Math Snap

STEP 1

Assumptions1. We are given two functions $q(x)=\frac{1}{\sqrt{x}}$ and $h(x)=x^{}-25$
. We are asked to find the domain of three different functions $\frac{q(x)}{h(x)}$, $q(h(x))$, and $h(q(x))$
3. The domain of a function is the set of all possible input values (x-values) that will output real numbers

STEP 2

Let's start with the first function $\frac{q(x)}{h(x)}$. The domain of this function is all x-values for which both $q(x)$ and $h(x)$ are defined and $h(x) \neq0$.

STEP 3

The function $q(x)$ is undefined when $x \leq0$ because we can't take the square root of a negative number or zero. So, the domain of $q(x)$ is $x >0$.

STEP 4

The function $h(x)$ is always defined for all real numbers. However, for the function $\frac{q(x)}{h(x)}$, we must exclude values of $x$ for which $h(x) =0$.

STEP 5

To find the values of $x$ for which $h(x) =0$, we solve the equation $x^{2}-25=0$.
$$x^{2}-25=0$$

STEP 6

olving the equation gives us two solutions $x=5$ and $x=-5$. Therefore, for the function $\frac{q(x)}{h(x)}$, we must exclude these two values.

STEP 7

Combining the domains of $q(x)$ and $h(x)$, we find that the domain of $\frac{q(x)}{h(x)}$ is $x >0$ and $x \neq5, -5$.

STEP 8

Now, let's find the domain of the second function $q(h(x))$. The domain of this function is all x-values for which $h(x)$ is defined and $h(x) >0$.

STEP 9

As we found earlier, $h(x)$ is always defined for all real numbers. However, for the function $q(h(x))$, we must exclude values of $x$ for which $h(x) \leq$.

STEP 10

To find the values of $x$ for which $h(x) \leq0$, we solve the inequality $x^{2}-25 \leq0$.
$$x^{2}-25 \leq0$$

STEP 11

olving the inequality gives us the solution $-5 \leq x \leq5$. Therefore, for the function $q(h(x))$, we must exclude these values.

STEP 12

So, the domain of $q(h(x))$ is $x < -5$ or $x >5$, which can also be written as $|x| >5$.

STEP 13

Finally, let's find the domain of the third function $h(q(x))$. The domain of this function is all x-values for which $q(x)$ is defined.

As we found earlier, $q(x)$ is undefined when $x \leq0$. So, the domain of $h(q(x))$ is $x >0$.
In conclusion,
a. The domain of $\frac{q(x)}{h(x)}$ is $x >0$ and $x \neq, -$
b. The domain of $q(h(x))$ is $|x| >$
c. The domain of $h(q(x))$ is $x >0$