mccartymichael
Answered

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)

Verified Answer :

Panoyin

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. \((\frac{f}{g})(x) = \frac{4x - 1}{2x^{2} + 3}\)
    Domain: 2x² + 3 ≠ 0
                         - 3  - 3
                        2x² ≠ 0
                         2      2
                          x² ≠ 0
                           x ≠ 0
                  (-∞, 0) ∨ (0, ∞)

6. \((\frac{g}{f})(x) = \frac{2x^{2} + 3}{4x - 1}\)
    Domain: 4x - 1 ≠ 0
                      + 1 + 1
                        4x ≠ 0
                         4     4
                         x ≠ 0
                (-∞, 0) ∨ (0, ∞)

MrRoyal

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

The values of the composite functions are:

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

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

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

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

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

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

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: \(\mathbf{(-\infty,\infty)}\)

\(\mathbf{(4)\ (f \times g)(x)}\)

This is calculated as:

\(\mathbf{(f \times g)(x) = f(x) \times g(x)}\)

So, we have:

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

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

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:

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

So, we have:

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

There are restrictions to the value of x.

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

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

This is calculated as:

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

So, we have:

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

There are restrictions to the value of x.

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

Read more about domain at:

https://brainly.com/question/21853810

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