SNHU MAT 225 Module 5 Problem Set Updated
2024 Solved Fully 100%!!
Question1: Score 3/6
Verify that Rolle's Theorem can be applied to the function
on the interval . Then find all
values of in the interval such that .
Enter the exact answers in increasing order.
To enter , type sqrt(a).
Your response Correct response
(10-sqrt(7))/3 (10-sqrt(7))/3
Auto graded Grade: 2/2.0
Your response Correct response
(10+sqrt(7))/3 (10+sqrt(7))/3
Auto graded Grade: 1/1.0
Show your work and explain, in your own words, how you arrived at your answers.
Verify that Rolle's Theorem can be applied:
f(2) = 2^3 - 10*2^2 + 31*2 - 30 = 0 f(5) = 5^3 -
10 * 5^2 + 31*5 - 30 = 0
Rolle's Theorem can be applied.
First I found the derivative of f(x) = x^3 - 10x^2 + 31x - 30:
d/dx(x^3 - 10x^2 + 31x - 30)
= d/dx(x^3) - d/dx(10x^2) + d/dx(31) - d/dx(30)
= 3x^2 - 20x + 31
Then, I solved for zero:
3x^2 - 20x + 31 = 0
Using the quadratic formula(-(-20) +- sqrt((-20)^2 - 4*3*31)) / (2*3)
=(-(-20) +- 2sqrt(7)) / (2*3)
= (10 +- sqrt(7)) / 3
Ungraded Grade: 0/3.0
Total grade: 1.0×2/6 + 1.0×1/6 + 0.0×3/6 = 33% + 17% + 0%
Feedback:
,The following is Mobius' explanation for a solution to this question. You can use this and other
online references as a guide, but your explanation should be in your own words.
The function is continuous on the interval and
differentiable on the interval
. Since , there exists at least one in such that .
The derivative of is
.
Solve .
Therefore, the values of such that are and .
Question2: Score 4/4
Verify that the Mean Value Theorem can be applied to the function on the interval . Then find
the value of in the interval that satisfies the conclusion of the Mean Value Theorem.
Enter the exact answer.
Your response Correct response
32768/3125 32768/3125
Auto graded Grade: 1/1.0
Total grade: 1.0×1/1 =
100% Feedback:
The function is continuous on the interval and differentiable on the interval . Therefore, by
the Mean Value Theorem, there exists a number in such that
, .
The derivative of is
.
Solve .
Therefore, the value of such that is .
Question3: Score 6/6
Consider the function .
(a) Find the domain of .
Note: Use the letter U for union. To enter , type infinity.
Domain:
Your response Correct response
(-infinity, -3) U (-3,
infinity)
Auto graded Grade: 1/1.0
(b) Give the horizontal and vertical asymptotes of , if any.
Enter the equations for the asymptotes. If there is no horizontal or vertical asymptote, enter NA in
the associated response area.
horizont
al
asymptot
e:
Your response Correct response
y=1 y=1
Auto graded Grade: 1/1.0
2024 Solved Fully 100%!!
Question1: Score 3/6
Verify that Rolle's Theorem can be applied to the function
on the interval . Then find all
values of in the interval such that .
Enter the exact answers in increasing order.
To enter , type sqrt(a).
Your response Correct response
(10-sqrt(7))/3 (10-sqrt(7))/3
Auto graded Grade: 2/2.0
Your response Correct response
(10+sqrt(7))/3 (10+sqrt(7))/3
Auto graded Grade: 1/1.0
Show your work and explain, in your own words, how you arrived at your answers.
Verify that Rolle's Theorem can be applied:
f(2) = 2^3 - 10*2^2 + 31*2 - 30 = 0 f(5) = 5^3 -
10 * 5^2 + 31*5 - 30 = 0
Rolle's Theorem can be applied.
First I found the derivative of f(x) = x^3 - 10x^2 + 31x - 30:
d/dx(x^3 - 10x^2 + 31x - 30)
= d/dx(x^3) - d/dx(10x^2) + d/dx(31) - d/dx(30)
= 3x^2 - 20x + 31
Then, I solved for zero:
3x^2 - 20x + 31 = 0
Using the quadratic formula(-(-20) +- sqrt((-20)^2 - 4*3*31)) / (2*3)
=(-(-20) +- 2sqrt(7)) / (2*3)
= (10 +- sqrt(7)) / 3
Ungraded Grade: 0/3.0
Total grade: 1.0×2/6 + 1.0×1/6 + 0.0×3/6 = 33% + 17% + 0%
Feedback:
,The following is Mobius' explanation for a solution to this question. You can use this and other
online references as a guide, but your explanation should be in your own words.
The function is continuous on the interval and
differentiable on the interval
. Since , there exists at least one in such that .
The derivative of is
.
Solve .
Therefore, the values of such that are and .
Question2: Score 4/4
Verify that the Mean Value Theorem can be applied to the function on the interval . Then find
the value of in the interval that satisfies the conclusion of the Mean Value Theorem.
Enter the exact answer.
Your response Correct response
32768/3125 32768/3125
Auto graded Grade: 1/1.0
Total grade: 1.0×1/1 =
100% Feedback:
The function is continuous on the interval and differentiable on the interval . Therefore, by
the Mean Value Theorem, there exists a number in such that
, .
The derivative of is
.
Solve .
Therefore, the value of such that is .
Question3: Score 6/6
Consider the function .
(a) Find the domain of .
Note: Use the letter U for union. To enter , type infinity.
Domain:
Your response Correct response
(-infinity, -3) U (-3,
infinity)
Auto graded Grade: 1/1.0
(b) Give the horizontal and vertical asymptotes of , if any.
Enter the equations for the asymptotes. If there is no horizontal or vertical asymptote, enter NA in
the associated response area.
horizont
al
asymptot
e:
Your response Correct response
y=1 y=1
Auto graded Grade: 1/1.0
