Asked by burnout
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)=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)=12x2−12
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 1 half x squared minus 1 half 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)=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)=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(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)=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)=12x2−12
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 1 half x squared minus 1 half 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)=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)=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 .
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To find \((f \cdot g \div h)(x)\), we first need to calculate \(f(x)\), \(g(x)\), and \(h(x)\):
1. \(f(x) = x^2 - 1\)
2. \(g(x) = x^2\)
3. \(h(x) = 2x^2\)
Now, let's multiply \(f(x)\) and \(g(x)\) first:
\[
f(x) \cdot g(x) = (x^2 - 1) \cdot (x^2) = x^4 - x^2
\]
Next, we will divide the result by \(h(x)\):
\[
\frac{f(x) \cdot g(x)}{h(x)} = \frac{x^4 - x^2}{2x^2}
\]
To simplify:
\[
\frac{x^4 - x^2}{2x^2} = \frac{x^2(x^2 - 1)}{2x^2} = \frac{x^2 - 1}{2} = \frac{1}{2}(x^2 - 1)
\]
So we have:
\[
(f \cdot g \div h)(x) = \frac{1}{2}(x^2 - 1)
\]
Now, let's determine the domain. The only point we need to be concerned about is where \(h(x)\) is zero, since we cannot divide by zero.
Setting \(h(x) = 0\):
\[
2x^2 = 0 \Rightarrow x^2 = 0 \Rightarrow x = 0
\]
Thus, the domain of \((f \cdot g \div h)(x)\) excludes \(x = 0\).
In conclusion:
\[
(f \cdot g \div h)(x) = \frac{1}{2}(x^2 - 1) \quad \text{and the domain is } x \neq 0.
\]
So the correct response is:
\((f \cdot g \div h)(x) = \frac{1}{2}(x^2 - 1)\) and the domain of \((f \cdot g \div h)(x)\) is \(x \neq 0\).
1. \(f(x) = x^2 - 1\)
2. \(g(x) = x^2\)
3. \(h(x) = 2x^2\)
Now, let's multiply \(f(x)\) and \(g(x)\) first:
\[
f(x) \cdot g(x) = (x^2 - 1) \cdot (x^2) = x^4 - x^2
\]
Next, we will divide the result by \(h(x)\):
\[
\frac{f(x) \cdot g(x)}{h(x)} = \frac{x^4 - x^2}{2x^2}
\]
To simplify:
\[
\frac{x^4 - x^2}{2x^2} = \frac{x^2(x^2 - 1)}{2x^2} = \frac{x^2 - 1}{2} = \frac{1}{2}(x^2 - 1)
\]
So we have:
\[
(f \cdot g \div h)(x) = \frac{1}{2}(x^2 - 1)
\]
Now, let's determine the domain. The only point we need to be concerned about is where \(h(x)\) is zero, since we cannot divide by zero.
Setting \(h(x) = 0\):
\[
2x^2 = 0 \Rightarrow x^2 = 0 \Rightarrow x = 0
\]
Thus, the domain of \((f \cdot g \div h)(x)\) excludes \(x = 0\).
In conclusion:
\[
(f \cdot g \div h)(x) = \frac{1}{2}(x^2 - 1) \quad \text{and the domain is } x \neq 0.
\]
So the correct response is:
\((f \cdot g \div h)(x) = \frac{1}{2}(x^2 - 1)\) and the domain of \((f \cdot g \div h)(x)\) is \(x \neq 0\).
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