# Calculus Exercise Relative Rates of Growth & Exploring The Graphs

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*timer*Asked: Feb 8th, 2019

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### Question Description

Attached is a file with questions pertaining to calculus, no work is necessary unless asked for.

Exploring the Graphs of f, f Prime, and f Double Prime

Relative Rates of Growth

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## Tutor Answer

I wrote answers and solutions. I gave explanations to most questions but ask for more if you want.Some questions are incomplete (mostly the last 5). I am ready to solve them when you'll give me the conditions.Some questions have no correct answers in list.docx and pdf files are identical

1.

The graph of f "(x) is continuous and decreasing with an x-intercept at x = 0. Which of the

following statements is true? (4 points)

The graph of f has a relative maximum at x = 0.

The graph of f has a relative minimum at x = 0.

The graph of f has an inflection point at x = 0.

The graph of f has an x-intercept at x = 0.

2.

The graph below shows the graph of f (x), its derivative f "(x), and its second derivative f

"(x). Which of the following is the correct statement?

(4 points)

A is f ", B is f '. C is f.

A is f ", B is f, C is f '.

A is f ', B is f, C is f ".

A is f, B is f ', C is f ".

There is no correct answer. A' must be negative for x1/3, neither B nor C.

B' must be positive, it could be A. C' must have a root at x=-1.5, neither A nor B.

3.

Below is the graph of f '(x), the derivative of f(x), and has x-intercepts at x = -3, x = 1 and

x = 2. There are horizontal tangents at x = -1.5 and x = 1.5. Which of the following

statements is true?

(4 points)

f has an inflection point at x = -1.5.

f is increasing on the interval from x = -3.2 to x = -4.5.

f has a relative minimum at x = 1.5.

All of these are true.

4.

The graph of f ' (x), the derivative of f of x, is continuous for all x and consists of five line

segments as shown below. Given f (-3) = 6, find the absolute maximum value of f (x) over

the interval [-3, 0].

(4 points)

3

4.5

6

10.5 (add the area of the triangle from -3 to 0)

5.

The graph of y = f '(x), the derivative of f(x), is shown below. Given f(4) = 6, evaluate f(0).

(4 points)

-2

2 (subtract the area of the triangle from 0 to 4)

4

10

1.

Which of the following functions grows the fastest as x goes to infinity? (4 points)

2x

3x (3>e)

ex

x20

2.

Compare the rates of growth of f(x) = x + sinx and g(x) = x as x approaches infinity. (4

points)

f(x) grows faster than g(x) as x goes to infinity.

g(x) grows faster than f(x) as x goes to infinity.

f(x) and g(x) grow at the same rate as x goes to infinity.

The rate of growth cannot be determined.

3.

What does

show? (4 points)

g(x) grows faster than f(x) as x goes to infinity.

f(x) and g(x) grow at the same rate as x goes to infinity.

f(x) grows faster than g(x) as x goes to infinity.

L'Hôpital's Rule must be used to determine the true limit value.

4.

Which of the following functions grows at the same rate as

? (4 points)

x

x2

x3

x4

5.

Which of the following functions grows the slowest as x goes to infinity? (4 points)

0.01x3

0.001x3

.00001x3

They all grow at the same rate.

1.

The function f is continuous on the interval [3, 13] with selected values of x and f(x) given

in the table below. Find the average rate of change of f(x) over the interval [3, 13]. (4

points)

x 3 4 7 10 13

f(x) 2 8 10 12 22

It is

𝑑𝑦

𝑑𝑥

=

22−2

13−3

= 𝟐.

2.

f is a differentiable function on the interval [0, 1] and g(x) = f(3x). The table below gives

values of f '(x). What is the value of g '(0.1)? (4 points)

x 0.1 0.2 0.3 0.4 0.5

f '(x) 1 2 3 -4 5

1

3

9 (𝑔′ (𝑥) = 3𝑓 ′ (3𝑥), 𝑔′ (0.1) = 3𝑓 ′ (0.3) = 9.)

Cannot be determined

3.

f(x) and g(x) are a differentiable function for all reals and h(x) = g[f(5x)]. The table below

gives selected values for f(x), g(x), f '(x), and g '(x). Find the value of h'(1). (4 points)

x 123456

f(x) 0 3 2 1 2 0

g(x) 1 3 2 6 5 0

f '(x) 3 2 1 4 0 2

g '(x) 1 5 4 3 2 0

ℎ′ (𝑥) = 5𝑓 ′ (5𝑥)𝑔′ (𝑓(5𝑥)), ℎ′ (1) = 5𝑓 ′ (5)𝑔′ (𝑓(5)) = 5 ∙ 0 ∙ 𝑔′ (𝑓(5)) = 𝟎.

4.

The table of values below shows the rate of water consumption in gallons per hour at

selected time intervals from t = 0 to t = 12.

Using a left Riemann sum with 5 subintervals, estimate the total amount of water consumed

in that time interval. (4 points)

x 0 2 5 7 11 12

f(x) 5.7 5.0 2.0 1.2 0.6 0.4

22.1

34.8

35.8 (5.7 ∙ 2 + 5 ∙ 3 + 2 ∙ 2 + 1.2 ∙ 4 + 0.6 ∙ 1 = 35.8)

None of these

5.

The function is continuous on the interval [10, 20] with some of its values given in the table

above. Estimate the average value of the function with a Trapezoidal Sum Approximation,

using the intervals between those given points. (4 points)

x 10 12 15 19 20

f(x) -2 -5 -9 -12 -16

It is

1

2∙10

(−7 ∙ 2 − 14 ∙ 3 − 21 ∙ 4 − 28 ∙ 1) = −

-8.750

-7.000

-8.400

7

10

∙ (1 + 3 + 6 + 2) = −8.4.

-5.500

Description

1.

Let

. Use your calculator to find F"(1). (4 points)

𝐹 ′ (𝑥) = 2 tan(4𝑥 2 ), 𝐹 ′′ (𝑥) = 16 sec 2 (4𝑥 2 ), 𝐹 ′′ (1) = 16 sec 2 (4) ≈ 37.448.

5.774

11.549

18.724

37.449

2.

Pumping stations deliver gasoline at the rate modeled by the function D, given

by

with t measured in hours and D(t) measured in gallons per hour. How much

oil will the pumping stations deliver during the 3-hour period from t = 0 to t = 3? Give 3

decimal places. (4 points)

3

3 6𝑡

It is ∫0 𝐷(𝑡)𝑑𝑡 = ∫0

1+2𝑡

3

𝑑𝑡 = ∫0 (3 −

3

1+2𝑡

3

3

2

2

) 𝑑𝑡 = 9 − (ln|1 + 2𝑡|)3𝑡=0 = 9 − ln 7 ≈ 𝟔. 𝟎𝟖𝟏 (𝑔𝑎𝑙𝑙𝑜𝑛𝑠).

3.

A particle moves along the x-axis with velocity v(t) = sin(2t), with t measured in seconds

and v(t) measured in feet per second. Find the total distance travelled by the particle from t

= 0 to t = π seconds. (4 points)

𝜋

𝜋

1

2𝜋

𝜋

It is ∫0 |𝑣(𝑡)|𝑑𝑡 = ∫0 |sin(2𝑡)|𝑑𝑡 = ∫0 |sin 𝑡|𝑑𝑡 = ∫0 sin 𝑡 𝑑𝑡 = cos 0 − cos 𝜋 = 𝟐.

2

("distance travelled", not "displacement")

2

1

0

4.

Find the range of the function

. (4 points)

[-4, 4]

[-4, 0]

[0, 4π]

[0, 8π]

5.

Use the graph of f(t) = 2t + 3 on the interval [-3, 6] to write the function F(x),

where

. (4 points)

F(x) = 2x2 + 6x

F(x) = 2x + 3

F(x) = x2 + 3x + 54

F(x) = x2 + 3x - 18

1.

The graph of f '(x) is continuous, positive, and has a relative maximum at x = 0. Which of

the following statements must be true? (5 points)

The graph of f is always concave down.

The graph of f is always increasing.

The graph of f has a relative maximum at x = 0.

The graph of f has a relative minimum at x = 0.

2.

Below is the graph of f '(x), the derivative of f(x), and has x-intercepts at x = -3, x = 1, and

x = 2 and a relative maximum at x = -1.5 and a relative minimum at x = 1.5. Which of the

following statement is false?

(5 points)

f is concave up from x = -1.5 to x = 1.5. (f' decreases, f'' negative, f concave down)

f has an inflection point at x = 1.5.

f has a relative minimum at x = 2.

All of these are false.

3.

The graph of y = f '(x), the derivative of f(x), is shown below. List the intervals where the

graph of f is concave down.

(5 points)

Where f'' is negative, i.e. f' decreases.

(-4, -2) U (2, 4)

(-2, 2)

(-4, 0)

(0, 4)

4.

Which of the following functions grows the fastest as x grows without bound? (5 points)

f(x) = ex (e>5/2)

g(x) = ecosx

h(x) =

They all grow at the same rate.

5.

Compare the growth rate of the functions f(x) = 4x and g(x) =

. (5 points)

f(x) grows faster than g(x).

g(x) grows faster than f(x).

f(x) and g(x) grow at the same rate.

It cannot be determined.

6.

f is a function that is differentiable for all reals. The value of f '(x) is given for several values

of x in the table below.

x -8 -3 0 3 8

f '(x) -4 -2 0 4 5

If f '(x) is always increasing, which statement about f(x) must be true? (5 points)

f(x) passes through the origin.

f(x) is concave downwards for all x.

f(x) has a relative minimum at x = 0.

f(x) has a point of inflection at x = 0.

7.

f is a differentiable function on the interval [0, 1] and g(x) = f(2x). The table below gives

values of f '(x). What is the value of g '(0.1)? (5 points)

x 0.1 0.2 0.3 0.4 0.5

f '(x) 1 2 3 -4 5

1

2

4 (𝑔′ (𝑥) = 2𝑓 ′ (2𝑥), 𝑔′ (0.1) = 2𝑓 ′ (0.2) = 4)

Cannot be determined

8.

Use the graph of f(t) = 2t + 2 on the interval [-1, 4] to write the function F(x),

where

. (5 points)

F(x) = x2 + 3x

F(x) = x2 + 2x - 12

F(x) = x2 + 2x - 3

F(x) = x2 + 4x - 8

9.

The velocity of a particle moving along the x-axis is v(t) = t2 + 2t + 1, with t measured in

minutes and v(t) measured in feet per minute. To the nearest foot find the total distance

travelled by the particle from t = 0 to t = 2 minutes. (5 points)

2

2

1

It is ∫0 |𝑣(𝑡)|𝑑𝑡 = ∫0 (𝑡 2 + 2𝑡 + 1)𝑑𝑡 = (3 𝑡 3 + 𝑡 2 + 𝑡)

2

𝑡=0

8

= 3 + 2 + 2 ≈ 𝟔.

10.

Find the range of the function

. (5 points)

[F is negative for negative x and positive for positive x so no correct answer strictly

speaking. If we suppose x>=0, then F(0)=0 and F increases, F(2)=π (1/4 area of a circle with

radius 2)]

[0, 4π]

[0, π]

[-4, 0]

[0, 4]

Must Show work

1. The figure below shows the graph of f ', the derivative of the function f, on the closed

interval from x = -2 to x = 6. The graph of the derivative has horizontal tangent lines at x =

2 and x = 4.

Find the x-coordinate of each of the points of inflection of the graph of f. Justify your

answer. (10 points)

f' increases from x=-2 to x=2, decreases from x=2 to x=4 and increases from x=4 to x=6.

Therefore, f''>0 from x=-2 to x=2, f''0 from x=4 to x=6.

So, f'' changes sign at x=2 and at x=4, so both points are inflection points of f.

2.

A car travels along a straight road for 30 seconds starting at time t = 0. Its acceleration in

ft/sec2 is given by the linear graph below for the time interval [0, 30]. At t = 0, the velocity

of the car is 0 and its position is 10.

What is the total distance the car travels in this 30 second interval? Your must show your

work but you may use your calculator to evaluate. Give 3 decimal places in your answer and

include units.

(10 points)

𝑡

1

Acceleration is 𝑎(𝑡) = 10 − 𝑡, velocity is 𝑣(𝑡) = ∫0 𝑎(𝑥)𝑑𝑥 = 10𝑥 − 2 𝑥 2 . The distance travelled is

30

30

20

30

1

1

1

∫ |𝑣(𝑡)|𝑑𝑡 = ∫ |10𝑡 − 𝑡 2 | 𝑑𝑡 = ∫ (10𝑡 − 𝑡 2 ) 𝑑𝑡 + ∫ ( 𝑡 2 − 10𝑡) 𝑑𝑡 =

2

2

0

0

0

20 2

1

1

1

= 5 ∙ 202 − ∙ 203 + ∙ 303 − ∙ 203 − 5 ∙ 302 + 5 ∙ 202 ...

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