If two pizzas cost as much as three cakes, and if one cake costs $6, how much do five pizzas cost?
2 answers:
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F(x) = 4x - 1
g(x) = 2x² + 3
1. (f + g)(x) = (4x - 1) + (2x² + 3)
(f + g)(x) = 2x² + 4x + (-1 + 3)
(f + g)(x) = 2x² + 4x + 2
Domain: {x| -∞ < x < ∞}, (-∞, ∞)
2. (f - g)(x) = (4x + 1) - (2x² + 3)
(f - g)(x) = 4x + 1 - 2x² - 3
(f - g)(x) = -2x² + 4x + 1 - 3
(f - g)(x) = -2x² + 4x - 2
Domain: {x|-∞ < x < ∞}, (-∞, ∞)
3. (g - f)(x) = (2x² + 3) - (4x - 1)
(g - f)(x) = 2x² + 3 - 4x + 1
(g - f)(x) = 2x² - 4x + 3 + 1
(g - f)(x) = 2x² - 4x + 4
Domain: {x| -∞ < x < ∞}, (-∞, ∞)
4. (f · g)(x) = (4x + 1)(2x² + 3)
(f · g)(x) = 4x(2x² + 3) + 1(2x² + 3)
(f · g)(x) = 4x(2x²) + 4x(3) + 1(2x²) + 1(3)
(f · g)(x) = 8x³ + 12x + 2x² + 3
(f · g)(x) = 8x³ + 2x² + 12x + 3
Domain: {x| -∞ < x < ∞}, (-∞, ∞)
5.
Domain: 2x² + 3 ≠ 0
- 3 - 3
2x² ≠ 0
2 2
x² ≠ 0
x ≠ 0
(-∞, 0) ∨ (0, ∞)
6.
Domain: 4x - 1 ≠ 0
+ 1 + 1
4x ≠ 0
4 4
x ≠ 0
(-∞, 0) ∨ (0, ∞)
g(x) = 2x² + 3
1. (f + g)(x) = (4x - 1) + (2x² + 3)
(f + g)(x) = 2x² + 4x + (-1 + 3)
(f + g)(x) = 2x² + 4x + 2
Domain: {x| -∞ < x < ∞}, (-∞, ∞)
2. (f - g)(x) = (4x + 1) - (2x² + 3)
(f - g)(x) = 4x + 1 - 2x² - 3
(f - g)(x) = -2x² + 4x + 1 - 3
(f - g)(x) = -2x² + 4x - 2
Domain: {x|-∞ < x < ∞}, (-∞, ∞)
3. (g - f)(x) = (2x² + 3) - (4x - 1)
(g - f)(x) = 2x² + 3 - 4x + 1
(g - f)(x) = 2x² - 4x + 3 + 1
(g - f)(x) = 2x² - 4x + 4
Domain: {x| -∞ < x < ∞}, (-∞, ∞)
4. (f · g)(x) = (4x + 1)(2x² + 3)
(f · g)(x) = 4x(2x² + 3) + 1(2x² + 3)
(f · g)(x) = 4x(2x²) + 4x(3) + 1(2x²) + 1(3)
(f · g)(x) = 8x³ + 12x + 2x² + 3
(f · g)(x) = 8x³ + 2x² + 12x + 3
Domain: {x| -∞ < x < ∞}, (-∞, ∞)
5.
Domain: 2x² + 3 ≠ 0
- 3 - 3
2x² ≠ 0
2 2
x² ≠ 0
x ≠ 0
(-∞, 0) ∨ (0, ∞)
6.
Domain: 4x - 1 ≠ 0
+ 1 + 1
4x ≠ 0
4 4
x ≠ 0
(-∞, 0) ∨ (0, ∞)