Which of the following accurately shows the range of the function defined below?

[tex]\[
f(x) = \left\{
\begin{array}{cl}
3 & \text{if } x \ \textless \ 0 \\
x^2 + 2 & \text{if } 0 \leq x \ \textless \ 2 \\
\frac{1}{2}x + 5 & \text{if } x \geq 2
\end{array}
\right.
\][/tex]

A. [tex]\([3, \infty)\)[/tex]

B. [tex]\([2, \infty)\)[/tex]

C. [tex]\([3, 6]\)[/tex]

D. [tex]\((-\infty, \infty)\)[/tex]



Answer :

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]