Question
The area of a rectangular backyard is given by the function f(x)=x2−16
. The length of a rectangular backyard is given by the function g(x)=x+4
. What function represents the width of the rectangular backyard? What is the domain?(1 point)
Responses
(fg)(x)=x−4
and the domain of (fg)(x)
is x>4
.
open paren f over g close paren of x is equal to x minus 4 and the domain of open paren f over g close paren of x is x is greater than 4 .
(fg)(x)=x−4
and the domain of (fg)(x)
is (−∞, ∞)
.
open paren f over g close paren of x is equal to x minus 4 and the domain of open paren f over g close paren of x is open paren negative infinity comma infinity close paren .
(fg)(x)=x2+x−12
and the domain of (fg)(x)
is (−∞, ∞)
.
open paren f over g close paren of x is equal to x squared plus x minus 12 and the domain of open paren f over g close paren of x is open paren negative infinity comma infinity close paren .
(fg)(x)=x2+x+12
and the domain of (fg)(x)
is x>4
.
open paren f over g close paren of x is equal to x squared plus x plus 12 and the domain of open paren f over g close paren of x is x is greater than 4 .
. The length of a rectangular backyard is given by the function g(x)=x+4
. What function represents the width of the rectangular backyard? What is the domain?(1 point)
Responses
(fg)(x)=x−4
and the domain of (fg)(x)
is x>4
.
open paren f over g close paren of x is equal to x minus 4 and the domain of open paren f over g close paren of x is x is greater than 4 .
(fg)(x)=x−4
and the domain of (fg)(x)
is (−∞, ∞)
.
open paren f over g close paren of x is equal to x minus 4 and the domain of open paren f over g close paren of x is open paren negative infinity comma infinity close paren .
(fg)(x)=x2+x−12
and the domain of (fg)(x)
is (−∞, ∞)
.
open paren f over g close paren of x is equal to x squared plus x minus 12 and the domain of open paren f over g close paren of x is open paren negative infinity comma infinity close paren .
(fg)(x)=x2+x+12
and the domain of (fg)(x)
is x>4
.
open paren f over g close paren of x is equal to x squared plus x plus 12 and the domain of open paren f over g close paren of x is x is greater than 4 .
Answers
GPT-4o mini
To find the width of the rectangular backyard, we start with the area and length functions given:
- Area function: \( f(x) = x^2 - 16 \)
- Length function: \( g(x) = x + 4 \)
The area of a rectangle is defined as \( \text{Area} = \text{Length} \times \text{Width} \). We can express the width function \( w(x) \) as follows:
\[
w(x) = \frac{f(x)}{g(x)}
\]
Now substituting our expressions for \( f(x) \) and \( g(x) \):
\[
w(x) = \frac{x^2 - 16}{x + 4}
\]
Next, we can factor \( f(x) = x^2 - 16 \) (which is a difference of squares):
\[
f(x) = (x - 4)(x + 4)
\]
So now we can rewrite the width function:
\[
w(x) = \frac{(x - 4)(x + 4)}{x + 4}
\]
Provided \( x + 4 \neq 0 \) (which happens when \( x \neq -4 \)), we can simplify this to:
\[
w(x) = x - 4
\]
Thus, the width function is:
\[
w(x) = x - 4
\]
Now, let's determine the domain of this function. The length function \( g(x) = x + 4 \) must be greater than 0 for the rectangle to have a valid dimension, which implies:
\[
x + 4 > 0 \implies x > -4
\]
However, we also need to consider that the area should be non-negative, meaning:
\[
f(x) = x^2 - 16 \geq 0
\]
Solving the inequality:
\[
x^2 \geq 16 \implies x \leq -4 \text{ or } x \geq 4
\]
Combining both conditions \( x > -4 \) and either \( x \leq -4 \) or \( x \geq 4 \), we see that the valid range for \( x \) is:
\[
x \geq 4
\]
Therefore, the width function \( w(x) = x - 4 \) is valid for \( x \geq 4 \).
Putting it all together, the correct response is:
**(fg)(x)=x−4 and the domain of (fg)(x) is x>4.**
- Area function: \( f(x) = x^2 - 16 \)
- Length function: \( g(x) = x + 4 \)
The area of a rectangle is defined as \( \text{Area} = \text{Length} \times \text{Width} \). We can express the width function \( w(x) \) as follows:
\[
w(x) = \frac{f(x)}{g(x)}
\]
Now substituting our expressions for \( f(x) \) and \( g(x) \):
\[
w(x) = \frac{x^2 - 16}{x + 4}
\]
Next, we can factor \( f(x) = x^2 - 16 \) (which is a difference of squares):
\[
f(x) = (x - 4)(x + 4)
\]
So now we can rewrite the width function:
\[
w(x) = \frac{(x - 4)(x + 4)}{x + 4}
\]
Provided \( x + 4 \neq 0 \) (which happens when \( x \neq -4 \)), we can simplify this to:
\[
w(x) = x - 4
\]
Thus, the width function is:
\[
w(x) = x - 4
\]
Now, let's determine the domain of this function. The length function \( g(x) = x + 4 \) must be greater than 0 for the rectangle to have a valid dimension, which implies:
\[
x + 4 > 0 \implies x > -4
\]
However, we also need to consider that the area should be non-negative, meaning:
\[
f(x) = x^2 - 16 \geq 0
\]
Solving the inequality:
\[
x^2 \geq 16 \implies x \leq -4 \text{ or } x \geq 4
\]
Combining both conditions \( x > -4 \) and either \( x \leq -4 \) or \( x \geq 4 \), we see that the valid range for \( x \) is:
\[
x \geq 4
\]
Therefore, the width function \( w(x) = x - 4 \) is valid for \( x \geq 4 \).
Putting it all together, the correct response is:
**(fg)(x)=x−4 and the domain of (fg)(x) is x>4.**