Math Snap

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 $x = -6$.

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