Find the function $f \cdot g$ and its domain if $f ( x ) = \sqrt { x + 7 }$ and
$g ( x ) = \sqrt { x - 7 }$.0
A) $\sqrt { x ^ { 2 } - 49 }$
$D = [ 7 , \infty )$
B) $\sqrt { x ^ { 2 } - 49 }$
$D = [ - 7 , \infty )$
C) $\sqrt { x ^ { 2 } + 49 }$
$D = [ - 7 , \infty )$
D) $\sqrt { x ^ { 2 } + 49 }$
$D = [ 7 , \infty )$

Question

Find the function $f \cdot g$ and its domain if $f ( x ) = \sqrt { x + 7 }$ and 
$g ( x ) = \sqrt { x - 7 }$.0
A) $\sqrt { x ^ { 2 } - 49 }$
$D = [ 7 , \infty )$
B) $\sqrt { x ^ { 2 } - 49 }$
$D = [ - 7 , \infty )$
C) $\sqrt { x ^ { 2 } + 49 }$
$D = [ - 7 , \infty )$
D) $\sqrt { x ^ { 2 } + 49 }$
$D = [ 7 , \infty )$

Quizplus · Accepted Answer

To find the product of two functions, we simply multiply them:
$$f \cdot g = \sqrt{(x+7)} \cdot \sqrt{(x-7)} = \sqrt{(x+7)(x-7)} = \sqrt{x^2-49}$$

However, we must also consider the domain of the product function. Since we are taking the square root of an expression, the expression inside the square root must be non-negative:
$$x^2 - 49 \ge 0$$
$$(x-7)(x+7) \ge 0$$

This inequality is satisfied when $x \in [-7,7] \cup (7, \infty)$. However, since the original functions $f$ and $g$ are defined with square roots, we need to exclude any values of $x$ that would make either of them negative.

From the function $f(x)$, we have $x+7 \ge 0$ or $x \ge -7$, and from $g(x)$, we have $x-7 \ge 0$ or $x \ge 7$.

Therefore, the domain of $f \cdot g$ is $[7, \infty)$.

Combining these results, we get:

$$f \cdot g = \sqrt{x^2 - 49}$$
$$D = [7, \infty)$$

Find the Function f⋅gf \cdot gf⋅g And Its Domain If f(x)=x+7f ( x ) = \sqrt { x + 7 }f(x)=x+7​

Find the Function $f \cdot g$ And Its Domain If $f ( x ) = \sqrt { x + 7 }$