Let's find [tex]\( f(-5) \)[/tex], [tex]\( f(-2) \)[/tex], and [tex]\( f(1) \)[/tex] using the given piecewise function:
[tex]\[ f(x)=\left\{\begin{array}{ll}
4 & \text{if } x < -2 \\
(x+1)^2 - 1 & \text{if } -2 \leq x \leq 2 \\
\frac{1}{2} x - 2 & \text{if } x > 2
\end{array}\right. \][/tex]
1. To find [tex]\( f(-5) \)[/tex]:
[tex]\(-5\)[/tex] is less than [tex]\(-2\)[/tex], so we use the first piece of the definition:
[tex]\[ f(x) = 4 \][/tex]
Thus,
[tex]\[ f(-5) = 4 \][/tex]
So,
[tex]\[ f(-5) = 4 \][/tex]
2. To find [tex]\( f(-2) \)[/tex]:
[tex]\(-2\)[/tex] is between [tex]\(-2\)[/tex] and [tex]\(2\)[/tex], inclusive, so we use the second piece of the definition:
[tex]\[ f(x) = (x+1)^2 - 1 \][/tex]
Thus, substituting [tex]\( x = -2 \)[/tex],
[tex]\[ f(-2) = ((-2)+1)^2 - 1 \][/tex]
[tex]\[ = (-1)^2 - 1 \][/tex]
[tex]\[ = 1 - 1 \][/tex]
[tex]\[ = 0 \][/tex]
So,
[tex]\[ f(-2) = 0 \][/tex]
3. To find [tex]\( f(1) \)[/tex]:
[tex]\(1\)[/tex] is between [tex]\(-2\)[/tex] and [tex]\(2\)[/tex], so again we use the second piece of the definition:
[tex]\[ f(x) = (x+1)^2 - 1 \][/tex]
Thus, substituting [tex]\( x = 1 \)[/tex],
[tex]\[ f(1) = (1+1)^2 - 1 \][/tex]
[tex]\[ = 2^2 - 1 \][/tex]
[tex]\[ = 4 - 1 \][/tex]
[tex]\[ = 3 \][/tex]
So,
[tex]\[ f(1) = 3 \][/tex]
In summary, the values are:
[tex]\[ f(-5) = 4 \][/tex]
[tex]\[ f(-2) = 0 \][/tex]
[tex]\[ f(1) = 3 \][/tex]
