Answer :
To determine the range of the inverse of the given function \( f(x) = \sqrt{x-2} \), let's start by understanding the original function and finding its inverse.
1. Original Function:
The function given is \( f(x) = \sqrt{x-2} \).
2. Domain of the Original Function:
Since the square root function is only defined for non-negative numbers, the expression inside the square root must be greater than or equal to zero:
[tex]\[ x - 2 \geq 0 \implies x \geq 2 \][/tex]
Therefore, the domain of \( f(x) \) is \( [2, \infty) \).
3. Finding the Inverse:
We now find the inverse of \( f(x) \). Let \( y = f(x) \), so
[tex]\[ y = \sqrt{x - 2} \][/tex]
To solve for \( x \) in terms of \( y \), we first square both sides of the equation:
[tex]\[ y^2 = x - 2 \][/tex]
Then, add 2 to both sides:
[tex]\[ x = y^2 + 2 \][/tex]
So, the inverse function \( f^{-1}(y) \) is:
[tex]\[ f^{-1}(y) = y^2 + 2 \][/tex]
4. Domain of the Inverse Function \( f^{-1}(y) \):
Next, we determine the domain of the inverse function \( y^2 + 2 \). Since the square root function \( f(x) \) must be non-negative (as square roots are only defined for non-negative numbers), \( y \) (which is \( \sqrt{x - 2} \)) must be in the range \( y \geq 0 \).
5. Range of the Inverse Function:
The range of the original function \( f(x) \) is the set of \( y \)-values that \( f(x) \) can take, which is \( [0, \infty) \) because \( f(x) \) is a square root function starting from 0.
Consequently, for the inverse function \( f^{-1}(y) = y^2 + 2 \):
- When \( y = 0 \), \( f^{-1}(0) = 0^2 + 2 = 2 \)
- As \( y \) increases beyond 0, \( y^2 + 2 \) increases without bound.
Therefore, the range of the inverse function \( f^{-1}(y) \) is \([2, \infty)\).
[tex]\[ \boxed{[2, \infty]} \][/tex]
1. Original Function:
The function given is \( f(x) = \sqrt{x-2} \).
2. Domain of the Original Function:
Since the square root function is only defined for non-negative numbers, the expression inside the square root must be greater than or equal to zero:
[tex]\[ x - 2 \geq 0 \implies x \geq 2 \][/tex]
Therefore, the domain of \( f(x) \) is \( [2, \infty) \).
3. Finding the Inverse:
We now find the inverse of \( f(x) \). Let \( y = f(x) \), so
[tex]\[ y = \sqrt{x - 2} \][/tex]
To solve for \( x \) in terms of \( y \), we first square both sides of the equation:
[tex]\[ y^2 = x - 2 \][/tex]
Then, add 2 to both sides:
[tex]\[ x = y^2 + 2 \][/tex]
So, the inverse function \( f^{-1}(y) \) is:
[tex]\[ f^{-1}(y) = y^2 + 2 \][/tex]
4. Domain of the Inverse Function \( f^{-1}(y) \):
Next, we determine the domain of the inverse function \( y^2 + 2 \). Since the square root function \( f(x) \) must be non-negative (as square roots are only defined for non-negative numbers), \( y \) (which is \( \sqrt{x - 2} \)) must be in the range \( y \geq 0 \).
5. Range of the Inverse Function:
The range of the original function \( f(x) \) is the set of \( y \)-values that \( f(x) \) can take, which is \( [0, \infty) \) because \( f(x) \) is a square root function starting from 0.
Consequently, for the inverse function \( f^{-1}(y) = y^2 + 2 \):
- When \( y = 0 \), \( f^{-1}(0) = 0^2 + 2 = 2 \)
- As \( y \) increases beyond 0, \( y^2 + 2 \) increases without bound.
Therefore, the range of the inverse function \( f^{-1}(y) \) is \([2, \infty)\).
[tex]\[ \boxed{[2, \infty]} \][/tex]
