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

\[
f(x) =
\begin{cases}
9 - 2x, & x < 0 \\
9, & 0 \leq x < 1 \\
8x + 1, & x \geq 1
\end{cases}
\]

We need to evaluate \( f(x) \) at the following values:

### (a) \( f(-4) \)

Since \( -4 < 0 \), we use the first case: \( f(x) = 9 - 2x \).

\[
f(-4) = 9 - 2(-4) = 9 + 8 = 17
\]

### (b) \( f(0) \)

Since \( 0 \leq x < 1 \), we use the second case: \( f(x) = 9 \).

\[
f(0) = 9
\]

### (c) \( f(1) \)

Since \( x = 1 \geq 1 \), we use the third case: \( f(x) = 8x + 1 \).

\[
f(1) = 8(1) + 1 = 8 + 1 = 9
\]

### Final Answers:
- (a) \( f(-4) = 17 \)
- (b) \( f(0) = 9 \)
- (c) \( f(1) = 9 \)

Would you like a detailed explanation of any step? 

Here are some related questions for practice:
1. What would \( f(x) \) be for \( x = 2 \)?
2. How would the value change for \( f(-1) \)?
3. What happens if \( x = 0.5 \)?
4. Can you define \( f(x) \) for all real numbers based on this function?
5. How does the slope of the function change in each piece?

**Tip:** Always check which part of the piecewise function applies based on the input range.

Evaluate the function f(x) at each specified value of the independent variable and simplify: (a) f(-4), (b) f(0), (c) f(1).

Learn how to evaluate a piecewise function at specific values of x, including f(-4), f(0), and f(1). This problem involves understanding different cases of a piecewise function, using formulas like f(x) = 9 - 2x, f(x) = 9, and f(x) = 8x + 1. Suitable for high school students in Grades 10-12.

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