Let us evaluate the function [tex]\( f \)[/tex] at the given points, using its definition:
1. Finding [tex]\( f(-2) \)[/tex]:
The function [tex]\( f(x) \)[/tex] has three different cases. Since [tex]\( -2 \leq -1 \)[/tex], we use the first case:
[tex]\[
f(x) = -\frac{1}{2}x - 1 \quad \text{if} \quad x \leq -1
\][/tex]
Substituting [tex]\( x = -2 \)[/tex]:
[tex]\[
f(-2) = -\frac{1}{2}(-2) - 1 = 1 - 1 = 0
\][/tex]
Therefore,
[tex]\[
f(-2) = 0
\][/tex]
2. Finding [tex]\( f(-1) \)[/tex]:
For [tex]\( x = -1 \)[/tex], we again use the first case because [tex]\( -1 \leq -1 \)[/tex]:
[tex]\[
f(x) = -\frac{1}{2}x - 1 \quad \text{if} \quad x \leq -1
\][/tex]
Substituting [tex]\( x = -1 \)[/tex]:
[tex]\[
f(-1) = -\frac{1}{2}(-1) - 1 = \frac{1}{2} - 1 = -\frac{1}{2}
\][/tex]
Therefore,
[tex]\[
f(-1) = -0.5
\][/tex]
3. Finding [tex]\( f(1) \)[/tex]:
Since [tex]\( -1 < 1 < 2 \)[/tex], we use the second case:
[tex]\[
f(x) = (x + 1)^2 - 3 \quad \text{if} \quad -1 < x < 2
\][/tex]
Substituting [tex]\( x = 1 \)[/tex]:
[tex]\[
f(1) = (1 + 1)^2 - 3 = 2^2 - 3 = 4 - 3 = 1
\][/tex]
Therefore,
[tex]\[
f(1) = 1
\][/tex]
Putting it all together, we have:
[tex]\[
\begin{array}{c}
f(-2) = 0 \\
f(-1) = -0.5 \\
f(1) = 1
\end{array}
\][/tex]
