MATHEMATICS HIGH SCHOOL

Let f(x)=4x-1 and g(x)=2x^2+3. Perform each function operations and then find the domain.1. (f+g)(x) 2. (f-g)(x) 3. (g-f)(x) 4. (f times g)(x) 5. f/g(x) 6. g/f(x)

Answers

Answer 1

Answer:

The domain of a function is the set of input values, the function can take.

The values of the composite functions are:

$\mathbf{(f + g)(x) = 2x^2 + 4x +2}$

$\mathbf{(f - g)(x) = -2x^2 + 4x - 4}$

$\mathbf{(g - f)(x) = 2x^2 - 4x + 4}$

$\mathbf{(f * g)(x) = 8x^3 - 2x^2 -12x + 4}$

$\mathbf{(f / g)(x) = ((4x - 1 ))/((2x^2 - 3))}$

$\mathbf{(g / f)(x) = (2x^2 - 3)/(4x - 1 )}$

The functions are given as:

$\mathbf{f(x) = 4x - 1}$

$\mathbf{g(x) = 2x^2 + 3}$

$\mathbf{(1)\ (f + g)(x)}$

This is calculated as:

$\mathbf{(f + g)(x) = f(x)+ g(x)}$

So, we have:

$\mathbf{(f + g)(x) = 4x - 1 + 2x^2 + 3}$

Collect like terms

$\mathbf{(f + g)(x) = 2x^2 + 4x - 1 + 3}$

$\mathbf{(f + g)(x) = 2x^2 + 4x +2}$

There is no restriction on the value of x.

So, the domain is: $\mathbf{(-\infty,\infty)}$

$\mathbf{(2)\ (f - g)(x)}$

This is calculated as:

$\mathbf{(f - g)(x) = f(x) - g(x)}$

So, we have:

$\mathbf{(f - g)(x) = 4x - 1 - 2x^2 - 3}$

Collect like terms

$\mathbf{(f - g)(x) = -2x^2 + 4x - 1 - 3}$

$\mathbf{(f - g)(x) = -2x^2 + 4x - 4}$

There is no restriction on the value of x.

So, the domain is: $\mathbf{(-\infty,\infty)}$

$\mathbf{(3)\ (g - f)(x)}$

This is calculated as:

$\mathbf{(g - f)(x) = -(f - g)(x) }$

So, we have:

$\mathbf{(g - f)(x) = 2x^2 - 4x + 4}$

There is no restriction on the value of x.

So, the domain is: $\mathbf{(-\infty,\infty)}$

$\mathbf{(4)\ (f * g)(x)}$

This is calculated as:

$\mathbf{(f * g)(x) = f(x) * g(x)}$

So, we have:

$\mathbf{(f * g)(x) = (4x - 1 )* (2x^2 - 3)}$

$\mathbf{(f * g)(x) = 8x^3 - 2x^2 -12x + 4}$

There is no restriction on the value of x.

So, the domain is: $\mathbf{(-\infty,\infty)}$

$\mathbf{(5)\ (f /g)(x)}$

This is calculated as:

$\mathbf{(f /g)(x) = (f(x) )/(g(x))}$

So, we have:

$\mathbf{(f / g)(x) = ((4x - 1 ))/((2x^2 - 3))}$

There are restrictions to the value of x.

So, the domain is: $\mathbf{(-\infty,-\sqrt{(3)/(2)} ) \ u\ ( -\sqrt{(3)/(2)},\sqrt{(3)/(2)}})\ u\ (\sqrt{(3)/(2)},\ \infty)}$

$\mathbf{(6)\ (g /f)(x)}$

This is calculated as:

$\mathbf{(g /f)(x) =1 / (f(x) )/(g(x))}$

So, we have:

$\mathbf{(g / f)(x) = (2x^2 - 3)/(4x - 1 )}$

There are restrictions to the value of x.

So, the domain is: $\mathbf{(-\infty, (1)/(4))\ u\ ((1)/(4),\infty)}$

Read more about domain at:

brainly.com/question/21853810

Answer 2

Answer: 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. $((f)/(g))(x) = (4x - 1)/(2x^(2) + 3)$
Domain: 2x² + 3 ≠ 0
- 3 - 3
2x² ≠ 0
2 2
  x² ≠ 0
x ≠ 0
(-∞, 0) ∨ (0, ∞)

6. $((g)/(f))(x) = (2x^(2) + 3)/(4x - 1)$
Domain: 4x - 1 ≠ 0
+ 1 + 1
4x ≠ 0
4   4
x ≠ 0
  (-∞, 0) ∨ (0, ∞)

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