# 17 Lecture need to Solve Problems On Calculus 1

*label*Mathematics

*timer*Asked: Dec 6th, 2017

*account_balance_wallet*$80

**Question description**

# 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

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

72

72

(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

82

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

40

40

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

80

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 x2 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 x2 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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