Math / Algebra

QuestionSuppose that the functions $g$ and $h$ are defined as follows. $\begin{array}{l} g(x)=x+7 \\ h(x)=(x-6)(x+6) \end{array}$ (a) Find $\left(\frac{g}{h}\right)$ (2). (b) Find all values that are NOT in the domain of $\frac{g}{h}$ .
If there is more than one value, separate them with commas. (a) $\left(\frac{g}{h}\right)(2)=$ $\square$ (b) Value(s) that are NOT in the domain of $\frac{g}{h}$ :

Studdy Solution

STEP 1

What is this asking? We've got two functions, $g(x)$ and $h(x)$ , and we need to figure out what $\frac{g(x)}{h(x)}$ is when $x = 2$ , and also what values of $x$ we can't plug in. Watch out! Don't forget: we can't divide by zero!
So, any $x$ that makes $h(x) = 0$ is a no-go.

STEP 2

1. Evaluate g(2)
2. Evaluate h(2)
3. Calculate g(2)/h(2)
4. Find the domain of g/h

STEP 3

Let's evaluate $g(x)$ at $x = 2$ .
We know that $g(x) = x + 7$ , so $g(2)$ is just $2 + 7$ .

STEP 4

$g(2) = 2 + 7 = \textbf{9}$ So, $g(2) = \textbf{9}$ !

STEP 5

Now let's find $h(2)$ .
We know $h(x) = (x - 6)(x + 6)$ , so we'll substitute $x = 2$ into the expression.

STEP 6

$h(2) = (2 - 6)(2 + 6)$

STEP 7

$h(2) = (-4)(8) = \textbf{-32}$ Great! $h(2) = \textbf{-32}$ .

STEP 8

Now for the main event: $\left(\frac{g}{h}\right)(2)$ , which is just another way of writing $\frac{g(2)}{h(2)}$ .
We already found $g(2) = \textbf{9}$ and $h(2) = \textbf{-32}$ , so let's put them together!

STEP 9

$\frac{g(2)}{h(2)} = \frac{\textbf{9}}{\textbf{-32}} = -\frac{9}{32}$ So, $\left(\frac{g}{h}\right)(2) = -\frac{9}{32}$ !

STEP 10

The domain of $\frac{g}{h}$ is all the $x$ values we can plug in without causing a divide-by-zero error.
Since $h(x)$ is in the denominator, we need to find any $x$ values that make $h(x) = 0$ .

STEP 11

We know $h(x) = (x - 6)(x + 6)$ .
If $h(x) = 0$ , then either $(x - 6) = 0$ or $(x + 6) = 0$ .

STEP 12

If $x - 6 = 0$ , then $x = 6$ .
If $x + 6 = 0$ , then $x = -6$ .

STEP 13

So, the values that are not in the domain of $\frac{g}{h}$ are $x = 6$ and $x = -6$ .

STEP 14

(a) $\left(\frac{g}{h}\right)(2) = -\frac{9}{32}$ (b) Values not in the domain: $6, -6$

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Studdy Solution

STEP 1

What is this asking? We've got two functions, $g(x)$ and $h(x)$ , and we need to figure out what $\frac{g(x)}{h(x)}$ is when $x = 2$ , and also what values of $x$ we can't plug in. Watch out! Don't forget: we can't divide by zero!
So, any $x$ that makes $h(x) = 0$ is a no-go.

STEP 2

1. Evaluate g(2)
2. Evaluate h(2)
3. Calculate g(2)/h(2)
4. Find the domain of g/h

STEP 3

Let's evaluate $g(x)$ at $x = 2$ .
We know that $g(x) = x + 7$ , so $g(2)$ is just $2 + 7$ .

STEP 4

$g(2) = 2 + 7 = \textbf{9}$ So, $g(2) = \textbf{9}$ !

STEP 5

Now let's find $h(2)$ .
We know $h(x) = (x - 6)(x + 6)$ , so we'll substitute $x = 2$ into the expression.

STEP 6

$h(2) = (2 - 6)(2 + 6)$

STEP 7

$h(2) = (-4)(8) = \textbf{-32}$ Great! $h(2) = \textbf{-32}$ .

STEP 8

Now for the main event: $\left(\frac{g}{h}\right)(2)$ , which is just another way of writing $\frac{g(2)}{h(2)}$ .
We already found $g(2) = \textbf{9}$ and $h(2) = \textbf{-32}$ , so let's put them together!

STEP 9

$\frac{g(2)}{h(2)} = \frac{\textbf{9}}{\textbf{-32}} = -\frac{9}{32}$ So, $\left(\frac{g}{h}\right)(2) = -\frac{9}{32}$ !

STEP 10

The domain of $\frac{g}{h}$ is all the $x$ values we can plug in without causing a divide-by-zero error.
Since $h(x)$ is in the denominator, we need to find any $x$ values that make $h(x) = 0$ .

STEP 11

We know $h(x) = (x - 6)(x + 6)$ .
If $h(x) = 0$ , then either $(x - 6) = 0$ or $(x + 6) = 0$ .

STEP 12

If $x - 6 = 0$ , then $x = 6$ .
If $x + 6 = 0$ , then $x = -6$ .

STEP 13

So, the values that are not in the domain of $\frac{g}{h}$ are $x = 6$ and $x = -6$ .

STEP 14

(a) $\left(\frac{g}{h}\right)(2) = -\frac{9}{32}$ (b) Values not in the domain: $6, -6$

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Studdy Solution

STEP 1

STEP 2

1. Evaluate g(2)2. Evaluate h(2)3. Calculate g(2)/h(2)4. Find the domain of g/h

STEP 3

Let's **evaluate** g(x)g(x)g(x) at x=2x = 2x=2.We know that g(x)=x+7g(x) = x + 7g(x)=x+7, so g(2)g(2)g(2) is just 2+72 + 72+7.

STEP 4

g(2)=2+7=9g(2) = 2 + 7 = \textbf{9}g(2)=2+7=9 So, g(2)=9g(2) = \textbf{9}g(2)=9!

STEP 5

Now let's **find** h(2)h(2)h(2).We know h(x)=(x−6)(x+6)h(x) = (x - 6)(x + 6)h(x)=(x−6)(x+6), so we'll substitute x=2x = 2x=2 into the expression.

STEP 6

h(2)=(2−6)(2+6)h(2) = (2 - 6)(2 + 6)h(2)=(2−6)(2+6)

STEP 7

h(2)=(−4)(8)=-32h(2) = (-4)(8) = \textbf{-32}h(2)=(−4)(8)=-32 Great! h(2)=-32h(2) = \textbf{-32}h(2)=-32.

STEP 8

Now for the main event: (gh)(2)\left(\frac{g}{h}\right)(2)(hg​)(2), which is just another way of writing g(2)h(2)\frac{g(2)}{h(2)}h(2)g(2)​.We already found g(2)=9g(2) = \textbf{9}g(2)=9 and h(2)=-32h(2) = \textbf{-32}h(2)=-32, so let's put them together!

STEP 9

g(2)h(2)=9-32=−932\frac{g(2)}{h(2)} = \frac{\textbf{9}}{\textbf{-32}} = -\frac{9}{32}h(2)g(2)​=-329​=−329​ So, (gh)(2)=−932\left(\frac{g}{h}\right)(2) = -\frac{9}{32}(hg​)(2)=−329​!

STEP 10

The **domain** of gh\frac{g}{h}hg​ is all the xxx values we can plug in *without* causing a divide-by-zero error.Since h(x)h(x)h(x) is in the denominator, we need to find any xxx values that make h(x)=0h(x) = 0h(x)=0.

STEP 11

We know h(x)=(x−6)(x+6)h(x) = (x - 6)(x + 6)h(x)=(x−6)(x+6).If h(x)=0h(x) = 0h(x)=0, then either (x−6)=0(x - 6) = 0(x−6)=0 or (x+6)=0(x + 6) = 0(x+6)=0.

STEP 12

If x−6=0x - 6 = 0x−6=0, then x=6x = 6x=6.If x+6=0x + 6 = 0x+6=0, then x=−6x = -6x=−6.

STEP 13

So, the values that are *not* in the domain of gh\frac{g}{h}hg​ are x=6x = 6x=6 and x=−6x = -6x=−6.

STEP 14

(a) (gh)(2)=−932\left(\frac{g}{h}\right)(2) = -\frac{9}{32}(hg​)(2)=−329​ (b) Values not in the domain: 6,−66, -66,−6

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1. Evaluate g(2)
2. Evaluate h(2)
3. Calculate g(2)/h(2)
4. Find the domain of g/h

Let's evaluate $g(x)$ at $x = 2$ .
We know that $g(x) = x + 7$ , so $g(2)$ is just $2 + 7$ .

$g(2) = 2 + 7 = \textbf{9}$ So, $g(2) = \textbf{9}$ !

Now let's find $h(2)$ .
We know $h(x) = (x - 6)(x + 6)$ , so we'll substitute $x = 2$ into the expression.

$h(2) = (2 - 6)(2 + 6)$

$h(2) = (-4)(8) = \textbf{-32}$ Great! $h(2) = \textbf{-32}$ .

Now for the main event: $\left(\frac{g}{h}\right)(2)$ , which is just another way of writing $\frac{g(2)}{h(2)}$ .
We already found $g(2) = \textbf{9}$ and $h(2) = \textbf{-32}$ , so let's put them together!

$\frac{g(2)}{h(2)} = \frac{\textbf{9}}{\textbf{-32}} = -\frac{9}{32}$ So, $\left(\frac{g}{h}\right)(2) = -\frac{9}{32}$ !

The domain of $\frac{g}{h}$ is all the $x$ values we can plug in without causing a divide-by-zero error.
Since $h(x)$ is in the denominator, we need to find any $x$ values that make $h(x) = 0$ .

We know $h(x) = (x - 6)(x + 6)$ .
If $h(x) = 0$ , then either $(x - 6) = 0$ or $(x + 6) = 0$ .

If $x - 6 = 0$ , then $x = 6$ .
If $x + 6 = 0$ , then $x = -6$ .

So, the values that are not in the domain of $\frac{g}{h}$ are $x = 6$ and $x = -6$ .

(a) $\left(\frac{g}{h}\right)(2) = -\frac{9}{32}$ (b) Values not in the domain: $6, -6$