To determine the domain and range of the function [tex]\(y = 2 \sqrt{x+2}\)[/tex], we need to consider the behavior and restrictions of the function.
### Domain
The domain of a function includes all values of [tex]\(x\)[/tex] for which the function is defined. For the function [tex]\(y = 2 \sqrt{x+2}\)[/tex], the expression under the square root, [tex]\(x+2\)[/tex], must be non-negative because the square root function is not defined for negative numbers.
So, we solve the inequality:
[tex]\[
x + 2 \geq 0
\][/tex]
[tex]\[
x \geq -2
\][/tex]
Thus, the domain of the function is:
[tex]\[
D = \{ x \mid x \geq -2 \}
\][/tex]
### Range
The range of the function includes all possible values of [tex]\(y\)[/tex]. Since [tex]\(y = 2 \sqrt{x+2}\)[/tex], we notice:
- When [tex]\(x = -2\)[/tex], [tex]\(y = 2 \sqrt{-2+2} = 2 \sqrt{0} = 0\)[/tex]
- As [tex]\(x\)[/tex] increases, [tex]\(y\)[/tex] increases because the square root of a larger positive number is larger, and multiplying by 2 doesn't change the fact that the function value is growing.
Thus, since [tex]\(y\)[/tex] can start from 0 and increase without bound as [tex]\(x\)[/tex] increases, the range of the function is:
[tex]\[
R = \{ y \mid y \geq 0 \}
\][/tex]
Summarizing:
- The domain [tex]\(D\)[/tex] is [tex]\(\{ x \mid x \geq -2 \}\)[/tex]
- The range [tex]\(R\)[/tex] is [tex]\(\{ y \mid y \geq 0 \}\)[/tex]
So, amongst the given options, the correct one is:
[tex]\[
\begin{array}{l}
D = \{ x \mid x \geq -2 \} \\
R = \{ y \mid y \geq 0 \}
\end{array}
\][/tex]
Here are the detailed steps of a few domain and range examples to give you further context:
For [tex]\(x = -2, -1, 0, 1, 2, 3, 4\)[/tex], the values of [tex]\(y\)[/tex] are:
- When [tex]\(x = -2\)[/tex]:
[tex]\[
y = 2 \sqrt{-2 + 2} = 2 \sqrt{0} = 0
\][/tex]
- When [tex]\(x = -1\)[/tex]:
[tex]\[
y = 2 \sqrt{-1 + 2} = 2 \sqrt{1} = 2
\][/tex]
- When [tex]\(x = 0\)[/tex]:
[tex]\[
y = 2 \sqrt{0 + 2} = 2 \sqrt{2} \approx 2.828
\][/tex]
- When [tex]\(x = 1\)[/tex]:
[tex]\[
y = 2 \sqrt{1 + 2} = 2 \sqrt{3} \approx 3.464
\][/tex]
- When [tex]\(x = 2\)[/tex]:
[tex]\[
y = 2 \sqrt{2 + 2} = 2 \sqrt{4} = 4
\][/tex]
- When [tex]\(x = 3\)[/tex]:
[tex]\[
y = 2 \sqrt{3 + 2} = 2 \sqrt{5} \approx 4.472
\][/tex]
- When [tex]\(x = 4\)[/tex]:
[tex]\[
y = 2 \sqrt{4 + 2} = 2 \sqrt{6} \approx 4.899
\][/tex]
This verifies our derived domain [tex]\(\{ x \mid x \geq -2 \}\)[/tex] and range [tex]\(\{ y \mid y \geq 0 \}\)[/tex].
