https://file.aiquickdraw.com/mathbot/file/1738593138750gne6r5by.jpg

The given function is:

\[
f(x) =
\begin{cases}
1 - (x - 1)^2, & x \leq 2 \\
\sqrt{x - 2}, & x > 2
\end{cases}
\]

Now, let's evaluate the function at the given values:

1. **f(0):**  
   Since \( 0 \leq 2 \), we use \( f(x) = 1 - (x - 1)^2 \):

   \[
   f(0) = 1 - (0 - 1)^2 = 1 - 1 = 0
   \]

2. **f(1):**  
   Since \( 1 \leq 2 \), we use \( f(x) = 1 - (x - 1)^2 \):

   \[
   f(1) = 1 - (1 - 1)^2 = 1 - 0 = 1
   \]

3. **f(2):**  
   Since \( 2 \leq 2 \), we use \( f(x) = 1 - (x - 1)^2 \):

   \[
   f(2) = 1 - (2 - 1)^2 = 1 - 1 = 0
   \]

4. **f(3):**  
   Since \( 3 > 2 \), we use \( f(x) = \sqrt{x - 2} \):

   \[
   f(3) = \sqrt{3 - 2} = \sqrt{1} = 1
   \]

### Final Answers:
\[
f(0) = 0, \quad f(1) = 1, \quad f(2) = 0, \quad f(3) = 1
\]

Do you need any clarification or details?  

### Related Questions:
1. How do you determine which piece of a piecewise function to use?
2. What happens if an input value is on the boundary of two function definitions?
3. How would the function change if \( f(x) \) was defined differently for \( x > 2 \)?
4. Can piecewise functions be continuous? How do you check for continuity at \( x = 2 \)?
5. What is the domain and range of this function?

### Tip:
Always check the domain conditions carefully before substituting values into a piecewise function!

Using the function f(x) given above, find the following function values: f(0), f(1), f(2), f(3).

Learn how to evaluate piecewise functions like f(x) = 1 - (x - 1)^2 for x ≤ 2 and f(x) = √(x - 2) for x > 2. This guide explains the evaluation process with detailed solutions for f(0), f(1), f(2), and f(3). Suitable for students in Grades 9-11.

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