QuestionGiven functions $q(x)=\frac{1}{\sqrt{x}}$ and $h(x)=x^{2}-25$ :
a. Domain of $\frac{q(x)}{h(x)}$ : $x>0, x \neq 5$
b. Domain of $q(h(x))$ : $|x|$ [Select] [Select]
c. Domain of $h(q(x))$ : [Select] [Select]
Please clarify parts b and c for completeness.

Studdy Solution

STEP 1

Assumptions1. We have two functions $q(x)=\frac{1}{\sqrt{x}}$ and $h(x)=x^{}-25$ . The domain of a function is the set of all possible input values (x-values) which will produce a valid output (y-values).
3. The domain of $\frac{q(x)}{h(x)}$ is given as $x>0, x \neq5$
4. We need to find the domains of $q(h(x))$ and $h(q(x))$ .

STEP 2

First, let's find the domain of $q(h(x))$ . The function $q(x)$ can only take positive values of x because of the square root in the denominator. Therefore, the inside function $h(x)$ must be greater than0.
$h(x) >0$

STEP 3

Now, plug in the given function $h(x)$ into the inequality.
$x^{2}-25 >0$

STEP 4

olving this inequality will give us the domain of $q(h(x))$ . To solve it, we can factor the quadratic expression.
$(x-)(x+) >0$

STEP 5

The solutions to the equation $(x-5)(x+5) =0$ are $x =5$ and $x = -5$ . These are the points where the sign of the inequality can change. We can test the intervals $(-\infty, -5)$ , $(-5,5)$ , and $(5, \infty)$ to determine where the inequality is true.

STEP 6

Testing the intervals, we find that the inequality is true for $x < -5$ and $x >5$ . Therefore, the domain of $q(h(x))$ is $x < -5$ or $x >5$ .

STEP 7

Now, let's find the domain of $h(q(x))$ . The function $h(x)$ can take any real number as input. However, the inside function $q(x)$ can only take positive values of x because of the square root in the denominator. Therefore, the domain of $h(q(x))$ is $x >0$ .

STEP 8

To summarize, the domains of the functions area. Domain of $\frac{q(x)}{h(x)}$ is $x>0, x \neq5$ b. Domain of $q(h(x))$ is $x < -5$ or $x >5$ c. Domain of $h(q(x))$ is $x >0$

STEP 9

For part c, the complete equation for $h(q(x))$ is $h(q(x)) = \left(\frac{}{\sqrt{x}}\right)^{2}-25 = \frac{}{x} -25$

STEP 10

Regarding the extracted text for parts b and c, there seems to be an error. The domain of $q(h(x))$ is not $|x|$ but $x < -5$ or $x >5$ . And for $h(q(x))$ , the domain is not [ Select ] [ Select ] but $x >0$ .

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QuestionGiven functions $q(x)=\frac{1}{\sqrt{x}}$ and $h(x)=x^{2}-25$ :
a. Domain of $\frac{q(x)}{h(x)}$ : $x>0, x \neq 5$
b. Domain of $q(h(x))$ : $|x|$ [Select] [Select]
c. Domain of $h(q(x))$ : [Select] [Select]
Please clarify parts b and c for completeness.

Studdy Solution

STEP 1

STEP 2

First, let's find the domain of $q(h(x))$ . The function $q(x)$ can only take positive values of x because of the square root in the denominator. Therefore, the inside function $h(x)$ must be greater than0.
$h(x) >0$

STEP 3

Now, plug in the given function $h(x)$ into the inequality.
$x^{2}-25 >0$

STEP 4

olving this inequality will give us the domain of $q(h(x))$ . To solve it, we can factor the quadratic expression.
$(x-)(x+) >0$

STEP 5

The solutions to the equation $(x-5)(x+5) =0$ are $x =5$ and $x = -5$ . These are the points where the sign of the inequality can change. We can test the intervals $(-\infty, -5)$ , $(-5,5)$ , and $(5, \infty)$ to determine where the inequality is true.

STEP 6

Testing the intervals, we find that the inequality is true for $x < -5$ and $x >5$ . Therefore, the domain of $q(h(x))$ is $x < -5$ or $x >5$ .

STEP 7

Now, let's find the domain of $h(q(x))$ . The function $h(x)$ can take any real number as input. However, the inside function $q(x)$ can only take positive values of x because of the square root in the denominator. Therefore, the domain of $h(q(x))$ is $x >0$ .

STEP 8

To summarize, the domains of the functions area. Domain of $\frac{q(x)}{h(x)}$ is $x>0, x \neq5$ b. Domain of $q(h(x))$ is $x < -5$ or $x >5$ c. Domain of $h(q(x))$ is $x >0$

STEP 9

For part c, the complete equation for $h(q(x))$ is $h(q(x)) = \left(\frac{}{\sqrt{x}}\right)^{2}-25 = \frac{}{x} -25$

STEP 10

Regarding the extracted text for parts b and c, there seems to be an error. The domain of $q(h(x))$ is not $|x|$ but $x < -5$ or $x >5$ . And for $h(q(x))$ , the domain is not [ Select ] [ Select ] but $x >0$ .

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Studdy Solution

STEP 1

STEP 2

First, let's find the domain of q(h(x))q(h(x))q(h(x)). The function q(x)q(x)q(x) can only take positive values of x because of the square root in the denominator. Therefore, the inside function h(x)h(x)h(x) must be greater than0.h(x)>0h(x) >0h(x)>0

STEP 3

Now, plug in the given function h(x)h(x)h(x) into the inequality.x2−25>0x^{2}-25 >0x2−25>0

STEP 4

olving this inequality will give us the domain of q(h(x))q(h(x))q(h(x)). To solve it, we can factor the quadratic expression.(x−)(x+)>0(x-)(x+) >0(x−)(x+)>0

STEP 5

STEP 6

Testing the intervals, we find that the inequality is true for x<−5x < -5x<−5 and x>5x >5x>5. Therefore, the domain of q(h(x))q(h(x))q(h(x)) is x<−5x < -5x<−5 or x>5x >5x>5.

STEP 7

Now, let's find the domain of h(q(x))h(q(x))h(q(x)). The function h(x)h(x)h(x) can take any real number as input. However, the inside function q(x)q(x)q(x) can only take positive values of x because of the square root in the denominator. Therefore, the domain of h(q(x))h(q(x))h(q(x)) is x>0x >0x>0.

STEP 8

To summarize, the domains of the functions area. Domain of q(x)h(x)\frac{q(x)}{h(x)}h(x)q(x)​ is x>0,x≠5x>0, x \neq5x>0,x=5 b. Domain of q(h(x))q(h(x))q(h(x)) is x<−5x < -5x<−5 or x>5x >5x>5 c. Domain of h(q(x))h(q(x))h(q(x)) is x>0x >0x>0

STEP 9

For part c, the complete equation for h(q(x))h(q(x))h(q(x)) ish(q(x))=(x)2−25=x−25h(q(x)) = \left(\frac{}{\sqrt{x}}\right)^{2}-25 = \frac{}{x} -25h(q(x))=(x​​)2−25=x​−25

STEP 10

Regarding the extracted text for parts b and c, there seems to be an error. The domain of q(h(x))q(h(x))q(h(x)) is not ∣x∣|x|∣x∣ but x<−5x < -5x<−5 or x>5x >5x>5. And for h(q(x))h(q(x))h(q(x)), the domain is not [ Select ] [ Select ] but x>0x >0x>0.

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First, let's find the domain of $q(h(x))$ . The function $q(x)$ can only take positive values of x because of the square root in the denominator. Therefore, the inside function $h(x)$ must be greater than0.
$h(x) >0$

Now, plug in the given function $h(x)$ into the inequality.
$x^{2}-25 >0$

olving this inequality will give us the domain of $q(h(x))$ . To solve it, we can factor the quadratic expression.
$(x-)(x+) >0$

Testing the intervals, we find that the inequality is true for $x < -5$ and $x >5$ . Therefore, the domain of $q(h(x))$ is $x < -5$ or $x >5$ .

Now, let's find the domain of $h(q(x))$ . The function $h(x)$ can take any real number as input. However, the inside function $q(x)$ can only take positive values of x because of the square root in the denominator. Therefore, the domain of $h(q(x))$ is $x >0$ .

To summarize, the domains of the functions area. Domain of $\frac{q(x)}{h(x)}$ is $x>0, x \neq5$ b. Domain of $q(h(x))$ is $x < -5$ or $x >5$ c. Domain of $h(q(x))$ is $x >0$

For part c, the complete equation for $h(q(x))$ is $h(q(x)) = \left(\frac{}{\sqrt{x}}\right)^{2}-25 = \frac{}{x} -25$

Regarding the extracted text for parts b and c, there seems to be an error. The domain of $q(h(x))$ is not $|x|$ but $x < -5$ or $x >5$ . And for $h(q(x))$ , the domain is not [ Select ] [ Select ] but $x >0$ .