The area of a rectangular backyard is given by the function f(x)=x2−16

Question

. 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 .

Answer 1

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.