Math / Algebra

QuestionFind the domain of the function. $g(x) = \frac{\sqrt{x-2}}{x-10}$ What is the domain of $g$ ? (Type your answer in interval notation.)

Studdy Solution

STEP 1

What is this asking? We need to find all the valid inputs ( $x$ values) for the function $g(x)$ , meaning all the $x$ values where $g(x)$ produces a real number. Watch out! Remember, we can't divide by zero and we can't take the square root of a negative number!

STEP 2

1. Numerator analysis
2. Denominator analysis
3. Combine restrictions
4. Write in interval notation

STEP 3

Let's look at the numerator: $\sqrt{x-2}$ .
Because we can't take the square root of a negative number, the expression inside the square root, $x-2$ , must be greater than or equal to zero.

STEP 4

So, we need $x-2 \ge 0$ .
To solve for $x$ , we can add $2$ to both sides of the inequality.
This gives us $x-2+2 \ge 0+2$ , which simplifies to $x \ge 2$ .
This is our first restriction!

STEP 5

Now, let's look at the denominator: $x-10$ .
Since we can't divide by zero, we must make sure that the denominator is not equal to zero.

STEP 6

So, we need $x-10 \ne 0$ .
Adding $10$ to both sides gives us $x-10+10 \ne 0+10$ , which simplifies to $x \ne 10$ .
This is our second restriction!

STEP 7

We have two restrictions: $x \ge 2$ and $x \ne 10$ .
Let's combine them!

STEP 8

The first restriction says $x$ must be greater than or equal to 2.
The second restriction says $x$ cannot be 10.
Since 10 is greater than 2, we need to exclude it from our allowed values.

STEP 9

We can write our combined restrictions in interval notation.
We start at 2 (including 2, so we use a square bracket) and go up to 10 (excluding 10, so we use a parenthesis).
Then, we continue from 10 (again excluding 10) to infinity (always excluding infinity).

STEP 10

This gives us the interval: $[2, 10) \cup (10, \infty)$ .
Awesome!

STEP 11

The domain of $g(x)$ is $[2, 10) \cup (10, \infty)$ .

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QuestionFind the domain of the function. $g(x) = \frac{\sqrt{x-2}}{x-10}$ What is the domain of $g$ ? (Type your answer in interval notation.)

Studdy Solution

STEP 1

What is this asking? We need to find all the valid inputs ( $x$ values) for the function $g(x)$ , meaning all the $x$ values where $g(x)$ produces a real number. Watch out! Remember, we can't divide by zero and we can't take the square root of a negative number!

STEP 2

1. Numerator analysis
2. Denominator analysis
3. Combine restrictions
4. Write in interval notation

STEP 3

Let's look at the numerator: $\sqrt{x-2}$ .
Because we can't take the square root of a negative number, the expression inside the square root, $x-2$ , must be greater than or equal to zero.

STEP 4

So, we need $x-2 \ge 0$ .
To solve for $x$ , we can add $2$ to both sides of the inequality.
This gives us $x-2+2 \ge 0+2$ , which simplifies to $x \ge 2$ .
This is our first restriction!

STEP 5

Now, let's look at the denominator: $x-10$ .
Since we can't divide by zero, we must make sure that the denominator is not equal to zero.

STEP 6

So, we need $x-10 \ne 0$ .
Adding $10$ to both sides gives us $x-10+10 \ne 0+10$ , which simplifies to $x \ne 10$ .
This is our second restriction!

STEP 7

We have two restrictions: $x \ge 2$ and $x \ne 10$ .
Let's combine them!

STEP 8

The first restriction says $x$ must be greater than or equal to 2.
The second restriction says $x$ cannot be 10.
Since 10 is greater than 2, we need to exclude it from our allowed values.

STEP 9

We can write our combined restrictions in interval notation.
We start at 2 (including 2, so we use a square bracket) and go up to 10 (excluding 10, so we use a parenthesis).
Then, we continue from 10 (again excluding 10) to infinity (always excluding infinity).

STEP 10

This gives us the interval: $[2, 10) \cup (10, \infty)$ .
Awesome!

STEP 11

The domain of $g(x)$ is $[2, 10) \cup (10, \infty)$ .

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QuestionFind the domain of the function. g(x)=x−2x−10g(x) = \frac{\sqrt{x-2}}{x-10}g(x)=x−10x−2​​ What is the domain of ggg? (Type your answer in interval notation.)

Studdy Solution

STEP 1

What is this asking? We need to find all the *valid* inputs (xxx values) for the function g(x)g(x)g(x), meaning all the xxx values where g(x)g(x)g(x) produces a real number. Watch out! Remember, we can't divide by **zero** and we can't take the square root of a **negative** number!

STEP 2

1. Numerator analysis2. Denominator analysis3. Combine restrictions4. Write in interval notation

STEP 3

Let's look at the numerator: x−2\sqrt{x-2}x−2​.Because we can't take the square root of a negative number, the expression inside the square root, x−2x-2x−2, must be greater than or equal to zero.

STEP 4

So, we need x−2≥0x-2 \ge 0x−2≥0.To solve for xxx, we can add 222 to both sides of the inequality.This gives us x−2+2≥0+2x-2+2 \ge 0+2x−2+2≥0+2, which simplifies to x≥2x \ge 2x≥2.This is our **first restriction**!

STEP 5

Now, let's look at the denominator: x−10x-10x−10.Since we can't divide by zero, we *must* make sure that the denominator is not equal to zero.

STEP 6

So, we need x−10≠0x-10 \ne 0x−10=0.Adding 101010 to both sides gives us x−10+10≠0+10x-10+10 \ne 0+10x−10+10=0+10, which simplifies to x≠10x \ne 10x=10.This is our **second restriction**!

STEP 7

We have two restrictions: x≥2x \ge 2x≥2 and x≠10x \ne 10x=10.Let's combine them!

STEP 8

The first restriction says xxx must be greater than or equal to **2**.The second restriction says xxx cannot be **10**.Since **10** is greater than **2**, we need to *exclude* it from our allowed values.

STEP 9

We can write our combined restrictions in interval notation.We start at **2** (including **2**, so we use a square bracket) and go up to **10** (excluding **10**, so we use a parenthesis).Then, we continue from **10** (again excluding **10**) to infinity (always excluding infinity).

STEP 10

This gives us the interval: [2,10)∪(10,∞)[2, 10) \cup (10, \infty)[2,10)∪(10,∞).Awesome!

STEP 11

The domain of g(x)g(x)g(x) is [2,10)∪(10,∞)[2, 10) \cup (10, \infty)[2,10)∪(10,∞).

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Start learning now

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QuestionFind the domain of the function. $g(x) = \frac{\sqrt{x-2}}{x-10}$ What is the domain of $g$ ? (Type your answer in interval notation.)

What is this asking? We need to find all the valid inputs ( $x$ values) for the function $g(x)$ , meaning all the $x$ values where $g(x)$ produces a real number. Watch out! Remember, we can't divide by zero and we can't take the square root of a negative number!

1. Numerator analysis
2. Denominator analysis
3. Combine restrictions
4. Write in interval notation

Let's look at the numerator: $\sqrt{x-2}$ .
Because we can't take the square root of a negative number, the expression inside the square root, $x-2$ , must be greater than or equal to zero.

So, we need $x-2 \ge 0$ .
To solve for $x$ , we can add $2$ to both sides of the inequality.
This gives us $x-2+2 \ge 0+2$ , which simplifies to $x \ge 2$ .
This is our first restriction!

Now, let's look at the denominator: $x-10$ .
Since we can't divide by zero, we must make sure that the denominator is not equal to zero.

So, we need $x-10 \ne 0$ .
Adding $10$ to both sides gives us $x-10+10 \ne 0+10$ , which simplifies to $x \ne 10$ .
This is our second restriction!

We have two restrictions: $x \ge 2$ and $x \ne 10$ .
Let's combine them!

The first restriction says $x$ must be greater than or equal to 2.
The second restriction says $x$ cannot be 10.
Since 10 is greater than 2, we need to exclude it from our allowed values.

We can write our combined restrictions in interval notation.
We start at 2 (including 2, so we use a square bracket) and go up to 10 (excluding 10, so we use a parenthesis).
Then, we continue from 10 (again excluding 10) to infinity (always excluding infinity).

This gives us the interval: $[2, 10) \cup (10, \infty)$ .
Awesome!

The domain of $g(x)$ is $[2, 10) \cup (10, \infty)$ .