Use the functions below.

Question

USE ONLY THE GIVEN ANSWERS PROVIDED.
f(x)=x−1−−−−√
g(x)=3x−1
What is(fg)(x)
? What is the domain of (fg)(x)
?

(1 point)
Responses

(fg)(x)=x√3
and the domain of (fg)(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 .

(fg)(x)=x√3
and the domain of (fg)(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 .

(fg)(x)=x−1√3x−1
and the domain of (fg)(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 .

(fg)(x)=x−1√3x−1
and the domain of (fg)(x)
is (−∞, ∞)
.

Answer 1

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

Given:

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

Now, we can substitute \(g(x)\) into \(f(x)\):

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

Now, let's determine the domain of \((fg)(x)\).

For the function \(f(x)\), \(x - 1\) must be non-negative:

\[ 3x - 2 \geq 0 \ 3x \geq 2 \ x \geq \frac{2}{3} \]

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

Among the given responses, none exactly match, but based on your options, "the domain of \((fg)(x)\) is \(x \geq 1\)" is not correct.

As the correct answer is: (fg)(x) = \sqrt{3x - 2} and the domain of (fg)(x) is \(x \geq \frac{2}{3}\).

Please remember there may have been an oversight in the original options you provided since the calculation doesn't lead to any of them being fully correct.