17 Lecture need to Solve Problems On Calculus 1

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17 Lecture need to Solve Problems On Calculus 1


1.4, 1.5, 1.6, 1.7, 1.8

2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.8, 2.9

3.1, 3.2, 3.3

Tutor Answer

Ace_Tutor
School: Boston College

Attached.

Question 1:
(a) The average velocity of the ball between 1 and 2 seconds is equal to



 



4.9  2   10  2   4.9 1  10 1
h  2   h 1

 4.7 m/s
2 1
2 1
2

2

(b) The average velocity of the ball between 1 and 1.1 seconds is equal to



 



4.9 1.1  10 1.1  4.9 1  10 1
h 1.1  h 1

 0.29 m/s
1.1  1
0.1
2

2

(c) The average velocity of the ball between 1 and 1.01 seconds is equal to



 



4.9 1.01  10 1.01  4.9 1  10 1
h 1.01  h 1

 0.0015 m/s
1.01  1
0.01
2

2

(d) The instantaneous velocity of the ball at 1 second is

h 1  9.8t  10 t 1  9.8 1  10  0.2 m/s
Question 2:
(a) The average velocity of the rock between 2 and 7 seconds is equal to

1 2 
1 2

10  7    7    10  2    2  

y  7   y  2 
5
5
 
  8.2 m/s

72
72
(b) The average velocity of the rock between 2 and 3 seconds is equal to

1 2 
1 2

10  3   3   10  2    2  

y  3  y  2  
5
5
 
  9 m/s

3 2
3 2
(c) The average velocity of the rock between 2 and 2.1 seconds is equal to

1
1 2
2 

10  2.1   2.1   10  2    2  

y  2.1  y  2  
5
5
 
  9.18 m/s

2.1  2
0.1
Question 3:
(a) The average velocity of the ball between 2 and 8 seconds after release is equal to

h  8  h  2  180  500

 53.33 m/s
82
6
(b) The speed of the ball at 4 seconds after release is equal to

h  4  420

 105 m/s
4
4
Question 4:
(a) The average velocity of the car between 0 and 4 seconds is equal to

h  4   h  0  270  0 270


 67.5 m/s
40
40
4
(b) The speed of the car at 9 seconds is equal to

1120
 124.4 m/s
9
Question 5:
(a) His average velocity from 0 to 15 seconds is equal to

80
 0.533 m/s
15  0
(b) His average velocity from 60 to 120 seconds is equal to

50  28
 0.366 m/s
120  60
(c) His instantaneous velocity at 30 seconds is estimated by

15
 0.5 m/s
30
Question 6:
(a) The slope is equal to

f   2    sin x x2   sin  2   0
(b) The slope is equal to

f      sin x x   sin    0

(c) The slope is equal to

 
 
f      sin x x     sin    1
2
2
2
(d) The slope is equal to

3
 
 
f      sin x x     sin    
2
3
3
3
(e) The slope is equal to

1
 
 
f      sin x x     sin    
2
6
6
6
(f) The slope of the line segment will get close to 0.
Question 7:
(a) The slope is equal to

f   2   cos x x2  cos  2   1
(b) The slope is equal to

f     cos x x  cos    1
(c) The slope is equal to

 
 
f     cos x x    cos    0
2
2
2
(d) The slope is equal to

 
  1
f     cos x x    cos   
3
3
3 2
(e) The slope is equal to

3
 
 
f     cos x x    cos   
6
6
6 2
(f) The slope of the line segment will get close to 1.
Question 8:

(a) True
(b) False
(c) Because the sequence can be denoted by an   0.1 where n is an integer. It can be
n

observed that lim an  0 ; hence, 0 is a better choice than -100.
n 

Question 9:
The graph is plotted below

The slope of the secant lines from P to (2,f(2)) is computed as
2
f  2   f 1  4  2   3
slope 

 3
2 1
1

The slope of the secant lines from P to (1.1,f(1.1)) is computed as

f 1.1  f 1  4  1.1   3

 2.1
1.1  1
0.1
2

slope 

The slope of the secant lines from P to (1.01,f(1.01)) is computed as

f 1.01  f 1  4  1.01   3
slope 

 2.01
1.01  1
0.01
2

Hence, the slope of the tangent line to the graph at point P is estimated to be -2.

Attached.

Question 1:

Question 2:

Question 3:
a) For x  0, 2 , we have

b) For x   ,   , no global extrema is found.
Question 4:

Question 5:
No absolute maximum and minimum values

Question 6:

x
f  x   x  cos  
2
d 
 x 
 f   x    x  cos   
dx 
 2 
1 x
 1  sin    0
2 2
  7 
 f is an increasing function on  ,
.
 4 4 

Question 7:
(a) The Extreme Value Theorem states that for continuous function f  x  

2 3 5 2
x  x  3x on
3
2

a closed and bounded interval  2, 2 , then f  x  attains either a maximum or a minimum, at
least once.
(b) The global maximum is

Question 8:
(a) The Extreme Value Theorem states that for continuous function f  x   

2 3 9 2
x  x  10 x
3
2

on a closed and bounded interval 0,1 , then f  x  attains either a maximum or a minimum, at
least once.
(b) The global maximum is

Question 9:
(a) The Extreme Value Theorem states that for continuous function sec

 x  on a closed and
3

bounded interval  2, 2 , then f  x  attains either a maximum or a minimum, at least once.
(b) The global minimum and maximum are

Question 10:

Attached.

Question 1:
(a) f  x   1

f  x  h  f  x
h 0
h
1 1
 lim
h 0 h
 0

f   x   lim

(b) f  x   3x

f  x  h  f  x
h 0
h
3  x  h   3x
 lim
h 0
h
3h
 lim
h 0 h
 3

f   x   lim

(c) f  x   x3

f   x   lim
h 0

f  x  h  f  x
h

 x  h
 lim

3

 x3

h 0

h
3 x h  3 xh 2  h3
 lim
h 0
h
2
 lim  3 x  3 xh  h 2 
2

h 0

 3x 2
3
(d) f  x   x  3x  1

f   x   lim
h 0

f  x  h  f  x
h

 x  h   3  x  h   1   x  3x  1
 lim
3

h 0

 3x2  3

3

h

Question 2:
Since f   0   lim

h 

f  0  h   f  0  h2 h

 h h  0 exists, it follows that f  x  is
h
h

differentiable at x  0 , as desired result.
Question 3:
We will compute f   0   lim

h 

f  0  h   f  0
to determine if the limit exists.
h

f  0  h   f  0
h 
h
2
h
h2
 lim
 lim  lim h  
h  h
h  h
h 

f   0   lim

Hence, f  x  is not differentiable at x = 0.

Attached.

Question 1:
(a) The derivative of f  x  is calculated by

f  x  h  f  x
h 0
h
1 x  h  1 x
 lim
h 0
h
f  x   lim

 lim



1 x  h  1 x

h 0

h

h



1 x  h  1 x

1 x  h  1 x

1  x  h   1  x 

 lim
h 0





1 x  h  1 x







1
h 0 1  x  h  1  x
1

1 x  0  1 x
 lim



1
2 1 x

(b) The domain of f(x) is

1 x  0
 x 1
The domain of f’(x) is

1 ...

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Anonymous
Awesome! Exactly what I wanted.

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