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Mathematics Midterm Exam, Exams of Differential and Integral Calculus

A midterm exam in mathematics with questions related to functions, limits, and asymptotes. The exam includes exercises on determining whether a relation is a function, evaluating limits, and finding vertical and horizontal asymptotes. The exam also includes an exercise on finding the values of x for which the derivative of a given function is zero. The document can be useful as study notes or exam preparation for a mathematics course covering these topics.

Typology: Exams

2022/2023

Available from 03/25/2023

Artemismd26
Artemismd26 ๐Ÿ‡ต๐Ÿ‡ญ

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MIDTERM EXAMINATION
ENDCAL30
1. Determine whether or not each of the following is a function or relation:
a. S = {(4,7), (5,8), (6,9), (7,10), (8,11)}
- FUNCTION
b. S = {(x,y) such that y=๎˜ x๎˜ ; x ฯต R}
- FUNCTION
c. y =
x2โˆ’5
- FUNCTION
d. ๎˜ y๎˜ = x
- RELATION
e. y =
2x
x+2
- FUNCTION
2. Evaluate the limit:
lim
xโ†’ โˆž
(xโˆ’2
โˆš
2x2โˆ’x+1
)
=
xโˆ’2
โˆš
2x2โˆ’x+1
(
1
x
1
โˆš
x2
)
=
x
xโˆ’2
x
โˆš
2x2
x2โˆ’x
x2+1
x2
=
1โˆ’2
x
โˆš
2โˆ’1
x+1
x2
=
1โˆ’2
x
โˆš
2โˆ’1
x+1
x2
pf3
pf4
pf5

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MIDTERM EXAMINATION

ENDCAL

  1. Determine whether or not each of the following is a function or relation:

a. S = {(4,7), (5,8), (6,9), (7,10), (8,11)}

- FUNCTION

b. S = {(x,y) such that y=โƒ“ xโƒ“ ; x ฯต R}

- FUNCTION

c. y = x

2

- FUNCTION

d. โƒ“yโƒ“ = x

- RELATION

e. y =

2 x

x + 2

- FUNCTION

  1. Evaluate the limit:

lim

xโ†’ โˆž

x โˆ’ 2

2 x

2

โˆ’ x + 1

x โˆ’ 2

2 x

2

โˆ’ x + 1

x

x

2

x

x

x

2 x

2

x

2

x

x

2

x

2

x

x

x

2

x

x

x

2

x

โˆš

x

x

2

โˆš

2

  1. Find all the asymptotes for f

x

5 + 2 x

2

2 โˆ’ x โˆ’ x

2

Vertical asymptote

2 โˆ’ x โˆ’ x

2

( 2 + x ) ( 1 โˆ’ x )= 0

2 + x = 0 1 โˆ’ x = 0

x =โˆ’ 2 โˆ’ x =โˆ’ 1

โˆ’ x

x = 1

x =โˆ’ 2 , 1

Horizontal asymptote

5 + 2 x

2

2 โˆ’ x โˆ’ x

2

x

2

x

2

dy =

2 dx + 6 dx

( 1 โˆ’ 2 x โˆ’ 2 dx )( 1 โˆ’ 2 x )

dy =

8 dx

( 1 โˆ’ 2 x โˆ’ 2 dx )( 1 โˆ’ 2 x )

Step 3:

dy =

8 dx

( 1 โˆ’ 2 x โˆ’ 2 dx )( 1 โˆ’ 2 x )

dx

dy

dx

( 1 โˆ’ 2 x โˆ’ 2 dx )( 1 โˆ’ 2 x )

Step 4:

lim

dx โ†’ 0

( 1 โˆ’ 2 x โˆ’ 2 dx )( 1 โˆ’ 2 x )

lim

dx โ†’ 0

( 1 โˆ’ 2 x โˆ’ 2 ( 0 ))( 1 โˆ’ 2 x )

lim

dx โ†’ 0

( 1 โˆ’ 2 x )( 1 โˆ’ 2 x )

( 1 โˆ’ 2 x )

2

  1. Using the four-step rule, find the derivative of: y =

4 x + 3

and verify your answer using the

differentiation formula.

y =

4 x + 3

Step 1:

y + dy =

โˆš

4 ( x + dx )+ 3

y + dy =

4 x + 4 dx + 3

Step 2:

y + dy โˆ’ y =

4 x + 4 dx + 3 โˆ’

4 x + 3

dy =โˆš 4 x + 4 dx + 3 โˆ’โˆš 4 x + 3

Step 3:

dy =

4 x + 4 dx + 3 โˆ’

4 x + 3 โˆ™

dx

dy

dx

4 x + 4 dx + 3 โˆ’

4 x + 3

dx

dy

dx

โˆš 4 x + 4 dx + 3 โˆ’โˆš 4 x + 3

dx

โˆš 4 x + 4 dx + 3 + โˆš 4 x + 3

4 x + 4 dx + 3 +

4 x + 3

dy

dx

4 x + 4 dx + 3 โˆ’( 4 x + 3 )

dx ( โˆš 4 x + 4 dx + 3 +โˆš 4 x + 3 )

dy

dx

4 x + 4 dx + 3 โˆ’ 4 x โˆ’ 3

dx (

4 x + 4 dx + 3 +

4 x + 3 )

dy

dx

4 x + 4 dx + 3 โˆ’ 4 x โˆ’ 3

dx ( โˆš 4 x + 4 dx + 3 + โˆš 4 x + 3 )

dy

dx

4 dx

dx (

4 x + 4 dx + 3 +

4 x + 3 )

dy

dx

4 x + 4 dx + 3 +

4 x + 3

Step 4:

lim

dx โ†’ 0

4 x + 4 dx + 3 +

4 x + 3

lim

dx โ†’ 0

โˆš

4 x + 4 ( 0 )+ 3 + โˆš 4 x + 3

lim

dx โ†’ 0

4 x + 3 +

4 x + 3

2 โˆš 4 x + 3

4 x + 3

4 x + 3

4 x + 3

4 x + 3

4 x + 3

4 x + 3

Check using the Derivative of a Radical

Note: d

dx

u =

du

dx

u

d

dx

โˆš 4 x + 3

d

dx

( 4 x + 3 )

2 โˆš 4 x + 3

2 โˆš 4 x + 3

4 x + 3