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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+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)=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)=−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 .

(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 .
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1 answer

To find \((f \cdot g \cdot h)(x)\), let's first express each function \(f(x)\), \(g(x)\), and \(h(x)\):

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

Now, let’s calculate the product \(f(x) \cdot g(x) \cdot h(x)\):

\[ (f \cdot g \cdot h)(x) = f(x) \cdot g(x) \cdot h(x) = \sqrt[3]{x - 1} \cdot \left(-\sqrt[3]{x - 1} + 1\right) \cdot 2\sqrt[3]{x - 1} \]

Now, simplify step-by-step:

  1. Start with \(f(x) \cdot g(x)\): \[ f(x) \cdot g(x) = \sqrt[3]{x - 1} \cdot (-\sqrt[3]{x - 1} + 1) = \sqrt[3]{x - 1} \cdot -\sqrt[3]{x - 1} + \sqrt[3]{x - 1} = -(\sqrt[3]{x - 1})^2 + \sqrt[3]{x - 1} \] This simplifies to: \[ -(x - 1)^{\frac{2}{3}} + (x - 1)^{\frac{1}{3}} \]

  2. Now, multiply by \(h(x) = 2\sqrt[3]{x - 1}\): \[ (f \cdot g)(x) \cdot h(x) = 2\sqrt[3]{x - 1} \left[-(x - 1)^{\frac{2}{3}} + (x - 1)^{\frac{1}{3}}\right] \] Distributing the \(2\sqrt[3]{x - 1}\): \[ = 2\sqrt[3]{x - 1} \cdot -(x - 1)^{\frac{2}{3}} + 2\sqrt[3]{x - 1} \cdot (x - 1)^{\frac{1}{3}} \] This simplifies to: \[ = -2(x - 1)^{1} + 2(x - 1)^{\frac{2}{3}} \]

So rounding everything up: \[ (f \cdot g \cdot h)(x) = -2(x - 1) + 2(x - 1)^{\frac{2}{3}} \]

The next step is to determine the domain of \((f \cdot g \cdot h)(x)\):

  • Each of the functions \(f(x)\), \(g(x)\), and \(h(x)\) are defined for all real numbers because \(\sqrt[3]{x - 1}\) is defined for all \(x\).

Thus, the domain of \((f \cdot g \cdot h)(x)\) is \((-∞, ∞)\).

Final Answers:

\((f \cdot g \cdot h)(x) = -2(x - 1) + 2(x - 1)^{\frac{2}{3}}\), and the domain of \((f \cdot g \cdot h)(x)\) is \((-∞, ∞)\).