To determine the range of the given piecewise function:
[tex]\[
f(x) = \left\{\begin{array}{cl}
3, & x < 0 \\
x^2 + 2, & 0 \leq x < 2 \\
\frac{1}{2} x + 5, & x \geq 2
\end{array}\right.
\][/tex]
we need to analyze the output values for each segment of the function separately.
1. For [tex]\(x < 0\)[/tex]:
[tex]\[
f(x) = 3
\][/tex]
Regardless of the value of [tex]\(x\)[/tex] (as long as [tex]\(x < 0\)[/tex]), the output is always [tex]\(3\)[/tex].
2. For [tex]\(0 \leq x < 2\)[/tex]:
[tex]\[
f(x) = x^2 + 2
\][/tex]
- When [tex]\(x = 0\)[/tex]:
[tex]\[
f(0) = 0^2 + 2 = 2
\][/tex]
- As [tex]\(x\)[/tex] approaches 2 (but is still less than 2):
[tex]\[
f(x) \approx 2^2 + 2 = 4 + 2 = 6
\][/tex]
Hence, for [tex]\(0 \leq x < 2\)[/tex], the function [tex]\( f(x) = x^2 + 2 \)[/tex] takes values from 2 up to, but not including, 6.
3. For [tex]\(x \geq 2\)[/tex]:
[tex]\[
f(x) = \frac{1}{2}x + 5
\][/tex]
- When [tex]\(x = 2\)[/tex]:
[tex]\[
f(2) = \frac{1}{2}(2) + 5 = 1 + 5 = 6
\][/tex]
- As [tex]\(x\)[/tex] increases beyond 2, [tex]\( f(x) \)[/tex] will continue to increase without bound. Therefore, for [tex]\(x \geq 2\)[/tex], [tex]\(f(x)\)[/tex] takes all values from 6 to [tex]\(\infty\)[/tex].
Now, combining the ranges of all the intervals:
- From the segment [tex]\( 0 \leq x < 2\)[/tex], the output of the function ranges from 2 to less than 6.
- From the segment [tex]\( x \geq 2 \)[/tex], the output starts at 6 and goes to [tex]\(\infty\)[/tex].
Thus, the overall range of the function is:
[tex]\[
\boxed{[2, \infty)}
\][/tex]
