Answer :
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.
