3/26/2021 Southern New Hampshire University - 4-2 Problem Set: Module Four
[PRINT]
MAT-136-H7409 21EW4 Intro to Quantitative Analysis, 4-2 Problem Set: Module Four
Briana Tattersall, 3/26/21 at 9:16:40 PM EDT
Question1: Score 5/5
f
Given f(x) = x 2 + 2x and g(x) = 1 − x 2, find f + g, f − g, fg, and g .
Enclose numerators and denominators in parentheses. For example, (a − b) / (1 + n).
(f + g)(x) =
Your response Correct response
2x+1 2*x+1
Auto graded Grade: 1/1.0 A+ 100%
(f − g)(x) =
Your response Correct response
2 x2 + 2 x − 1 2*x^2+2*x-1
Auto graded Grade: 1/1.0 A+ 100%
fg(x) =
Your response Correct response
− x4 − 2 x3 + 1 x2 + 2 x -x^4-2*x^3+x^2+2*x
Auto graded Grade: 1/1.0 A+ 100%
f
g
(x) =
Your response Correct response
(x +2 x )
2
(x^2+2*x)/(1-x^2)
(1−x )
2
Auto graded Grade: 1/1.0 A+ 100%
Total grade: 1.0×1/4 + 1.0×1/4 + 1.0×1/4 + 1.0×1/4 = 25% + 25% + 25% + 25%
Feedback:
( ) (
f + g = x 2 + 2x + 1 − x 2 )
= 2x + 1
Since both f(x) and g(x) have domains of (− ∞, ∞), the domain of f + g is (− ∞, ∞).
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, 3/26/2021 Southern New Hampshire University - 4-2 Problem Set: Module Four
(
f − g = x 2 + 2x − 1 − x 2 ) ( )
= 2x 2 + 2x − 1
Since both f(x) and g(x) have domains of (− ∞, ∞), the domain of f − g is (− ∞, ∞).
(
fg = x 2 + 2x )(1 − x )
2
= 1x 2 − x 4 + 2x − 2x 3
= − x 4 − 2x 3 + 1x 2 + 2x
Since both f(x) and g(x) have domains of (− ∞, ∞), the domain of fg is (− ∞, ∞).
f ( x + 2x )
2
= where x 2 ≠ 1
g
(1−x ) 2
Since the denominator equals zero whenever x 2 = 1, we must exclude ± 1 from the domain. Thus, the domain
f
of is (− ∞, − 1) ∪ (− 1, 1) ∪ (1, ∞).
g
Question2: Score 2.5/5
Use the pair of functions to find f(g(x)) and g(f(x)). Simplify your answers.
f(x) = √x + 4, g(x) = x 2 + 9
Reminder, to use sqrt(() to enter a square root.
f(g(x)) =
Your response Correct response
√x 2 + 9 + 4 sqrt(x^2+9)+4
Auto graded Grade: 0/1.0 F 0%
g(f(x)) =
Your response Correct response
x + 8√x + 25 x+8*sqrt(x)+25
https://snhu.mobius.cloud/modules/gradeProctoredTest.Login 2/13
[PRINT]
MAT-136-H7409 21EW4 Intro to Quantitative Analysis, 4-2 Problem Set: Module Four
Briana Tattersall, 3/26/21 at 9:16:40 PM EDT
Question1: Score 5/5
f
Given f(x) = x 2 + 2x and g(x) = 1 − x 2, find f + g, f − g, fg, and g .
Enclose numerators and denominators in parentheses. For example, (a − b) / (1 + n).
(f + g)(x) =
Your response Correct response
2x+1 2*x+1
Auto graded Grade: 1/1.0 A+ 100%
(f − g)(x) =
Your response Correct response
2 x2 + 2 x − 1 2*x^2+2*x-1
Auto graded Grade: 1/1.0 A+ 100%
fg(x) =
Your response Correct response
− x4 − 2 x3 + 1 x2 + 2 x -x^4-2*x^3+x^2+2*x
Auto graded Grade: 1/1.0 A+ 100%
f
g
(x) =
Your response Correct response
(x +2 x )
2
(x^2+2*x)/(1-x^2)
(1−x )
2
Auto graded Grade: 1/1.0 A+ 100%
Total grade: 1.0×1/4 + 1.0×1/4 + 1.0×1/4 + 1.0×1/4 = 25% + 25% + 25% + 25%
Feedback:
( ) (
f + g = x 2 + 2x + 1 − x 2 )
= 2x + 1
Since both f(x) and g(x) have domains of (− ∞, ∞), the domain of f + g is (− ∞, ∞).
https://snhu.mobius.cloud/modules/gradeProctoredTest.Login 1/13
, 3/26/2021 Southern New Hampshire University - 4-2 Problem Set: Module Four
(
f − g = x 2 + 2x − 1 − x 2 ) ( )
= 2x 2 + 2x − 1
Since both f(x) and g(x) have domains of (− ∞, ∞), the domain of f − g is (− ∞, ∞).
(
fg = x 2 + 2x )(1 − x )
2
= 1x 2 − x 4 + 2x − 2x 3
= − x 4 − 2x 3 + 1x 2 + 2x
Since both f(x) and g(x) have domains of (− ∞, ∞), the domain of fg is (− ∞, ∞).
f ( x + 2x )
2
= where x 2 ≠ 1
g
(1−x ) 2
Since the denominator equals zero whenever x 2 = 1, we must exclude ± 1 from the domain. Thus, the domain
f
of is (− ∞, − 1) ∪ (− 1, 1) ∪ (1, ∞).
g
Question2: Score 2.5/5
Use the pair of functions to find f(g(x)) and g(f(x)). Simplify your answers.
f(x) = √x + 4, g(x) = x 2 + 9
Reminder, to use sqrt(() to enter a square root.
f(g(x)) =
Your response Correct response
√x 2 + 9 + 4 sqrt(x^2+9)+4
Auto graded Grade: 0/1.0 F 0%
g(f(x)) =
Your response Correct response
x + 8√x + 25 x+8*sqrt(x)+25
https://snhu.mobius.cloud/modules/gradeProctoredTest.Login 2/13
