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Literature & Language

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# Mathematics Questions on Functions (Coursework Sample)

Instructions:

This task involves solving various algebraic problems related to functions, composite functions, and quadratic equations. The problems require performing operations like addition, subtraction, and multiplication on functions, as well as evaluating functions at specific points. Additionally, it includes solving composite functions and quadratic equations to find dimensions and unknown values in word problems. Specifically, the sample covers:
Function operations (addition, subtraction, multiplication) and evaluation at given values.
Composite functions and calculating their values for specific inputs.
Solving a quadratic equation related to the dimensions of an envelope.
Finding consecutive even integers based on a condition involving their squares.
This type of task typically appears in algebra or pre-calculus coursework, focusing on the manipulation and evaluation of functions and solving real-world word problems through algebraic techniques. source..

Content:

Student’s Name
Professor’s Name
Subject
Date
Solving Problems
1.
A.
(f+g)(x)
(f+g)(x) = f(x) + g(x)
(f+g)(x) = (2x2 - x - 3) + (2x - 6)
(f+g)(x) = 2x2 + x - 9
B.
(f-g)(x)
(f-g)(x) = f(x) - g(x)
(f-g)(x) = (2x2 - x - 3) - (2x - 6)
(f-g)(x) = 2x2 - 3x + 3
C.
(g-f)(x)
(g-f)(x) = g(x) - f(x)
(g-f)(x) = (2x - 6) - (2x2 - x - 3)
(g-f)(x) = -2x2 + 3x - 3
D.
(g × f)(-2)
(g × f)(-2) = g(-2) × f(-2)
First, find g(-2):
g(-2) = 2(-2) - 6 = -4 - 6 = -10
Now, find f(-2):
f(-2) = 2(-2)2 - (-2) - 3
f(-2) = 2(4) + 2 - 3 = 8 - 1 = 7
(g × f)(-2) = (-10) × 7 = -70
E.
(fg)(2)
(fg)(2) = f(2) × g(2)
First, find f(2):
f(2) = 2(2)2 - 2 - 3
f(2) = 2(4) - 2 - 3 = 8 - 5 = 3
Now, find g(2):
g(2) = 2(2) - 6 = 4 - 6 = -2
(fg)(2) = 3 × (-2) = -6
F.
(f / g)(x)
(f / g)(x) = f(x) / g(x)
(f / g)(x) = 2x2−x−32x−6
Factor the numerator:
(f / g)(x) = (2x+1)(x−3)2(x−3)
(f / g)(x) = 2x+12
2
A.
(f+g)(x)
(f+g)(x) = f(x) + g(x)
(f+g)(x) = (2x + 5) + (x2 + x - 6)
(f+g)(x) = x2 + 3x - 1
B.
(g-f)(x)
(g-f)(x) = g(x) - f(x)
(g-f)(x) = (x2 + x - 6) - (2x + 5)
(g-f)(x) = x2 - x - 11
C.
(g+f)(3)
(g+f)(3) = g(3) + f(3)
g(3) = 32 + 3 - 6 = 9 + 3 - 6 = 6
f(3) = 2(3) + 5 = 6 + 5 = 11
(g+f)(3) = 6 + 11 = 17
D.
(fg)(0)
(fg)(0) = f(0) × g(0)
f(0) = 2(0) + 5 = 5
g(0) = 02 + 0 - 6 = -6
(fg)(0) = 5 × (-6) = -30
E.
(fg)(-1)
(fg)(-1) = f(-1) × g(-1)
f(-1) = 2(-1) + 5 = -2 + 5 = 3
g(-1) = (-1)2 - 1 - 6 = 1 - 1 - 6 = -6
(fg)(-1) = 3 × (-6) = -18
F.
(g/f)(1)
(g/f)(1) = g(1) / f(1)
g(1) = 12 + 1 - 6 = 1 + 1 - 6 = -4
f(1) = 2(1) + 5 = 2 + 5 = 7
(g/f)(1) = −47
G.
(f/g)(-3)
f(-3) and g(-3):
f(-3) = 2(-3) + 5 = -1
g(-3) = (-3)2 - 3 - 6 = 6
(f / g)(-3) = f(-3)g(-3) = −16
3
A.
Composite (f ∘ g)(x)
(f ∘ g)(x) = f(g(x))
(f ∘ g)(x) = f(5x + 4)
(f ∘ g)(x) = 2(5x + 4) - 3
(f ∘ g)(x) = 10x + 8 - 3
(f ∘ g)(x) = 10x + 5
B.
Composite (g ∘ f)(x)
(g ∘ f)(x) = g(f(x))
(g ∘ f)(x) = g(2x - 3)
(g ∘ f)(x) = 5(2x - 3) + 4
(g ∘ f)(x) = 10x - 15 + 4
(g ∘ f)(x) = 10x - 11
C.
Composite (f ∘ g)(-3)
(f ∘ g)(-3) = f(g(-3))
First, find g(-3):
g(-3) = 5(-3) + 4 = -15 + 4 = -11
Now, substitute into f(x):
(f ∘ g)(-3) = f(-11)
(f ∘ g)(-3) = 2(-11) - 3
(f ∘ g)(-3) = -22 - 3
(f ∘ g)(-3) = -25
4
A.
Composite (f ∘ g)(0)
(f ∘ g)(x) = f(g(x))
(f ∘ g)(x) = f(x2 + 6x + 5)
(f ∘ g)(0) = f(02 + 6(0) + 5)
(f ∘ g)(0) = f(5)
(f ∘ g)(0) = 52 - 8(5) - 9
(f ∘ g)(0) = 25 - 40 - 9
(f ∘ g)(0) = -24
B.
Composite (f ∘ g)(1)
(f ∘ g)(x) = f(g(x))
(f ∘ g)(x) = f(x2 + 6x + 5)
(f ∘ g)(1) = f(12 + 6(1) + 5)
(f ∘ g)(1) = f(12)
(f ∘ g)(1) = 122 - 8(12) - 9
(f ∘ g)(1) = 144 - 96...

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