Question

Use the functions below.

f(x)=x−1−−−−−√3
g(x)=−x−1−−−−−√3+1
h(x)=2x−1−−−−−√3
What is (f⋅g⋅h)(x)
? What is the domain of (f⋅g⋅h)(x)
?

(1 point)
Responses

(f⋅g⋅h)(x)=2x−1−−−−−√3+1
and the domain of (f⋅g⋅h)(x)
is x≥1
.
open paren f times g times h close paren times x is equal to 2 cube root of x minus 1 plus 1 and the domain of open paren f times g times h close paren times x is x is greater than or equal to 1 .

(f⋅g⋅h)(x)=2x−1−−−−−√3+1
and the domain of (f⋅g⋅h)(x)
is (−∞, ∞)
.
open paren f times g times h close paren times x is equal to 2 cube root of x minus 1 plus 1 and the domain of open paren f times g times h close paren times x is open paren negative infinity comma infinity close paren .

(f⋅g⋅h)(x)=−2x+2+2(x−1)23
and the domain of (f⋅g⋅h)(x)
is (−∞, ∞)
.
open paren f times g times h close paren times x is equal to negative 2 x plus 2 plus 2 times open paren x minus 1 close paren raised to the 2 thirds power and the domain of open paren f times g times h close paren times x is open paren negative infinity comma infinity close paren .

(f⋅g⋅h)(x)=−2(x−1)23x−1−−−−−√3+2(x−1)23
and the domain of (f⋅g⋅h)(x)
is x≥1
.
open paren f times g times h close paren times x is equal to negative 2 open paren x minus 1 close paren raised to the 2 thirds power cube root of x minus 1 plus 2 times open paren x minus 1 close paren raised to the 2 thirds power and the domain of open paren f times g times h close paren times x is x is greater than or equal to 1 .
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Answer 1

Let's analyze each of the functions given:

1. **Functions**:
- \( f(x) = \sqrt[3]{x - 1} \)
- \( g(x) = -\sqrt[3]{x - 1} + 1 \)
- \( h(x) = 2\sqrt[3]{x - 1} \)

2. **Calculating \( (f \cdot g \cdot h)(x) \)**:
First, we need to evaluate \( f(x) \), \( g(x) \), and \( h(x) \):
- \( f(x) = \sqrt[3]{x - 1} \)
- \( g(x) = -\sqrt[3]{x - 1} + 1 \)
- \( h(x) = 2\sqrt[3]{x - 1} \)

Now combining them:
\[
(f \cdot g \cdot h)(x) = f(x) \cdot g(x) \cdot h(x)
\]
First, compute \( f(x) \cdot g(x) \):
\[
f(x) \cdot g(x) = \sqrt[3]{x - 1} \cdot (-\sqrt[3]{x - 1} + 1)
\]
Simplifying this:
\[
= \sqrt[3]{x - 1} \cdot (-\sqrt[3]{x - 1}) + \sqrt[3]{x - 1} \cdot 1
= -(\sqrt[3]{x - 1})^2 + \sqrt[3]{x - 1}
\]

Next, let's multiply this result with \( h(x) \):
\[
(f \cdot g)(x) \cdot h(x) = (- (\sqrt[3]{x - 1})^2 + \sqrt[3]{x - 1}) \cdot 2\sqrt[3]{x - 1}
\]
This gives us:
\[
= -2 (\sqrt[3]{x - 1})^3 + 2 (\sqrt[3]{x - 1})^2
= -2(x - 1) + 2\sqrt[3]{(x - 1)^2}
\]
Thus, the expression simplifies to:
\[
(f \cdot g \cdot h)(x) = -2(x - 1) + 2(x - 1)^{\frac{2}{3}}
\]

3. **Finding the Domain**:
The expressions \( f(x) \), \( g(x) \), and \( h(x) \) involve cube roots, which are defined for all real numbers. Therefore, the domain of \( (f \cdot g \cdot h)(x) \) is:
\[
(-\infty, \infty)
\]

4. **Final Result**:
Thus, we can conclude:
- \( (f \cdot g \cdot h)(x) = -2(x - 1) + 2(x - 1)^{\frac{2}{3}} \)
- The domain of \( (f \cdot g \cdot h)(x) \) is \( (-\infty, \infty) \).

**Final selection**:
The correct answer is:
\[
(f \cdot g \cdot h)(x) = -2x + 2 + 2(x - 1)^{\frac{2}{3}}
\]
and the domain of \( (f \cdot g \cdot h)(x) \) is \( (-\infty, \infty) \).