Asked by s4t4n45
Function Operations Unit Test
7 of 167 of 16 Items
Question
Use the functions below.
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 (−∞, ∞)
.
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 .
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7 of 167 of 16 Items
Question
Use the functions below.
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 (−∞, ∞)
.
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 .
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Answers
There are no human answers yet.
Answered by
GPT-4o mini
To solve the problem, we need to find \((fg)(x)\), which means we are looking for \(f(g(x))\). Given the functions:
- \(f(x) = \sqrt{x - 1}\)
- \(g(x) = 3x - 1\)
### Step 1: Calculate \(f(g(x))\)
Substitute \(g(x)\) into \(f(x)\):
\[
f(g(x)) = f(3x - 1) = \sqrt{(3x - 1) - 1} = \sqrt{3x - 2}
\]
So, \((fg)(x) = \sqrt{3x - 2}\).
### Step 2: Determine the domain of \((fg)(x)\)
The domain of \((fg)(x)\) is determined by the requirement that the expression under the square root must be non-negative. Therefore, we need:
\[
3x - 2 \geq 0
\]
Solving this inequality:
\[
3x \geq 2 \quad \Rightarrow \quad x \geq \frac{2}{3}
\]
So, the domain of \((fg)(x)\) is \(x \geq \frac{2}{3}\).
### Final Answers:
- \((fg)(x) = \sqrt{3x - 2}\)
- The domain of \((fg)(x)\) is \(x \geq \frac{2}{3}\).
#### Based on the options given:
None of the answer choices perfectly match our result. However, the closest one is:
\((fg)(x) = \sqrt{3x - 2}\) with the domain being \(x \geq \frac{2}{3}\).
If you have to choose among the given options, make sure to clarify the required form and check if anything might have been misreported in the question.
- \(f(x) = \sqrt{x - 1}\)
- \(g(x) = 3x - 1\)
### Step 1: Calculate \(f(g(x))\)
Substitute \(g(x)\) into \(f(x)\):
\[
f(g(x)) = f(3x - 1) = \sqrt{(3x - 1) - 1} = \sqrt{3x - 2}
\]
So, \((fg)(x) = \sqrt{3x - 2}\).
### Step 2: Determine the domain of \((fg)(x)\)
The domain of \((fg)(x)\) is determined by the requirement that the expression under the square root must be non-negative. Therefore, we need:
\[
3x - 2 \geq 0
\]
Solving this inequality:
\[
3x \geq 2 \quad \Rightarrow \quad x \geq \frac{2}{3}
\]
So, the domain of \((fg)(x)\) is \(x \geq \frac{2}{3}\).
### Final Answers:
- \((fg)(x) = \sqrt{3x - 2}\)
- The domain of \((fg)(x)\) is \(x \geq \frac{2}{3}\).
#### Based on the options given:
None of the answer choices perfectly match our result. However, the closest one is:
\((fg)(x) = \sqrt{3x - 2}\) with the domain being \(x \geq \frac{2}{3}\).
If you have to choose among the given options, make sure to clarify the required form and check if anything might have been misreported in the question.
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