To solve the given problem, we need to use the piecewise definition of the function [tex]\( f(x) \)[/tex], which is:
[tex]\[ f(x)=\left\{
\begin{array}{ll}
-3x - 3, & \text{if } x < -1 \\
x^2 + 2x - 1, & \text{if } x \geq -1
\end{array}
\right. \][/tex]
We will evaluate the function for the given values of [tex]\( x \)[/tex].
### (a) [tex]\( f(-2) \)[/tex]
Given that [tex]\( x = -2 \)[/tex], we determine which branch of the piecewise function to use. Since [tex]\( -2 < -1 \)[/tex], we use the first part of the function:
[tex]\[ f(x) = -3x - 3 \][/tex]
Substitute [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = -3(-2) - 3 \][/tex]
[tex]\[ f(-2) = 6 - 3 \][/tex]
[tex]\[ f(-2) = 3 \][/tex]
So, [tex]\( f(-2) = 3 \)[/tex].
### (b) [tex]\( f(-1) \)[/tex]
Given that [tex]\( x = -1 \)[/tex], we determine which branch of the piecewise function to use. Since [tex]\( -1 \geq -1 \)[/tex], we use the second part of the function:
[tex]\[ f(x) = x^2 + 2x - 1 \][/tex]
Substitute [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = (-1)^2 + 2(-1) - 1 \][/tex]
[tex]\[ f(-1) = 1 - 2 - 1 \][/tex]
[tex]\[ f(-1) = -2 \][/tex]
So, [tex]\( f(-1) = -2 \)[/tex].
### (c) [tex]\( f(1) \)[/tex]
Given that [tex]\( x = 1 \)[/tex], we determine which branch of the piecewise function to use. Since [tex]\( 1 \geq -1 \)[/tex], we use the second part of the function:
[tex]\[ f(x) = x^2 + 2x - 1 \][/tex]
Substitute [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 1^2 + 2(1) - 1 \][/tex]
[tex]\[ f(1) = 1 + 2 - 1 \][/tex]
[tex]\[ f(1) = 2 \][/tex]
So, [tex]\( f(1) = 2 \)[/tex].
### Summary
- [tex]\( f(-2) = 3 \)[/tex]
- [tex]\( f(-1) = -2 \)[/tex]
- [tex]\( f(1) = 2 \)[/tex]
These results provide the values for the function [tex]\( f(x) \)[/tex] at [tex]\( x = -2 \)[/tex], [tex]\( x = -1 \)[/tex], and [tex]\( x = 1 \)[/tex] respectively.
