Use the functions below.

Question

f(x)=
√
x−1
g(x)=3x−1
What is(
f
g

)(x)? What is the domain of (
f
g

)(x)?

(1 point)
Responses

(
f
g

)(x)=
√
x
3

and the domain of (
f
g

)(x) is x≥1.
open paren f over g close paren of x is equal to the fraction with numerator square root of x and denominator 3 and the domain of open paren f over g close paren times x times is x is greater than or equal to 1 .

(
f
g

)(x)=
√
x
3

and the domain of (
f
g

)(x) is (−∞, ∞).
open paren f over g close paren of x is equal to the fraction with numerator square root of x and denominator 3 and the domain of open paren f over g close paren of x is open paren negative infinity comma infinity close paren .

(
f
g

)(x)=
√
x−1
3x−1

and the domain of (
f
g

)(x) is x≥1.
open paren f over g close paren of x is equal to the fraction with numerator square root of x minus 1 end root and denominator 3 x minus 1 and the domain of open paren f over g close paren times x times is x is greater than or equal to 1 .

(
f
g

)(x)=
√
x−1
3x−1

and the domain of (
f
g

)(x) is (−∞, ∞).

Answer 1

To find \((f \circ g)(x)\), we need to evaluate \(f(g(x))\).

Given the functions:

• \(f(x) = \sqrt{x - 1}\)
• \(g(x) = 3x - 1\)

Now we compute \(f(g(x))\):

\[ f(g(x)) = f(3x - 1) = \sqrt{(3x - 1) - 1} = \sqrt{3x - 2} \]

So we have:

\[ (f \circ g)(x) = \sqrt{3x - 2} \]

Next, we need to determine the domain of \((f \circ g)(x)\).

The expression \(\sqrt{3x - 2}\) is defined when:

\[ 3x - 2 \geq 0 \]

Solving this inequality:

\[ 3x \geq 2 \ x \geq \frac{2}{3} \]

Thus, the domain of \((f \circ g)(x)\) is \(x \geq \frac{2}{3}\).

Summarizing:

\[ (f \circ g)(x) = \sqrt{3x - 2} \quad \text{and the domain is } x \geq \frac{2}{3}. \]

Given the options provided, none of them match the correct answer exactly. If you are looking for a precise match in an answer sheet, please check for \( \sqrt{3x - 2} \) with domain \( x \geq \frac{2}{3} \).