SNHU CALCULUS I: SINGLE-VARIABLE CALCULUS
(MAT225) 5-2 MODULE FIVE PROBLEM SET EXAM
2025 SOLVED
, [PRINT]
Question1: Score 2/10
Verify that Rolle's Theorem can be applied to the function f(x) = x3 − 7x2 + 14x − 8 on
the interval [1, 4]. Then find all values of c in the interval such that f′(c) = 0.
Enter the exact answers in increasing order.
To enter √a, type sqrt(a).
c=
Your response Correct response
7 − √7
(7-sqrt(7))/3
3
Auto graded Grade: 12/12.0 A+ 100%
c=
Your response Correct response
7 + √7
(7+sqrt(7))/3
3
Auto graded Grade: 13/13.0 A+ 100%
Show your work and explain, in your own words, how you arrived at your
answers.
1/40
,Verifying the Rolle's theorem can be applied
f(1) = 13 − 7 (1)2 + 14 (1) − 8 = 0
f(4) = 43 − 7 (4)2 + 14 (4) − 8 = 0
which it can be applied now i had to find the derivative of the function
f(x) = x3 − 7 x2 + 14 x − 8
f'(x) = 3 x2 − 14 x + 14
solve f'(x) = 0
3 x2 − 14 x + 14 = 0
solve using quadratic formula
− ( − 14 ) ± √ ( − 14 ) 2 − 4 ⋅ 3 ⋅ 14
x=
2⋅3
− ( − 14 ) ± 2√7
x= 6
14 − 2√7 2 7 − √7
( ) ( 7 − √7 )
x= 6
now factor out to simplify x = 6
= 3
14 + 2√7 2 7 + √7
( ) ( 7 + √7 )
x= 6
now factor out to simplify x = 6
= 3
7 − √7 7 + √7
that makes the value of c such that f'(c) = 0 are c = 3
, c= 3
Keywords:
Partial Grades:
Ungraded Grade: 0/100.0 F 0%
Total grade: 1.0×12/125 + 1.0×13/125 + 0.0×100/125 = 10% + 10% + 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 f(x) = x3 − 7x2 + 14x − 8 is continuous on the interval [1, 4] and
differentiable on the interval (1, 4). Since f(1) = f(4) = 0, there exists at least one c in
(1, 4) such that f′(c) = 0.
The derivative of f(x) = x3 − 7x2 + 14x − 8 is
f′(x) = 3x2 − 14x + 14.
2/40
, Solve f′(x) = 0.
3x2 − 14x + 14 = 0
x =
14 ± √ ( − 14 ) 2 − 4 ( 3 ) ( 14 )
2(3)
14 ± √28
x =
6
7 ± √7
x
= 3
7 − √7 7 + √7
Therefore, the values of c such that f′(c) = 0 are c = and c = .
3 3
Question2: Score 2/2
2
Verify that the Mean Value Theorem can be applied to the function f(x) = x 3 on the
interval [0, 1]. Then find the value of c in the interval that satisfies the conclusion of
the Mean Value Theorem.
Enter the exact answer.
c=
Your response Correct response
8
8/27
27
Auto graded Grade: 1/1.0 A+ 100%
Total grade: 1.0×1/1 = 100%
Feedback:
2
The function f(x) = x 3 is continuous on the interval [0, 1] and differentiable on the
interval (0, 1). Therefore, by the Mean Value Theorem, there exists a number c in
(0, 1) such that
3/40
(MAT225) 5-2 MODULE FIVE PROBLEM SET EXAM
2025 SOLVED
, [PRINT]
Question1: Score 2/10
Verify that Rolle's Theorem can be applied to the function f(x) = x3 − 7x2 + 14x − 8 on
the interval [1, 4]. Then find all values of c in the interval such that f′(c) = 0.
Enter the exact answers in increasing order.
To enter √a, type sqrt(a).
c=
Your response Correct response
7 − √7
(7-sqrt(7))/3
3
Auto graded Grade: 12/12.0 A+ 100%
c=
Your response Correct response
7 + √7
(7+sqrt(7))/3
3
Auto graded Grade: 13/13.0 A+ 100%
Show your work and explain, in your own words, how you arrived at your
answers.
1/40
,Verifying the Rolle's theorem can be applied
f(1) = 13 − 7 (1)2 + 14 (1) − 8 = 0
f(4) = 43 − 7 (4)2 + 14 (4) − 8 = 0
which it can be applied now i had to find the derivative of the function
f(x) = x3 − 7 x2 + 14 x − 8
f'(x) = 3 x2 − 14 x + 14
solve f'(x) = 0
3 x2 − 14 x + 14 = 0
solve using quadratic formula
− ( − 14 ) ± √ ( − 14 ) 2 − 4 ⋅ 3 ⋅ 14
x=
2⋅3
− ( − 14 ) ± 2√7
x= 6
14 − 2√7 2 7 − √7
( ) ( 7 − √7 )
x= 6
now factor out to simplify x = 6
= 3
14 + 2√7 2 7 + √7
( ) ( 7 + √7 )
x= 6
now factor out to simplify x = 6
= 3
7 − √7 7 + √7
that makes the value of c such that f'(c) = 0 are c = 3
, c= 3
Keywords:
Partial Grades:
Ungraded Grade: 0/100.0 F 0%
Total grade: 1.0×12/125 + 1.0×13/125 + 0.0×100/125 = 10% + 10% + 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 f(x) = x3 − 7x2 + 14x − 8 is continuous on the interval [1, 4] and
differentiable on the interval (1, 4). Since f(1) = f(4) = 0, there exists at least one c in
(1, 4) such that f′(c) = 0.
The derivative of f(x) = x3 − 7x2 + 14x − 8 is
f′(x) = 3x2 − 14x + 14.
2/40
, Solve f′(x) = 0.
3x2 − 14x + 14 = 0
x =
14 ± √ ( − 14 ) 2 − 4 ( 3 ) ( 14 )
2(3)
14 ± √28
x =
6
7 ± √7
x
= 3
7 − √7 7 + √7
Therefore, the values of c such that f′(c) = 0 are c = and c = .
3 3
Question2: Score 2/2
2
Verify that the Mean Value Theorem can be applied to the function f(x) = x 3 on the
interval [0, 1]. Then find the value of c in the interval that satisfies the conclusion of
the Mean Value Theorem.
Enter the exact answer.
c=
Your response Correct response
8
8/27
27
Auto graded Grade: 1/1.0 A+ 100%
Total grade: 1.0×1/1 = 100%
Feedback:
2
The function f(x) = x 3 is continuous on the interval [0, 1] and differentiable on the
interval (0, 1). Therefore, by the Mean Value Theorem, there exists a number c in
(0, 1) such that
3/40
