Test 3
MATH 1040
Fall 2015
2.7, 2.8, 3.1 – 3.4
Version A
Student’s Printed Name: __Key_&_Grading Guidelines__ CUID:___________________
Instructor: ______________________
Section # :_________
You are not permitted to use a calculator on any portion of this test. You are not allowed to use
any textbook, notes, cell phone, laptop, PDA, or any technology on either portion of this test. All
devices must be turned off while you are in the testing room.
During this test, any communication with any person (other than the instructor or his designated
proctor) in any form, including written, signed, verbal, or digital, is understood to be a violation
of academic integrity.
No part of this test may be removed from the testing room.
Read each question very carefully. In order to receive full credit for the free response portion of
the test, you must:
1.
Show legible and logical (relevant) justification which supports your final answer.
2.
Use complete and correct mathematical notation.
3.
Include proper units, if necessary.
4.
Give exact numerical values whenever possible.
You have 90 minutes to complete the entire test.
On my honor, I have neither given nor received inappropriate or unauthorized information
at any time before or during this test.
Student’s Signature: ________________________________________________
Do not write below this line.
Free
Response
Problem
Possible
Points
9
6c
5
2
6
6d
3
3
7
7
7
4
8
8
7
5
6
1
6a
4
66
6b
3
9
Free
Response
Multiple
Choice
Test Total
100
Free Response
Problem
Possible
Points
1
Points
Earned
Points
Earned
34
Page 1 of 12
Test 3
MATH 1040
Fall 2015
Version A
2.7, 2.8, 3.1 – 3.4
Multiple Choice. There are 12 multiple choice questions. Each question is worth 2 – 3
points and has one correct answer. The multiple choice problems will count as 34% of the
total grade. Use a number 2 pencil and bubble in the letter of your response on the
scantron sheet for problems 1 – 12. For your own record, also circle your choice on your
test since the scantron will not be returned to you. Only the responses recorded on your
scantron sheet will be graded. You are NOT permitted to use a calculator on any portion of
this test.
1.
(3 pts.)
Given that f (π ) = 4 , f ′ (π ) = −1 , g (π ) = −3 , and g ′ (π ) = 2 .
If h ( x ) = f ( x ) i g ( x ) − f ( x ) , find h ′ (π ) .
a) -15
b) -3
c) -1
d) 12
e) 11
2.
(3 pts.)
Given that f ( −2 ) = 4 and f ′ ( −2 ) = −3 , find an equation for the normal line to the
graph of f ( x ) at x = −2 .
1
14
a) y = − x +
3
3
b) y = −3x − 2
c) y =
1
14
x−
3
3
d) y =
1
14
x+
3
3
e) y = −3x + 4
3.
(2 pts.)
The following limit describes the derivative of some function f at some number a.
3
−27 + h + 3
lim
h
h→0
What are f ( x ) and a?
a) f (x) = 3 x , a = −27
b) f (x) = 3 x 3 , a = 3
c) f (x) = 3 −x , a = −3
d) f (x) = 3 x 3 , a = −3
e) f (x) = 3 −x , a = 27
Page 2 of 12
Test 3
MATH 1040
Fall 2015
2.7, 2.8, 3.1 – 3.4
Version A
4.
(3 pts.)
5.
(3 pts.)
Solve the following equation on [0,2π ] .
d2
d
sin x ] + [ cos x ] = 1
2[
dx
dx
a) x =
7π 11π
,
6 6
b) x =
4π 5π
,
3 3
c) x =
π 2π
,
3 3
d) x =
π 5π
,
6 6
e) x =
5π 7π
,
4 4
Find the x-value where the tangent line to f ( x ) =
y = −2x + π .
7x − e x x
is parallel to the line
x
a) x = e2
b) x = ln 2
⎛ 1⎞
c) x = ln ⎜ ⎟
⎝ 2⎠
d) x = −
1
e2
e) x = ln 9
6.
(3 pts.)
Compute
d tan x
(2 ) .
dx
a) ln ( 2 ) 2 tan x sec x tan x
b) ln ( 2 ) 2 tan x sec 2 x
c) 2 tan x sec 2 x
d) 2 sec
2
x
e) ln ( 2 ) 2 sec x tan x
Page 3 of 12
Test 3
MATH 1040
7.
(3 pts.)
Fall 2015
Version A
2.7, 2.8, 3.1 – 3.4
The graph of the function f ( x ) is shown below. Which of the following is the graph
of the derivative function f ′ ( x ) ?
4
f ( x)
3
2
1
-2
-1
1
a)
b)
4
2
3
-2
-2
-1
2
-2
1
-4
-1
1
1
2
1
2
-6
2
d)
c)
-2
2
4
4
2
3
-1
1
2
2
1
-2
-4
-2
-1
e)
4
2
-2
-1
1
2
-2
-4
Page 4 of 12
Test 3
MATH 1040
Fall 2015
2.7, 2.8, 3.1 – 3.4
Version A
8.
(2 pts.)
Which of the following functions is NOT differentiable everywhere, that is, on the
interval ( −∞,∞ ) ?
a) f (x) = e x
b) f ( x ) = 2
c) f (x) = sin x
d) f (x) = x 2 + x + 1
e) f ( x ) =
9.
(3 pts.)
Let y =
a) y ′ =
c) y ′ =
cos(x)
and assume f ( x ) is a differentiable function. What is y ′ ?
f (x)
sin(x)
⎡⎣ f '(x) ⎤⎦
10.
b) y′ =
2
sin(x) f (x) − cos(x) f '(x)
e) y ′ = −
(3 pts.)
1
x2
⎡⎣ f (x) ⎤⎦
cos x ⋅ f ′ ( x ) + f ( x ) sin x
cos 2 x
d) y ′ = −
2
sin(x)
f '(x)
sin(x) f (x) + cos(x) f '(x)
⎡⎣ f (x) ⎤⎦
2
Given the information in the table about the functions f ( x ) , g(x) , and h(x) ,
(
)
compute the derivative of f g ( h ( x )) when x = 2 .
x
2
3
f ( x)
4
7
f ′( x)
6
1
g( x)
2
3
g′ ( x )
5
7
a) 28
b) 42
c) 252
d) 1
h( x)
3
2
h′ ( x )
4
6
e) 120
Page 5 of 12
MATH 1040
Test 3
Version A
11.
(3 pts.)
Fall 2015
2.7, 2.8, 3.1 – 3.4
The tangent line to a function f ( x ) at the point (2, 2) goes through (0, 1).
What is f ′ ( 2 ) ?
a) 1
b) 2
c) 1/2
d) 0
e) –1/2
12.
(3 pts.)
If the function f ( x ) is differentiable at the point ( a, f ( a )) , then
a) The Intermediate Value Theorem guarantees that f ′ ( x ) has a root at x = a .
b) f ( x ) must NOT be continuous at ( a, f ( a )) .
c) f ( x ) must be continuous at ( a, f ( a )) .
d) f ′ ( a ) must be positive.
e) f ( x ) must have a horizontal tangent line at ( a, f ( a )) .
The Free Response section follows.
PLEASE TURN OVER YOUR SCANTRON while you work on the Free Response questions. You
are welcome to return to the Multiple Choice section at any time.
Page 6 of 12
MATH 1040
Test 3
Version A
Fall 2015
2.7, 2.8, 3.1 – 3.4
Free Response. The Free Response questions will count as 66% of the total grade. Read
each question carefully. In order to receive full credit you must show legible and logical
(relevant) justification which supports your final answer. Give answers as exact answers.
You are NOT permitted to use a calculator on any portion of this test.
1. The Mars rover Curiosity has cruising speeds of up to 300 ft/hour, but to ensure its safety
while gathering data, it usually travels much slower. Suppose Curiosity’s next mission is to
travel out to a particular point, turn around, and come back to its starting position. We can
describe the rover’s position from the starting position using the position function
9 3
81 1
and t is measured in hours.
s(t) = t − t 3 feet, where 0 ≤ t ≤
2
4
3
Be sure to include units with answer on each part.
a) (5 pts.) Find when the velocity is 0 ft/hour.
81
Work on Problem
Points Awarded
v(t) = − t 2 ft/s
4
Velocity function (units not
2
required)
81 2
Velocity = 0
1
−t = 0
4
Solve
1.5
Correct answer w units
0.5
81
t2 =
Notes:
4
-0.5 for missing or incorrect units on final answer
OK never shows negative solution (and thus does not eliminate)
9 9
-1 per instance of notation equating s and v, or v and a
t= ,−
2 2 nid
-4 answers without supporting work
9
hr
Work on Problem
Points Awarded
2
b) (2 pts.) Find the acceleration function.
Acceleration function
1.5
proper units
0.5
Notes:
-0.5 for missing or incorrect units on final answer
a ( t ) = −2t ft/hr 2
c) (2 pts.) What is the rover’s acceleration when the velocity is 0 ft/hour?
Work on Problem
⎛ 9⎞
⎛ 9⎞
a ⎜ ⎟ = −2 ⎜ ⎟ = −9 ft/hr 2
Proper
evaluation
of part b
⎝ 2⎠
⎝ 2⎠
proper units
Points Awarded
1.5
0.5
Notes:
-0.5 for missing or incorrect units on final answer
(if units on b and c are the same but incorrect both should
be marked off)
-1.5 if a(t) in part b is constant and thus evaluation is trivial
-0.5 absolute value applied to acceleration answer
Page 7 of 12
Test 3
MATH 1040
Fall 2015
Version A
2.7, 2.8, 3.1 – 3.4
f ( x) − f (a)
2x
2. (6 pts.) Using lim
, find the slope of the tangent line at (1, 1) for f (x) =
.
x→a
1+ x
x−a
Detailed limit work must be shown. Use of any “shortcut” derivatives rules will receive no
credit.
2x
2x 1+ x
−1
−
f ( x ) − f (1)
x −1
f ′ (1) = lim
= lim 1+ x
= lim 1+ x 1+ x = lim
÷ ( x − 1)
x→1
x→1
x→1
x→1 1+ x
x −1
x −1
x −1
x −1 1
1
1
= lim
⋅
=lim
=
x→1 1+ x x − 1
x→1 1+ x
2
Work on Problem
Points Awarded
Set up
1
Common denominator in numerator
2
Simplify numerator
1
Get rid of complex fraction
1
Correct evaluation
1
Notes:
-1 including 0/0 in work
-1 missing equals
-1 missing limits
-0.5 limit notation carried too far
up to -1 for notational errors (such as omitting the parentheses in the
denominator or poor limit notation)
-3 Bad cancellation
-6 using quotient rule or any derivative rule besides the definition
3. (7 pts.) For what x-value(s) in the interval [0,2π ] does the function y = 2sin 2 (x) have a
horizontal tangent line?
horizontal tangents when y′ = 0
Work on Problem
Points Awarded
y′ = 2 ⋅ 2 ( sin x ) ⋅ cos x = 4 sin x cos x
1
4 sin x cos x = 0
sin x cos x = 0
sin x = 0
cos x = 0
π 3π
x = 0, π ,2π x = ,
2 2
Correct derivative
3
Derivative = 0
1
Solves for x
3
Notes:
-0.5 to -1 algebra mistakes or not simplifying answer
-7 setting function = 0
only received last point of the 3 points for solving if fully
correct
-0.5 each extra “solution”
-0.5 each missing solutions
-2.5 only presenting one solution
if some attempt at chain rule but not correct, derivative 1.5/3
OR
4 sin x cos x = 2sin 2x = 0
2x = 0, π ,2π , 3π , 4π
π
3π
x = 0, , π , ,2π
2
2
Page 8 of 12
Test 3
MATH 1040
Version A
4. Let y = f (x) ⋅ x where f ( x ) is a twice differentiable function.
a. (2 pts.) Find y′ in terms of x, f (x), and f ′ ( x ) .
Fall 2015
2.7, 2.8, 3.1 – 3.4
2
y ′ = f ( x ) ⋅ 2x + f ′ ( x ) ⋅ x 2
Work on Problem
Points Awarded
Product Rule
0.5
Derivative of first term
0.5
Derivative of second term
0.5
Correct
0.5
Notes:
-1 incorrectly labeling derivative or not labeling derivative
-2 incorrect “product rule”
-1 for multiplication instead of addition
b. (3 pts.) Find y ′′ in terms of x, f (x), f ′ ( x ) , and f ′′ ( x ) .
y′′ = f ( x ) ⋅ 2 + f ′ ( x ) ⋅ 2x + f ′ ( x ) ⋅ 2x + f ′′ ( x ) ⋅ x 2
= 2 f ( x ) + 4xf ′ ( x ) + x f ′′ ( x )
2
Work on Problem
Points Awarded
First Product Rule
1.5
Second Product Rule
1.5
Notes:
-1 incorrectly labeling derivative or not labeling derivative
-1.5 incorrect “product rule” each
does not have to be simplified
c. (3 pts.) Given that f ( −2 ) = 2 and f ′ ( −2 ) = 1 , what must the value of f ′′ ( −2 ) be so that
y′′ ( −2 ) = 0 .
y′′ ( −2 ) = 2 f ( −2 ) + 4xf ′ ( −2 ) + x 2 f ′′ ( −2 ) = 0
2 ⋅ 2 + 4 ⋅ ( −2 ) ⋅1+ ( −2 ) f ′′ ( −2 ) = 0
2
4 − 8 + 4 f ′′ ( −2 ) = 0
−4 + 4 f ′′ ( −2 ) = 0
4 f ′′ ( −2 ) = 4
f ′′ ( −2 ) = 1
Work on Problem
Points Awarded
Proper “plug-in”
0.5
Algebra
1.5
Correct answer
1
Notes:
-1 incorrectly labeling derivative or not labeling derivative
-0.5 function = evaluation
last point awarded only if correct
-1 answer not explicitly stated
-3 no work
-0.5 u=f”(-2) but not stated properly
-2 f”(-2)=0 and solved for y”(-2)
-3 f”(-2) given in terms of x
Page 9 of 12
Test 3
MATH 1040
Version A
Fall 2015
2.7, 2.8, 3.1 – 3.4
tan 4x
x→0
3x
Full limit work must be shown. No credit will be given for use of L’Hopital’s Rule.
tan 4 x 1
tan 4 x 1
sin 4 x
1
sin 4 x 1
lim
= lim
= lim
÷ x = lim
⋅
x→0
3x
3 x→0 x
3 x→0 cos 4 x
3 x→0 cos 4 x x
1
sin 4 x
1
= lim
⋅ lim
x→0 cos 4 x
3 x→0 x
Work on Problem
Proper
constants
separated from
4
sin 4 x
1
tangent
= lim
⋅ lim
3 x→0 4 x x→0 cos 4 x
Strategically grouping temrs
Placing k/k where needed
4
1
4
= ⋅1⋅
=
Limit
3 cos 0 3
Notes:
5. (6 pts.) Evaluate the following limit. lim
Points Awarded
1
2
1
2
-6 sinkt=ksint or tankt=ktant
-6 (sinkt)/t = sink or tankt=ktant
-0.5 to -1 algebra mistakes or not simplifying answer
-0.5 poor notation
only received last point of the 2 points for limit if correct
-5 subbed tanx=sinx/cosx but nothing else was correct
-3 used limtankx/kx=1 (not proved in text)
6. Find the first derivative of the following functions. Use appropriate notation to denote the
derivative. You are NOT required to simplify the derivatives.
2
a. (4 pts.) y(t) = 2 + π 6 − 5 t 4 + t 4 π
3t
Work on Problem
Points Awarded
4
2 −2
Derivative of first term
1
6
4π
5
y(t) = t + π − t + t
Derivative
of
second
term
1
3
Derivative of third term
1
1
4 −3
4 −5
Derivative of fourth term
1
4 π −1
y ′ ( t ) = − t + 0 − t + 4π t
Notes:
3
5
-2 incorrectly labeling derivative or not labeling derivative
simplified answers were not checked for correctness
b. (3 pts.) c(x) = e5 cot 7 3x
c ( x ) = e5 ⋅ cot 7 3x = e5 ⋅ ( cot 3x )
7
c ′ ( x ) = 7e5 ( cot 3x ) ⋅ ( − csc 2 3x ) ⋅ 3
6
Work on Problem
Points Awarded
Hold the multiplier
0.5
Derivative of outside, keep the inside
1
Derivative of first inside
1
Derivative of second inside
0.5
Notes:
-1.5 incorrectly labeling derivative or not labeling
derivative
-3 for derivative of outside at derivative of inside
-0.5 missing ( ) each up to -1
simplified answers were not checked for correctness
Page 10 of 12
Test 3
MATH 1040
Fall 2015
2.7, 2.8, 3.1 – 3.4
Version A
c. (5 pts.) f ( x ) =
x2
where x > 0
2x + 1
y=
1⎛ x ⎞
f ′( x) = ⎜
2 ⎝ 2x + 1 ⎟⎠
2
−
1
2
⋅
( 2x + 1)( 2x ) − x ( 2 )
( 2x + 1)2
2
x
1
( 2x + 1) 2
OR
y′ =
1
1
( 2x + 1)− 2 ⋅ 2
2
2
2x + 1
2x + 1 ⋅1− x ⋅
(
)
Work on Problem
Points Awarded
Derivative of outside, keep the inside
2
Derivative of inside w quotient rule
3
Notes:
-2 incorrectly labeling derivative or not labeling derivative
-5 for derivative of outside at derivative of inside
-0.5 missing ( ) each up to -1
simplified answers were not checked for correctness
3
⎛ 3
⎞
x4
d. (3 pts.) m(x) = csc ⎜ 4x − 2e ⎟
⎝
⎠
3
3
3
⎛
⎛
⎞
4 ⎞
4 ⎞ ⎛
4
m ′ ( x ) = − csc ⎜ 4x 3 − 2e x ⎟ cot ⎜ 4x 3 − 2e x ⎟ ⋅ ⎜ 12x 2 − 2e x ⋅ ( −12x −5 )⎟
⎝
⎠
⎝
⎠ ⎝
⎠
Work on Problem
Points Awarded
Derivative of outside, keep the inside
1
Derivative of first inside
1
Derivative of second inside
1
Notes:
-1.5 incorrectly labeling derivative or not labeling
derivative
-3 for derivative of outside at derivative of inside
-1 extra incorrect parts to derivative
-0.5 missing ( ) each up to -1
simplified answers were not checked for correctness
Page 11 of 12
Test 3
MATH 1040
Fall 2015
Version A
2.7, 2.8, 3.1 – 3.4
7. (7 pts.) Let f (x) = 3 − x . Use the limit definition of the derivative to show that
1
f ′( x) = −
.
Work on Problem
Points
2 3− x
Awarded
f ′ ( x ) = lim
h→0
= lim
h→0
= lim
h→0
= lim
h→0
=
h
h
(
3 − ( x + h) − 3 − x 3 − ( x + h) + 3 − x
⋅
h
3 − ( x + h) + 3 − x
3 − x − h − (3 − x)
(
3 − ( x + h) + 3 − x
−h
3 − ( x + h) + 3 − x
)
)
−1
3 − ( x + h) + 3 − x
−1
2 3− x
Set up
1
Conjugate
1
Simplify numerator
2
Cancel h
1
Correct evaluation
2
Notes:
-1 including 0/0 in work
-1 missing equals
-2 missing limits
-0.5 limit notation carried too far
up to -1 for notational errors (such as
omitting the parentheses in the
denominator or poor limit notation)
-1 for stating the conjugate correctly but
not multiplying the denominator by the
conjugate (-2 total if remainder of problem
is correct, additional -1 for incorrect
solution)
-2 Bad cancellation
-7 using derivative in place of f(x)
8. (7 pts.) Find the equation of the tangent line to the curve y = (4x − 4)3 + e2 x−2 at the point (1,1) .
y′ ( x ) = 3( 4x − 4 ) ⋅ 4 + e2 x−2 ⋅ 2
2
y′ (1) = 3⋅ 0 2 ⋅ 4 + e0 ⋅ 2 = 2
so the equation of the tangent to y at (1,1) is
y − 1 = 2(x − 1)
Work on Problem
Derivative first term
Lack of use of chain rule 1st term
Derivative of second term
Lack of use of chain on 2nd term
Derivative evaluation
Equation of line
(slope 1, x ½, y ½)
Points Awarded
2
-1
2
-1
1
2
Notes:
-0.5 arithmetic error
-1 extra terms in derivative
-0.5 lost exponent in first term
-0.5 to -1 poor notation
-0.5 failure to label y’(1)
-1 y’(x)=y’(1)
-4 tangent line is not linear (i.e. they used derivative instead
of derivative evaluation)
-1 slope incorrect
-0.5 point/slope correct but slope/intercept incorrect
9. (1 pt.) Check to make sure your Scantron form meets the following criteria. If any of the items
are NOT satisfied when your Scantron is handed in and/or when your Scantron is processed one
point will be subtracted from your test total.
My scantron:
□ …
□ has Test No. 3 written at the top;
□ has Test Version A both written at the top and bubbled in below my CUID;
□ and shows my correct CUID both written and bubbled in (bubble in a 0 in place of the C).
Page 12 of 12