Question

The volume of a box in the shape of a rectangular prism is given by the function f(x)=x3−8
. The height of the box is given by the function g(x)=x−2
.

What is the area of the base of the box? What is the domain?

(1 point)
Responses

The function (fg)(x)=1x2+2x+4
represents the volume of the chest. The domain is (−∞,∞)
.
The function open paren f over g close paren of x is equal to the fraction with numerator 1 and denominator x squared plus 2 x plus 4 represents the volume of the chest. The domain is open paren negative infinity comma infinity close paren .

The function (fg)(x)=1x2+2x+4
represents the area of the base of the box. The domain is x>2
.
The function open paren f over g close paren of x is equal to the fraction with numerator 1 and denominator x squared plus 2 x plus 4 represents the area of the base of the box. The domain is x is greater than 2 .

The function (fg)(x)=x2+2x+4
represents the area of the base of the box. The domain is x>2
.
The function open paren f over g close paren of x is equal to x squared plus 2 x plus 4 represents the area of the base of the box. The domain is x is greater than 2 .

The function (fg)(x)=x2+2x+4
represents the area of the base of the box. The domain is (−∞,∞)
.

Answer 1

To find the area of the base of the box and the domain, we need to analyze the given functions:

1. **Volume of the box:** The volume \( V \) is represented by the function:
\[
f(x) = x^3 - 8
\]

2. **Height of the box:** The height \( h \) is given by the function:
\[
g(x) = x - 2
\]

The volume \( V \) of a box (rectangular prism) is calculated by multiplying the area of the base \( A \) by the height \( h \):
\[
V = A \times h
\]

From this, we can express the area of the base \( A \) as:
\[
A = \frac{V}{h} = \frac{f(x)}{g(x)} = \frac{x^3 - 8}{x - 2}
\]

To simplify \( A = \frac{x^3 - 8}{x - 2} \), we can perform polynomial long division or factor the numerator. \( x^3 - 8 \) can be factored as:
\[
x^3 - 8 = (x - 2)(x^2 + 2x + 4)
\]

Thus,
\[
A = \frac{(x - 2)(x^2 + 2x + 4)}{x - 2}
\]

Assuming \( x - 2 \neq 0 \) (i.e., \( x \neq 2 \)), we can cancel the \( x - 2 \) terms:
\[
A = x^2 + 2x + 4
\]

**Area and Domain:**
- The area of the base of the box is given by:
\[
A = x^2 + 2x + 4
\]

- The domain of \( A \) is restricted by the condition \( g(x) = x - 2 \neq 0 \):
\[
x - 2 > 0 \implies x > 2
\]

Thus, the area of the base of the box is \( x^2 + 2x + 4 \), and the domain is \( x > 2 \).

So the correct response based on your options appears to be:

**The function \( (fg)(x) = x^2 + 2x + 4 \) represents the area of the base of the box. The domain is \( x > 2 \).**