Use the functions below.

Question

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

Answer 1

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

Given:

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

Now, calculating \(f(g(x))\):

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

Next, we need to find the domain of \((fg)(x) = \sqrt{3x - 2}\). The expression under the square root must be non-negative for \(f(g(x))\) to be defined, so we set up the following inequality:

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

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

None of the options you provided matches this result accurately. However, if we check for errors and missing options, it seems none precisely match, especially regarding the domain requirement of \(x \geq \frac{2}{3}\).

In conclusion:

• \((fg)(x) = \sqrt{3x - 2}\)
• Domain: \(x \geq \frac{2}{3}\) (not represented in your options).