Solution to Mathematical Problems Involving Functions (Essay Sample)

Student’s Name
Professor’s Name
Subject
Date
Task 3
1
f(x) = 2x^2 - x - 3
g(x) = 2x - 6
A
(f + g)(x) = (2x^2 - x - 3) + (2x - 6) = 2x^2 + x - 9
B
(f - g)(x) = (2x^2 - x - 3) - (2x - 6) = 2x^2 - 3x + 3
C
(g - f)(x) = (2x - 6) - (2x^2 - x - 3) = -2x^2 + 3x - 3
D
(g - f)(-2) = -2 ×( - 2)^2 + 3 × (- 2) - 3 = - 17
E
(fg)(x) = (2x^2 - x - 3) × (2x - 6)
(fg)(2) = (2×(2)^2 - 2 - 3) × (2(2) - 6) = - 6
F
(f / g)(x) = (2x^2 - x - 3) / (2x - 6) = (2x^2 - 3x + 2x - 3) / (2x - 6)
(f / g)(x) = (2x - 3)(x + 1) / (2x - 6)
G
(f / g)(2) = (2(2) - 3)(2 + 1) / (2(2) - 6) = - 1.5
Domain : R - {1.5}
2
f(x) = 2x + 5
g(x) = x^2 + x - 6
A
(f + g)(x) = (2x + 5) + (x^2 + x - 6) = x^2 + 3x - 1
B
(g - f)(x) = (x^2 + x - 6) - (2x + 5) = x^2 - x - 11
C
(g + f)(x) = (x^2 + x - 6) + (2x + 5) = x^2 + 3x - 1
(g + f)(3) = (3^2 + 3 - 6) + (2(3) + 5) = 3^2 + 3(3) - 1 = 17
D
(fg)(0) = (2(0) + 5) × (0^2 + 0 - 6) = - 30
E
(fg)(-1) = (2(- 1) + 5) × (- 1^2 + (- 1) - 6) = - 18
F
(g / f)(1) = (1^2 + 1 - 6) / (2(1) + 5) = - 0.571
G
(f / g)(-3) = (2(- 3)+ 5) / ((- 3)^2 + (-3) - 6) = Not defined
3
f(x) = 2x – 3
g(x) = 5x + 4
(f º g)(x) = f(g(x))
A
(f º g)(x) = 2( 5x + 4) - 3 = 10x + 5
B
(g º f)(x) = 5( 2x - 3) + 4 = 10x - 11
C
(f º g)(-3) = 10 × (- 3) + 5 = -25
4
f(x) = x^2 – 8x - 9
g(x) = x^2 + 6x + 5
(f º g)(x) = f(g(x)) = (x^2 + 6x + 5)^2 - 8(x^2 + 6x + 5) - 9
A
(f º g)(0) = (0^2 + 6(0) + 5)^2 - 8(0^2 + 6(0) + 5) - 9 = -24
B
(f º g)(1)