Asked by s4t4n45
Use the functions below.
f(x)=x2−1
g(x)=x2
h(x)=2x2
What is (f⋅g÷h)(x)
? What is the domain of (f⋅g÷h)(x)
?
(1 point)
Responses
(f⋅g÷h)(x)=2x6−2x4
and the domain of (f⋅g÷h)(x)
is x≠0
.
open paren f times g divided by h close paren times x is equal to 2 x to the 6th power minus 2 x to the 4th power and the domain of open paren f times g divided by h close paren times x is x is not equal to 0 .
(f⋅g÷h)(x)=12x2−12
and the domain of (f⋅g÷h)(x)
is (−∞, ∞)
.
open paren f times g divided by h close paren times x is equal to 1 half x squared minus 1 half and the domain of open paren f times g divided by h close paren times x is open paren negative infinity comma infinity close paren .
(f⋅g÷h)(x)=2x6−2x4
and the domain of (f⋅g÷h)(x)
is (−∞, ∞)
.
open paren f times g divided by h close paren times x is equal to 2 x to the 6th power minus 2 x to the 4th power and the domain of open paren f times g divided by h close paren times x is open paren negative infinity comma infinity close paren .
(f⋅g÷h)(x)=12x2−12
and the domain of (f⋅g÷h)(x)
is x≠0
.
f(x)=x2−1
g(x)=x2
h(x)=2x2
What is (f⋅g÷h)(x)
? What is the domain of (f⋅g÷h)(x)
?
(1 point)
Responses
(f⋅g÷h)(x)=2x6−2x4
and the domain of (f⋅g÷h)(x)
is x≠0
.
open paren f times g divided by h close paren times x is equal to 2 x to the 6th power minus 2 x to the 4th power and the domain of open paren f times g divided by h close paren times x is x is not equal to 0 .
(f⋅g÷h)(x)=12x2−12
and the domain of (f⋅g÷h)(x)
is (−∞, ∞)
.
open paren f times g divided by h close paren times x is equal to 1 half x squared minus 1 half and the domain of open paren f times g divided by h close paren times x is open paren negative infinity comma infinity close paren .
(f⋅g÷h)(x)=2x6−2x4
and the domain of (f⋅g÷h)(x)
is (−∞, ∞)
.
open paren f times g divided by h close paren times x is equal to 2 x to the 6th power minus 2 x to the 4th power and the domain of open paren f times g divided by h close paren times x is open paren negative infinity comma infinity close paren .
(f⋅g÷h)(x)=12x2−12
and the domain of (f⋅g÷h)(x)
is x≠0
.
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To find \((f \cdot g \div h)(x)\), we will first compute \(f(x) \cdot g(x)\) and then divide by \(h(x)\).
Given:
- \(f(x) = x^2 - 1\)
- \(g(x) = x^2\)
- \(h(x) = 2x^2\)
1. **Calculate \(f(x) \cdot g(x)\)**:
\[
f(x) \cdot g(x) = (x^2 - 1) \cdot x^2 = x^4 - x^2
\]
2. **Now divide by \(h(x)\)**:
\[
(f \cdot g)(x) \div h(x) = \frac{x^4 - x^2}{2x^2}
\]
3. **Simplify the expression**:
\[
= \frac{x^2(x^2 - 1)}{2x^2}
\]
For \(x^2 \neq 0\) (i.e., \(x \neq 0\)):
\[
= \frac{x^2 - 1}{2}
\]
Thus, we have:
\[
(f \cdot g \div h)(x) = \frac{1}{2}(x^2 - 1) = \frac{1}{2}x^2 - \frac{1}{2}
\]
4. **Determine the domain**:
The expression \((f \cdot g \div h)(x)\) is undefined when \(h(x) = 2x^2 = 0\), which occurs at \(x = 0\). Therefore, the domain excludes this point.
The domain of \((f \cdot g \div h)(x)\) is:
\[
(-\infty, 0) \cup (0, \infty)
\]
**Final Answer**:
\((f \cdot g \div h)(x) = \frac{1}{2}x^2 - \frac{1}{2}\) and the domain of \((f \cdot g \div h)(x)\) is \(x \neq 0\).
Given:
- \(f(x) = x^2 - 1\)
- \(g(x) = x^2\)
- \(h(x) = 2x^2\)
1. **Calculate \(f(x) \cdot g(x)\)**:
\[
f(x) \cdot g(x) = (x^2 - 1) \cdot x^2 = x^4 - x^2
\]
2. **Now divide by \(h(x)\)**:
\[
(f \cdot g)(x) \div h(x) = \frac{x^4 - x^2}{2x^2}
\]
3. **Simplify the expression**:
\[
= \frac{x^2(x^2 - 1)}{2x^2}
\]
For \(x^2 \neq 0\) (i.e., \(x \neq 0\)):
\[
= \frac{x^2 - 1}{2}
\]
Thus, we have:
\[
(f \cdot g \div h)(x) = \frac{1}{2}(x^2 - 1) = \frac{1}{2}x^2 - \frac{1}{2}
\]
4. **Determine the domain**:
The expression \((f \cdot g \div h)(x)\) is undefined when \(h(x) = 2x^2 = 0\), which occurs at \(x = 0\). Therefore, the domain excludes this point.
The domain of \((f \cdot g \div h)(x)\) is:
\[
(-\infty, 0) \cup (0, \infty)
\]
**Final Answer**:
\((f \cdot g \div h)(x) = \frac{1}{2}x^2 - \frac{1}{2}\) and the domain of \((f \cdot g \div h)(x)\) is \(x \neq 0\).
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