QuestionFind the domain of $\left(\frac{v}{s}\right)(x)=\frac{x^{2} \sqrt{x+6}-36 \sqrt{x+6}}{x-3}$ given $s(x)=\frac{x-3}{x^{2}-36}$ and $v(x)=\sqrt{x+6}$ .

Studdy Solution

STEP 1

Assumptions1. The function $s(x)=\frac{x-3}{x^{}-36}$ is defined for all $x$ except $x = \pm6$ because the denominator cannot be zero. . The function $v(x)=\sqrt{x+6}$ is defined for all $x \geq -6$ because the square root of a negative number is not a real number.
3. The function $\left(\frac{v}{s}\right)(x)=\frac{x^{} \sqrt{x+6}-36 \sqrt{x+6}}{x-3}$ is defined for all $x$ that makes both $v(x)$ and $s(x)$ defined, excluding any $x$ that makes $s(x) =0$ .

STEP 2

First, we need to find the domain of $s(x)$ .
The denominator of $s(x)$ is $x^{2}-36$ . This can be factored into $(x-6)(x+6)$ .
So, $s(x)$ is undefined when $x^{2}-36=0$ .

STEP 3

olve the equation $x^{2}-36=0$ to find the values of $x$ that make $s(x)$ undefined.
$x^{2}-36=0$

STEP 4

Factor the equation to find the roots.
$(x-6)(x+6)=0$

STEP 5

Set each factor equal to zero and solve for $x$ .
$x-=0 \quad \text{or} \quad x+=0$

STEP 6

olving the equations gives $x=6$ or $x=-6$ . These are the values that make $s(x)$ undefined.

STEP 7

Next, we need to find the domain of $v(x)$ .
The function $v(x)$ is defined for all $x \geq -6$ because the square root of a negative number is not a real number.

STEP 8

Finally, we need to find the values of $x$ that make $s(x) =0$ because these values need to be excluded from the domain of $\left(\frac{v}{s}\right)(x)$ .
The numerator of $s(x)$ is $x-3$ , so $s(x) =0$ when $x-3=0$ .

STEP 9

olve the equation $x-3=$ to find the values of $x$ that make $s(x) =$ .
$x-3=$

STEP 10

olving the equation gives $x=3$ . This is the value that makes $s(x) =0$ .

STEP 11

Combining the results from steps6,7, and10, we find that the domain of $\left(\frac{v}{s}\right)(x)$ is all $x$ such that $x \geq -6$ , excluding $x = -6$ , $x =6$ , and $x =3$ .
The domain of $\left(\frac{v}{s}\right)(x)$ is $(-6,3) \cup (3,6) \cup (6, \infty)$ .

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