Use the functions below.

Question

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
.

Answer 1

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 \(x \neq 0\).