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The piecewise function \( f(x) \) is defined as follows:

\[
f(x) = 
\begin{cases} 
x^2 - 1 & \text{if } x \leq 0 \\
2x - 1 & \text{if } 0 < x \leq 5 \\
3 & \text{if } x > 5 
\end{cases}
\]

We are tasked with answering the following:

### 1. Is \( f(x) \) a function?
To determine whether this is a function, we check if each input \( x \) gives exactly one output \( f(x) \). Since for each range of \( x \), the function gives a single value, \( f(x) \) **is a function**.

### 2. Evaluate \( f(-2) \), \( f(0) \), and \( f(5) \)

- For \( f(-2) \):
  Since \( -2 \leq 0 \), use the first piece: 
  \[
  f(-2) = (-2)^2 - 1 = 4 - 1 = 3
  \]

- For \( f(0) \):
  Since \( 0 \leq 0 \), use the first piece again: 
  \[
  f(0) = (0)^2 - 1 = 0 - 1 = -1
  \]

- For \( f(5) \):
  Since \( 0 < 5 \leq 5 \), use the second piece:
  \[
  f(5) = 2(5) - 1 = 10 - 1 = 9
  \]

### Summary:
- \( f(-2) = 3 \)
- \( f(0) = -1 \)
- \( f(5) = 9 \)

Would you like further clarification, or help graphing the function?

Here are some related questions:
1. How would you plot the piecewise function on the given grid?
2. What happens to \( f(x) \) for values like \( x = 6 \) or \( x = -3 \)?
3. How do you find the domain and range of \( f(x) \)?
4. What is the continuity of \( f(x) \) at \( x = 0 \) and \( x = 5 \)?
5. How would the function change if the second piece was \( 2x + 1 \) instead of \( 2x - 1 \)?

**Tip:** When evaluating piecewise functions, always check which condition the input \( x \) satisfies to choose the correct expression.

The piecewise function f(x) is defined as follows:

f(x) = 
{
  x^2 - 1   if x <= 0,
  2x - 1   if 0 < x <= 5,
  3       if x > 5
}
Determine if it is a function. Additionally, evaluate f(-2), f(0), and f(5).

Learn how to evaluate a piecewise function f(x) with different expressions for different values of x. This example covers f(x) = x^2 - 1, f(x) = 2x - 1, and f(x) = 3, along with determining whether this is a valid function. Suitable for students in Grades 9-11.

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