Asked by s4t4n45
Use the functions below.
f(x)=x−2
g(x)=2x
h(x)=x+7−−−−√
What is (f⋅g⋅h)(x)
? What is the domain of (f⋅g⋅h)(x)
?
(1 point)
Responses
(f⋅g⋅h)(x)=2xxx+7−−−−−√−2x+1x+7−−−−−√
and the domain of (f⋅g⋅h)(x)
is (−∞,∞)
.
open paren f times g times h close paren times x is equal to 2 to the x th power x square root of x plus 7 end root minus 2 raised to the x plus 1 power square root of x plus 7 end root 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)=2xxx+7−−−−−√−2x+1x+7−−−−−√
and the domain of (f⋅g⋅h)(x)
is x≥−7
.
open paren f times g times h close paren times x is equal to 2 to the x th power x square root of x plus 7 end root minus 2 raised to the x plus 1 power square root of x plus 7 end root and the domain of open paren f times g times h close paren times x is x is greater than or equal to negative 7 .
(f⋅g⋅h)(x)=x−2+2x+x+7−−−−−√
and the domain of (f⋅g⋅h)(x)
is x≥−7
.
open paren f times g times h close paren times x is equal to x minus 2 plus 2 to the x th power plus square root of x plus 7 end root and the domain of open paren f times g times h close paren times x is x is greater than or equal to negative 7 .
(f⋅g⋅h)(x)=x−2+2x+x+7−−−−−√
and the domain of (f⋅g⋅h)(x)
is (−∞,∞)
.
f(x)=x−2
g(x)=2x
h(x)=x+7−−−−√
What is (f⋅g⋅h)(x)
? What is the domain of (f⋅g⋅h)(x)
?
(1 point)
Responses
(f⋅g⋅h)(x)=2xxx+7−−−−−√−2x+1x+7−−−−−√
and the domain of (f⋅g⋅h)(x)
is (−∞,∞)
.
open paren f times g times h close paren times x is equal to 2 to the x th power x square root of x plus 7 end root minus 2 raised to the x plus 1 power square root of x plus 7 end root 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)=2xxx+7−−−−−√−2x+1x+7−−−−−√
and the domain of (f⋅g⋅h)(x)
is x≥−7
.
open paren f times g times h close paren times x is equal to 2 to the x th power x square root of x plus 7 end root minus 2 raised to the x plus 1 power square root of x plus 7 end root and the domain of open paren f times g times h close paren times x is x is greater than or equal to negative 7 .
(f⋅g⋅h)(x)=x−2+2x+x+7−−−−−√
and the domain of (f⋅g⋅h)(x)
is x≥−7
.
open paren f times g times h close paren times x is equal to x minus 2 plus 2 to the x th power plus square root of x plus 7 end root and the domain of open paren f times g times h close paren times x is x is greater than or equal to negative 7 .
(f⋅g⋅h)(x)=x−2+2x+x+7−−−−−√
and the domain of (f⋅g⋅h)(x)
is (−∞,∞)
.
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To find \((f \cdot g \cdot h)(x)\), we will first express it in terms of the given functions and then determine the domain.
**Given Functions:**
- \( f(x) = x - 2 \)
- \( g(x) = 2x \)
- \( h(x) = \sqrt{x + 7} \)
**Calculating \((f \cdot g \cdot h)(x)\):**
\[
(f \cdot g \cdot h)(x) = f(x) \cdot g(x) \cdot h(x)
\]
Substituting the functions:
\[
= (x - 2) \cdot (2x) \cdot \sqrt{x + 7}
\]
Now, we simplify it:
1. First, calculate \(f(x) \cdot g(x)\):
\[
f(x) \cdot g(x) = (x - 2)(2x) = 2x^2 - 4x
\]
2. Now multiply by \(h(x)\):
\[
(2x^2 - 4x) \cdot \sqrt{x + 7}
= (2x^2 - 4x) \sqrt{x + 7}
\]
Thus, we have:
\[
(f \cdot g \cdot h)(x) = (2x^2 - 4x) \sqrt{x + 7}
\]
**Finding the Domain:**
The domain of \((f \cdot g \cdot h)(x)\) is determined by the domain restrictions of the functions involved:
- For \(f(x) = x - 2\), there are no restrictions: it is defined for all \(x \in (-\infty, \infty)\).
- For \(g(x) = 2x\), there are also no restrictions: it is defined for all \(x \in (-\infty, \infty)\).
- For \(h(x) = \sqrt{x + 7}\), we require:
\[
x + 7 \geq 0 \implies x \geq -7
\]
Therefore, the domain of \((f \cdot g \cdot h)(x)\) is:
\[
x \geq -7 \text{ or } [-7, \infty)
\]
**Final Answer:**
\((f \cdot g \cdot h)(x) = (2x^2 - 4x) \sqrt{x + 7}\)
And the domain of \((f \cdot g \cdot h)(x)\) is \(x \geq -7\) or \( [-7, \infty) \).
**Given Functions:**
- \( f(x) = x - 2 \)
- \( g(x) = 2x \)
- \( h(x) = \sqrt{x + 7} \)
**Calculating \((f \cdot g \cdot h)(x)\):**
\[
(f \cdot g \cdot h)(x) = f(x) \cdot g(x) \cdot h(x)
\]
Substituting the functions:
\[
= (x - 2) \cdot (2x) \cdot \sqrt{x + 7}
\]
Now, we simplify it:
1. First, calculate \(f(x) \cdot g(x)\):
\[
f(x) \cdot g(x) = (x - 2)(2x) = 2x^2 - 4x
\]
2. Now multiply by \(h(x)\):
\[
(2x^2 - 4x) \cdot \sqrt{x + 7}
= (2x^2 - 4x) \sqrt{x + 7}
\]
Thus, we have:
\[
(f \cdot g \cdot h)(x) = (2x^2 - 4x) \sqrt{x + 7}
\]
**Finding the Domain:**
The domain of \((f \cdot g \cdot h)(x)\) is determined by the domain restrictions of the functions involved:
- For \(f(x) = x - 2\), there are no restrictions: it is defined for all \(x \in (-\infty, \infty)\).
- For \(g(x) = 2x\), there are also no restrictions: it is defined for all \(x \in (-\infty, \infty)\).
- For \(h(x) = \sqrt{x + 7}\), we require:
\[
x + 7 \geq 0 \implies x \geq -7
\]
Therefore, the domain of \((f \cdot g \cdot h)(x)\) is:
\[
x \geq -7 \text{ or } [-7, \infty)
\]
**Final Answer:**
\((f \cdot g \cdot h)(x) = (2x^2 - 4x) \sqrt{x + 7}\)
And the domain of \((f \cdot g \cdot h)(x)\) is \(x \geq -7\) or \( [-7, \infty) \).
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