Answer :
Let's solve the problem step by step.
### Given Function:
[tex]\[ h(x) = x^2 - 4 \quad \text{where } x \geq 0 \][/tex]
### Part (a): Find the inverse [tex]\( h^{-1}(x) \)[/tex]
To find the inverse [tex]\( h^{-1}(x) \)[/tex], we need to express [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
1. Start with the equation:
[tex]\[ y = x^2 - 4 \][/tex]
2. Solve for [tex]\( x \)[/tex]:
[tex]\[ y + 4 = x^2 \][/tex]
[tex]\[ x = \sqrt{y + 4} \][/tex]
Since [tex]\( x \geq 0 \)[/tex], we only consider the positive square root.
Therefore, the inverse function is:
[tex]\[ h^{-1}(x) = \sqrt{x + 4} \][/tex]
### Part (b): Find the attributes
#### 1. Range of [tex]\( h(x) \)[/tex]:
To find the range of [tex]\( h(x) = x^2 - 4 \)[/tex] for [tex]\( x \geq 0 \)[/tex]:
- Minimum value of [tex]\( x^2 \)[/tex] when [tex]\( x = 0 \)[/tex] is 0:
[tex]\[ h(0) = 0^2 - 4 = -4 \][/tex]
- As [tex]\( x \)[/tex] increases from 0 to [tex]\(\infty\)[/tex], [tex]\( x^2 \)[/tex] also increases from 0 to [tex]\(\infty\)[/tex]. Thus, [tex]\( x^2 - 4 \)[/tex] increases from [tex]\(-4\)[/tex] to [tex]\(\infty\)[/tex].
Therefore, the range of [tex]\( h(x) \)[/tex] is:
[tex]\[ [-4, \infty) \][/tex]
#### 2. Domain of [tex]\( h^{-1}(x) \)[/tex]:
The domain of the inverse function [tex]\( h^{-1}(x) = \sqrt{x + 4} \)[/tex]:
- The expression under the square root must be non-negative:
[tex]\[ x + 4 \geq 0 \][/tex]
[tex]\[ x \geq -4 \][/tex]
Therefore, the domain of [tex]\( h^{-1}(x) \)[/tex] is:
[tex]\[ [-4, \infty) \][/tex]
#### 3. Range of [tex]\( h^{-1}(x) \)[/tex]:
To find the range of [tex]\( h^{-1}(x) = \sqrt{x + 4} \)[/tex]:
- Since [tex]\( h^{-1}(x) \)[/tex] involves taking the positive square root, the smallest value of [tex]\( h^{-1}(x) \)[/tex] is when [tex]\( x + 4 = 0 \)[/tex], which gives:
[tex]\[ h^{-1}(-4) = \sqrt{0} = 0 \][/tex]
- As [tex]\( x \)[/tex] increases to [tex]\(\infty\)[/tex], [tex]\( \sqrt{x + 4} \)[/tex] also increases to [tex]\(\infty\)[/tex].
Therefore, the range of [tex]\( h^{-1}(x) \)[/tex] is:
[tex]\[ [0, \infty) \][/tex]
### Summary of Results
- The inverse function [tex]\( h^{-1}(x) = \sqrt{x + 4} \)[/tex]
- Range of [tex]\( h(x) \)[/tex]: [tex]\([-4, \infty)\)[/tex]
- Domain of [tex]\( h^{-1}(x) \)[/tex]: [tex]\([-4, \infty)\)[/tex]
- Range of [tex]\( h^{-1}(x) \)[/tex]: [tex]\([0, \infty)\)[/tex]
So the detailed answers are:
a. The inverse function [tex]\( h^{-1}(x) = \sqrt{x + 4} \)[/tex].
b.
- Range of [tex]\( h(x) \)[/tex]: [tex]\([-4, \infty)\)[/tex]
- Domain of [tex]\( h^{-1}(x) \)[/tex]: [tex]\([-4, \infty)\)[/tex]
- Range of [tex]\( h^{-1}(x) \)[/tex]: [tex]\([0, \infty)\)[/tex]
### Given Function:
[tex]\[ h(x) = x^2 - 4 \quad \text{where } x \geq 0 \][/tex]
### Part (a): Find the inverse [tex]\( h^{-1}(x) \)[/tex]
To find the inverse [tex]\( h^{-1}(x) \)[/tex], we need to express [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
1. Start with the equation:
[tex]\[ y = x^2 - 4 \][/tex]
2. Solve for [tex]\( x \)[/tex]:
[tex]\[ y + 4 = x^2 \][/tex]
[tex]\[ x = \sqrt{y + 4} \][/tex]
Since [tex]\( x \geq 0 \)[/tex], we only consider the positive square root.
Therefore, the inverse function is:
[tex]\[ h^{-1}(x) = \sqrt{x + 4} \][/tex]
### Part (b): Find the attributes
#### 1. Range of [tex]\( h(x) \)[/tex]:
To find the range of [tex]\( h(x) = x^2 - 4 \)[/tex] for [tex]\( x \geq 0 \)[/tex]:
- Minimum value of [tex]\( x^2 \)[/tex] when [tex]\( x = 0 \)[/tex] is 0:
[tex]\[ h(0) = 0^2 - 4 = -4 \][/tex]
- As [tex]\( x \)[/tex] increases from 0 to [tex]\(\infty\)[/tex], [tex]\( x^2 \)[/tex] also increases from 0 to [tex]\(\infty\)[/tex]. Thus, [tex]\( x^2 - 4 \)[/tex] increases from [tex]\(-4\)[/tex] to [tex]\(\infty\)[/tex].
Therefore, the range of [tex]\( h(x) \)[/tex] is:
[tex]\[ [-4, \infty) \][/tex]
#### 2. Domain of [tex]\( h^{-1}(x) \)[/tex]:
The domain of the inverse function [tex]\( h^{-1}(x) = \sqrt{x + 4} \)[/tex]:
- The expression under the square root must be non-negative:
[tex]\[ x + 4 \geq 0 \][/tex]
[tex]\[ x \geq -4 \][/tex]
Therefore, the domain of [tex]\( h^{-1}(x) \)[/tex] is:
[tex]\[ [-4, \infty) \][/tex]
#### 3. Range of [tex]\( h^{-1}(x) \)[/tex]:
To find the range of [tex]\( h^{-1}(x) = \sqrt{x + 4} \)[/tex]:
- Since [tex]\( h^{-1}(x) \)[/tex] involves taking the positive square root, the smallest value of [tex]\( h^{-1}(x) \)[/tex] is when [tex]\( x + 4 = 0 \)[/tex], which gives:
[tex]\[ h^{-1}(-4) = \sqrt{0} = 0 \][/tex]
- As [tex]\( x \)[/tex] increases to [tex]\(\infty\)[/tex], [tex]\( \sqrt{x + 4} \)[/tex] also increases to [tex]\(\infty\)[/tex].
Therefore, the range of [tex]\( h^{-1}(x) \)[/tex] is:
[tex]\[ [0, \infty) \][/tex]
### Summary of Results
- The inverse function [tex]\( h^{-1}(x) = \sqrt{x + 4} \)[/tex]
- Range of [tex]\( h(x) \)[/tex]: [tex]\([-4, \infty)\)[/tex]
- Domain of [tex]\( h^{-1}(x) \)[/tex]: [tex]\([-4, \infty)\)[/tex]
- Range of [tex]\( h^{-1}(x) \)[/tex]: [tex]\([0, \infty)\)[/tex]
So the detailed answers are:
a. The inverse function [tex]\( h^{-1}(x) = \sqrt{x + 4} \)[/tex].
b.
- Range of [tex]\( h(x) \)[/tex]: [tex]\([-4, \infty)\)[/tex]
- Domain of [tex]\( h^{-1}(x) \)[/tex]: [tex]\([-4, \infty)\)[/tex]
- Range of [tex]\( h^{-1}(x) \)[/tex]: [tex]\([0, \infty)\)[/tex]
