To evaluate the piecewise function [tex]\( f(x) \)[/tex] at the points [tex]\( x = 3 \)[/tex], [tex]\( x = 0 \)[/tex], and [tex]\( x = 5 \)[/tex], let's analyze each case step-by-step.
### Case 1: Evaluating [tex]\( f(x) \)[/tex] at [tex]\( x = 3 \)[/tex]
For [tex]\( x = 3 \)[/tex], we use the part of the function that applies when [tex]\( x \geq 3 \)[/tex], which is [tex]\( f(x) = -2x + 5 \)[/tex].
Substituting [tex]\( x = 3 \)[/tex]:
[tex]\[ f(3) = -2(3) + 5 \][/tex]
[tex]\[ f(3) = -6 + 5 \][/tex]
[tex]\[ f(3) = -1 \][/tex]
Hence,
[tex]\[ f(3) = -1 \][/tex]
### Case 2: Evaluating [tex]\( f(x) \)[/tex] at [tex]\( x = 0 \)[/tex]
For [tex]\( x = 0 \)[/tex], we use the part of the function that applies when [tex]\( x < 3 \)[/tex], which is [tex]\( f(x) = x + 2 \)[/tex].
Substituting [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 0 + 2 \][/tex]
[tex]\[ f(0) = 2 \][/tex]
Hence,
[tex]\[ f(0) = 2 \][/tex]
### Case 3: Evaluating [tex]\( f(x) \)[/tex] at [tex]\( x = 5 \)[/tex]
For [tex]\( x = 5 \)[/tex], we use the part of the function that applies when [tex]\( x \geq 3 \)[/tex], which is [tex]\( f(x) = -2x + 5 \)[/tex].
Substituting [tex]\( x = 5 \)[/tex]:
[tex]\[ f(5) = -2(5) + 5 \][/tex]
[tex]\[ f(5) = -10 + 5 \][/tex]
[tex]\[ f(5) = -5 \][/tex]
Hence,
[tex]\[ f(5) = -5 \][/tex]
### Final Results
By summarizing the results for each evaluated point, we have:
[tex]\[
f(3) = -1, \quad f(0) = 2, \quad f(5) = -5
\][/tex]
So the evaluations of the function [tex]\( f(x) \)[/tex] at the specified points are:
[tex]\[
\boxed{(-1, 2, -5)}
\][/tex]
