Westonci.ca is the premier destination for reliable answers to your questions, provided by a community of experts. Explore thousands of questions and answers from a knowledgeable community of experts ready to help you find solutions. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
Certainly! Let's explore the process step-by-step for restricting the domain of the function [tex]\( f(x) = (x + 4)^2 - 6 \)[/tex] and finding its inverse for each restricted domain.
### Step 1: Understanding the Function
We start with the function:
[tex]\[ f(x) = (x + 4)^2 - 6 \][/tex]
This is a quadratic function, which is not one-to-one because it is symmetric around its vertex. To find the inverse, we need to restrict the domain so that it becomes one-to-one.
### Step 2: Finding the Inverse for Different Domains
#### 1. Restricted Domain: [tex]\( x \leq -4 \)[/tex]
First, we need to restrict the domain to [tex]\( x \leq -4 \)[/tex] to make the function one-to-one on this interval.
1. [tex]\[ x \leq -4 \][/tex]
2. Solve for the inverse:
[tex]\[ y = (x + 4)^2 - 6 \][/tex]
First, solve for [tex]\(x\)[/tex]:
[tex]\[ y + 6 = (x + 4)^2 \][/tex]
[tex]\[ \sqrt{y + 6} = x + 4 \][/tex]
Since [tex]\( x \leq -4 \)[/tex], we take the negative square root to solve for [tex]\(x\)[/tex]:
[tex]\[ x = -\sqrt{y + 6} - 4 \][/tex]
3. Therefore, the inverse function is:
[tex]\[ f^{-1}(x) = -\sqrt{x + 6} - 4 \text{ for } x \leq -4 \][/tex]
#### 2. Restricted Domain: [tex]\( x \geq -4 \)[/tex]
Second, we need to restrict the domain to [tex]\( x \geq -4 \)[/tex] to make the function one-to-one on this interval.
1. [tex]\[ x \geq -4 \][/tex]
2. Solve for the inverse:
[tex]\[ y = (x + 4)^2 - 6 \][/tex]
First, solve for [tex]\(x\)[/tex]:
[tex]\[ y + 6 = (x + 4)^2 \][/tex]
[tex]\[ \sqrt{y + 6} = x + 4 \][/tex]
Since [tex]\( x \geq -4 \)[/tex], we take the positive square root to solve for [tex]\(x\)[/tex]:
[tex]\[ x = \sqrt{y + 6} - 4 \][/tex]
3. Therefore, the inverse function is:
[tex]\[ f^{-1}(x) = \sqrt{x + 6} - 4 \text{ for } x \geq -4 \][/tex]
#### 3. Restricted Domain: [tex]\( x \leq 4 \)[/tex]
Third, we need to restrict the domain to [tex]\( x \leq 4 \)[/tex] to make the function one-to-one on this interval.
1. [tex]\[ x \leq 4 \][/tex]
2. Solve for the inverse:
[tex]\[ y = (x + 4)^2 - 6 \][/tex]
First, solve for [tex]\(x\)[/tex]:
[tex]\[ y + 6 = (x + 4)^2 \][/tex]
[tex]\[ \sqrt{y + 6} = x + 4 \][/tex]
Since [tex]\( x \leq 4 \)[/tex], we take the square root to solve for [tex]\(x\)[/tex]:
[tex]\[ x = \sqrt{y - 4} + 6 \][/tex]
3. Therefore, the inverse function is:
[tex]\[ f^{-1}(x) = \sqrt{x - 4} + 6 \text{ for } x \leq 4 \][/tex]
#### 4. Restricted Domain: [tex]\( x \geq -6 \)[/tex]
Finally, we need to restrict the domain to [tex]\( x \geq -6 \)[/tex] to make the function one-to-one on this interval.
1. [tex]\[ x \geq -6 \][/tex]
2. Solve for the inverse:
[tex]\[ y = (x + 4)^2 - 6 \][/tex]
First, solve for [tex]\(x\)[/tex]:
[tex]\[ y + 6 = (x + 4)^2 \][/tex]
[tex]\[ \sqrt{y + 6} = x + 4 \][/tex]
Since [tex]\( x \geq -6 \)[/tex], we again take the positive square root to solve for [tex]\(x\)[/tex]:
[tex]\[ x = \sqrt{y + 6} - 4 \][/tex]
3. Therefore, the inverse function is:
[tex]\[ f^{-1}(x) = \sqrt{x + 6} - 4 \text{ for } x \geq -6 \][/tex]
### Summary:
1. Restricted domain [tex]\( x \leq -4 \)[/tex]: [tex]\( f^{-1}(x) = -\sqrt{x+6} - 4 \)[/tex]
2. Restricted domain [tex]\( x \geq -4 \)[/tex]: [tex]\( f^{-1}(x) = \sqrt{x+6} - 4 \)[/tex]
3. Restricted domain [tex]\( x \leq 4 \)[/tex]: [tex]\( f^{-1}(x) = \sqrt{x-4} + 6 \)[/tex]
4. Restricted domain [tex]\( x \geq -6 \)[/tex]: [tex]\( f^{-1}(x) = \sqrt{x+6} - 4 \)[/tex]
These are the appropriate inverses for the function [tex]\( f(x) = (x + 4)^2 - 6 \)[/tex] given the specified restricted domains.
### Step 1: Understanding the Function
We start with the function:
[tex]\[ f(x) = (x + 4)^2 - 6 \][/tex]
This is a quadratic function, which is not one-to-one because it is symmetric around its vertex. To find the inverse, we need to restrict the domain so that it becomes one-to-one.
### Step 2: Finding the Inverse for Different Domains
#### 1. Restricted Domain: [tex]\( x \leq -4 \)[/tex]
First, we need to restrict the domain to [tex]\( x \leq -4 \)[/tex] to make the function one-to-one on this interval.
1. [tex]\[ x \leq -4 \][/tex]
2. Solve for the inverse:
[tex]\[ y = (x + 4)^2 - 6 \][/tex]
First, solve for [tex]\(x\)[/tex]:
[tex]\[ y + 6 = (x + 4)^2 \][/tex]
[tex]\[ \sqrt{y + 6} = x + 4 \][/tex]
Since [tex]\( x \leq -4 \)[/tex], we take the negative square root to solve for [tex]\(x\)[/tex]:
[tex]\[ x = -\sqrt{y + 6} - 4 \][/tex]
3. Therefore, the inverse function is:
[tex]\[ f^{-1}(x) = -\sqrt{x + 6} - 4 \text{ for } x \leq -4 \][/tex]
#### 2. Restricted Domain: [tex]\( x \geq -4 \)[/tex]
Second, we need to restrict the domain to [tex]\( x \geq -4 \)[/tex] to make the function one-to-one on this interval.
1. [tex]\[ x \geq -4 \][/tex]
2. Solve for the inverse:
[tex]\[ y = (x + 4)^2 - 6 \][/tex]
First, solve for [tex]\(x\)[/tex]:
[tex]\[ y + 6 = (x + 4)^2 \][/tex]
[tex]\[ \sqrt{y + 6} = x + 4 \][/tex]
Since [tex]\( x \geq -4 \)[/tex], we take the positive square root to solve for [tex]\(x\)[/tex]:
[tex]\[ x = \sqrt{y + 6} - 4 \][/tex]
3. Therefore, the inverse function is:
[tex]\[ f^{-1}(x) = \sqrt{x + 6} - 4 \text{ for } x \geq -4 \][/tex]
#### 3. Restricted Domain: [tex]\( x \leq 4 \)[/tex]
Third, we need to restrict the domain to [tex]\( x \leq 4 \)[/tex] to make the function one-to-one on this interval.
1. [tex]\[ x \leq 4 \][/tex]
2. Solve for the inverse:
[tex]\[ y = (x + 4)^2 - 6 \][/tex]
First, solve for [tex]\(x\)[/tex]:
[tex]\[ y + 6 = (x + 4)^2 \][/tex]
[tex]\[ \sqrt{y + 6} = x + 4 \][/tex]
Since [tex]\( x \leq 4 \)[/tex], we take the square root to solve for [tex]\(x\)[/tex]:
[tex]\[ x = \sqrt{y - 4} + 6 \][/tex]
3. Therefore, the inverse function is:
[tex]\[ f^{-1}(x) = \sqrt{x - 4} + 6 \text{ for } x \leq 4 \][/tex]
#### 4. Restricted Domain: [tex]\( x \geq -6 \)[/tex]
Finally, we need to restrict the domain to [tex]\( x \geq -6 \)[/tex] to make the function one-to-one on this interval.
1. [tex]\[ x \geq -6 \][/tex]
2. Solve for the inverse:
[tex]\[ y = (x + 4)^2 - 6 \][/tex]
First, solve for [tex]\(x\)[/tex]:
[tex]\[ y + 6 = (x + 4)^2 \][/tex]
[tex]\[ \sqrt{y + 6} = x + 4 \][/tex]
Since [tex]\( x \geq -6 \)[/tex], we again take the positive square root to solve for [tex]\(x\)[/tex]:
[tex]\[ x = \sqrt{y + 6} - 4 \][/tex]
3. Therefore, the inverse function is:
[tex]\[ f^{-1}(x) = \sqrt{x + 6} - 4 \text{ for } x \geq -6 \][/tex]
### Summary:
1. Restricted domain [tex]\( x \leq -4 \)[/tex]: [tex]\( f^{-1}(x) = -\sqrt{x+6} - 4 \)[/tex]
2. Restricted domain [tex]\( x \geq -4 \)[/tex]: [tex]\( f^{-1}(x) = \sqrt{x+6} - 4 \)[/tex]
3. Restricted domain [tex]\( x \leq 4 \)[/tex]: [tex]\( f^{-1}(x) = \sqrt{x-4} + 6 \)[/tex]
4. Restricted domain [tex]\( x \geq -6 \)[/tex]: [tex]\( f^{-1}(x) = \sqrt{x+6} - 4 \)[/tex]
These are the appropriate inverses for the function [tex]\( f(x) = (x + 4)^2 - 6 \)[/tex] given the specified restricted domains.
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.
