2

5x2as much as possible.

2x−6

5x2=

25x8

(b) Find the derivative of g(x) = 4x3 − 10x2 + 5.

g′(x) = 12x2 − 20x

(c) Write 45 = 1024 in logarithmic form.

log4 1024 = 5

(d) Write log5(x+ 1) = y in exponential form.

5y = x+ 1

(e) Place the following complex numbers on the plane:

3 + i, −2 + 4i, 2i(−1 + i), (1− i) + (−4 + 2i)

1 2 3 4 5−1−2−3−4−5

1

2

3

4

5

−1

−2

−3

−4

−5

3 + i

−2 + 4i

2i(−1 + i)

(1− i) + (−4 + 2i)

2

3. Solve each of the following equations for x.

x3 − 2x2 − 3x = 0.

x(x2 − 2x− 3) = 0x(x− 3)(x+ 1) = 0

x = 0, 3,−1

22x−1 = 16

22x−1 = 24

2x− 1 = 4

x =52

ln(1/e) = x

ln e−1 = x

x = −1

log2 4 + log2 x = 5

log2 4x = 5

4x = 25 = 32

x = 8

ln 2x = ln(x+ 2)

2x = x+ 2

x = 2

3

4. (a) Find an equation for the straight line through the points (1, 2) and (3, 8).

First, we calculate the slope of the line:8− 23− 1

=62

= 3. Now we can use either point to

find an equation in point-slope form, and if we wish we can convert it to slope-interceptform; any of the following are acceptable as answers:

y − 2 = 3(x− 1)

y − 8 = 3(x− 3)

y = 3x− 1

(b) Find an equation for the tangent line to the graph of y = x3−2x2−x+7 at the point (2, 5).

First, we calculate the slope of the line: let f(x) = x3 − 2x2 − x+ 7. Then the derivativeis f ′(x) = 3x2−4x−1, which at x = 2 becomes f ′(2) = 3(2)2−4(2)−1 = 12−8−1 = 3.Now we can use the point (2, 5) to find an equation in point-slope form, and if we wishwe can convert it to slope-intercept form; either of the following is acceptable:

y − 5 = 3(x− 2)

y = 3x− 1

5. (a) Find the exact value of cosπ

3+ cos

5π4

.

12

+

(−√

22

)=

1−√

22

(b) Find the exact value of cos19π12

(use19π12

=π

3+

5π4

).

Using the angle sum formula for cosine, we find

cos(π

3+

5π4

)= cos

π

3cos

5π4− sin

π

3sin

5π4

=(

12

)(−√

22

)−

(√3

2

)(−√

22

)

=√

6−√

24

4

6. Let (2,−1) be the center of a circle with radius√

2.

(a) Write an equation for this circle.

(x− 2)2 + (y + 1)2 = 2

(b) Find an equation for the tangent line to the circle at the point (3, 0). (Remember: eachtangent line to a circle is perpendicular to the radius at the point of tangency.)

The slope of the radius from (2,−1) to (3, 0) is0− (−1)

3− 2= 1. Because the tangent line

is perpendicular to this segment, its slope is −1. An equation for the tangent line istherefore y = −(x− 3), or y = −x+ 3 .

(c) Draw the circle from (a) and the line from (b) together.

1 2 3 4 5−1

1

2

3

4

−1

−2

5

7. Let f(x) =12x2 − 2

(a) Find the roots of f(x).

We want to solve f(x) = 0, that is, 12x

2 − 2 = 0. This is the same as x2 − 4 = 0, or(x+ 2)(x− 2) = 0. The solutions to this equation, which are the same as the roots of f ,are x = 2,−2 .

(b) Find the roots of f ′(x), and make a table of variations for f .

The derivative of f is f ′(x) = x, which has only 0 as a root , is negative for x < 0, andis positive for x > 0. The table of variations is below.

Table of variations of fInterval (−∞, 0) 0 (0,∞)sign of f ′ − 0 +f is decr. −2 incr.

(c) What is the shape of the graph of f? Draw the graph of f below.

The graph is a parabola .

-15 -10 -5 0 5 10 15 20

-5

5

10

15

6

8. Sketch the graph of each of the following equations. Find at least four points on each graph.

y = 2x + 1

Some possible points to include are(0, 2), (1, 3), (2, 5), (−1, 3

2), and (−2, 54).

-20 -15 -10 -5 0 5 10 15

-5

5

10

15

x2

49+y2

16= 1

Some possible points to include are(7, 0), (−7, 0), (0, 4), and (0,−4).

-15 -10 -5 0 5 10 15 20

-10

-5

5

10

7

y = cosx

Some possible points to include are(0, 1), (π,−1), (2π, 1), (−π,−1),(π2 , 0), (3π

2 , 0), and (−π2 , 0).

-15 -10 -5 0 5 10 15 20

-10

-5

5

10

9. Matching: place the label of each graph next to the description of the function it represents.

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

(e) (f) (g) (h)

exponential with base > 1 (d)

exponential with base < 1 (f)

absolute value (b)

trigonometric (e)

logarithmic (a)

first-degree polynomial (h)

second-degree polynomial (g)

cubic polynomial (c)

8

On this page, you should simplify your answers as much as possible,but you do not need to give decimal approximations.

10. Suppose a certain credit card charges 18% annual interest. You make a single charge of $200on one of these cards. What is the total debt on this amount after six months, if the interestis compounded monthly and no payments are made in that time?

Answer: Using the formula for compound interest, with P = 200, r = .18, n = 12, and t = .5,we find

debt after six months = 200(

1 +.1812

)12(.5)

= 200(1 + .015)6

= $200(1.015)6

(Afterwards, using a calculator, we find this to be about $218.69.)

11. To make yogurt at home, first the milk must be heated to 200◦, then it is set in a room at 65◦.The cooling is modeled by the function

f(t) = 135× 4−t + 65,

where t is measured in hours. How long will it take for the milk to reach 110◦?

Answer: We want to solve the equation 135 · 4−t + 65 = 110:

135× 4−t = 45

4−t =45135

=13

log 4−t = log13

−t log 4 = − log 3

t =log 3log 4

hours

(An answer using any base other than 10 is also acceptable, as long as the same base is usedin all places. The above answer can also be written as log4 3, but no other simplification ispossible. In particular, the answer is not 3

4 , although it’s not far off, because it becomes about47.5 minutes.)

9

12. (a) What angle θ corresponds to the point

(−1

2,

√3

2

)on the unit circle?

2π3, or 120◦

An answer in either degree or radians is acceptable.

(b) Give the following values, where θ is as in part (a):

sin θ =√

32

cos θ = −12

tan θ = −√

3

cot θ = − 1√3

= −√

33

sec θ = −2

csc θ =2√3

=2√

33

(c) If sin θ =35

andπ

2< θ < π, find tan θ.

If sin θ = 35 , then using a 3–4–5 triangle, we find that tan θ = ±3

4 . To determine thesign, we observe that in the second quadrant (which is where π

2 < θ < π), the tangent

is negative, so the answer is tan θ = −34

.

13. What was your favorite topic that we covered this summer?

My answer: I’ve always been a big fan of complex numbers and conic sections.

Thanks to everyone in the class for all your hard work this summer!

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