To find the formula for the inverse function $f^{-1}(x)$ of the function $f(x) = \sqrt{x - 5}$, we can follow these steps:

1. Express the function $f(x)$:

$f(x) = \sqrt{x - 5}$

2. Replace $f(x)$ with $y$:

$y = \sqrt{x - 5}$

3. Swap $x$ and $y$ to find the inverse:

$x = \sqrt{y - 5}$

4. Solve for $y$ to find $f^{-1}(x)$:

First, square both sides to eliminate the square root: $x^2 = y - 5$ Now, solve for $y$: $y = x^2 + 5$

So, the inverse function is: $f^{-1}(x) = x^2 + 5$

5. Determine the domain of $f^{-1}(x)$:

Since $f(x) = \sqrt{x - 5}$ requires $x - 5 \geq 0$, the domain of $f(x)$ is $x \geq 5$. This restriction is inherited by the range of the inverse function $f^{-1}(x)$, meaning $f^{-1}(x)$ can only output values $y \geq 5$.

The domain of $f^{-1}(x) = x^2 + 5$ is all real numbers $\mathbb{R}$, but due to the range restriction inherited from $f(x)$, the output of $f^{-1}(x)$ must be greater than or equal to 5.

Final Answer:

• The inverse function is $f^{-1}(x) = x^2 + 5$.
• The domain of $f^{-1}(x)$ is all real numbers $\mathbb{R}$, but $f^{-1}(x)$ produces values in the range $[5, \infty)$.

Would you like more details or have any questions?

Related Questions:

1. How does the graph of $f(x) = \sqrt{x - 5}$ compare to its inverse $f^{-1}(x)$?
2. Can the inverse function be restricted further based on additional conditions?
3. What is the general process for finding the inverse of radical functions?
4. How do you find the domain of more complex inverse functions?
5. How does the composition of a function and its inverse demonstrate that they are inverses?

Tip:

Always pay attention to the domain restrictions of the original function when finding the inverse, as these restrictions will affect the range of the inverse function.