Let's evaluate each statement one by one, using the given piecewise function:
[tex]\[
f(x) = \begin{cases}
-x + 1 & \text{if } x < 0 \\
-2 & \text{if } x = 0 \\
x^2 - 1 & \text{if } x > 0
\end{cases}
\][/tex]
### Statement A: \( f(-1) = 2 \)
For \( x = -1 \):
- Since \(-1 < 0\), we use the first case: \( f(x) = -x + 1 \)
- Calculate \( f(-1) \):
[tex]\[
f(-1) = -(-1) + 1 = 1 + 1 = 2
\][/tex]
So, \( f(-1) = 2 \) is true.
### Statement B: \( f(4) = 7 \)
For \( x = 4 \):
- Since \( 4 > 0 \), we use the third case: \( f(x) = x^2 - 1 \)
- Calculate \( f(4) \):
[tex]\[
f(4) = 4^2 - 1 = 16 - 1 = 15
\][/tex]
So, \( f(4) = 7 \) is false. The correct value is \( f(4) = 15 \).
### Statement C: \( f(1) = 0 \)
For \( x = 1 \):
- Since \( 1 > 0 \), we use the third case: \( f(x) = x^2 - 1 \)
- Calculate \( f(1) \):
[tex]\[
f(1) = 1^2 - 1 = 1 - 1 = 0
\][/tex]
So, \( f(1) = 0 \) is true.
### Statement D: \( f(-2) = 0 \)
For \( x = -2 \):
- Since \(-2 < 0\), we use the first case: \( f(x) = -x + 1 \)
- Calculate \( f(-2) \):
[tex]\[
f(-2) = -(-2) + 1 = 2 + 1 = 3
\][/tex]
So, \( f(-2) = 0 \) is false. The correct value is \( f(-2) = 3 \).
### Summary
- Statement A: \( f(-1) = 2 \) is true.
- Statement B: \( f(4) = 7 \) is false (correct value is 15).
- Statement C: \( f(1) = 0 \) is true.
- Statement D: \( f(-2) = 0 \) is false (correct value is 3).
Therefore, the true statements are A and C.
