Answer :
To determine the domain and range of the inequality [tex]\( y < \sqrt{x + 3} + 1 \)[/tex], let's analyze it step by step:
1. Domain:
- The expression inside the square root, [tex]\( x + 3 \)[/tex], must be non-negative because the square root function is only defined for non-negative numbers.
- So, we set up the inequality:
[tex]\[ x + 3 \geq 0 \][/tex]
- Solving for [tex]\( x \)[/tex]:
[tex]\[ x \geq -3 \][/tex]
- Since we are only considering [tex]\( y \)[/tex] values that actually satisfy [tex]\( y < \sqrt{x + 3} + 1 \)[/tex], [tex]\( x \)[/tex] must be strictly greater than [tex]\(-3\)[/tex].
- Therefore, the domain is:
[tex]\[ x > -3 \][/tex]
2. Range:
- Consider the smallest value [tex]\( y \)[/tex] can take. This happens when [tex]\( x \)[/tex] is at its lowest permissible value from the domain, which is as [tex]\( x \)[/tex] approaches [tex]\(-3\)[/tex].
- When [tex]\( x = -3 \)[/tex]:
[tex]\[ y = \sqrt{x + 3} + 1 = \sqrt{-3 + 3} + 1 = \sqrt{0} + 1 = 1 \][/tex]
- However, the inequality specifies [tex]\( y \)[/tex] must be less than [tex]\( \sqrt{x + 3} + 1 \)[/tex], so [tex]\( y \)[/tex] can never actually be 1.
- Therefore, [tex]\( y \)[/tex] must be greater than 1 to satisfy the inequality.
- Hence, the range is:
[tex]\[ y > 1 \][/tex]
Combining both parts, we get:
- The domain is [tex]\( x > -3 \)[/tex]
- The range is [tex]\( y > 1 \)[/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{\text{A. } x > -3 \text{ and } y > 1 } \][/tex]
1. Domain:
- The expression inside the square root, [tex]\( x + 3 \)[/tex], must be non-negative because the square root function is only defined for non-negative numbers.
- So, we set up the inequality:
[tex]\[ x + 3 \geq 0 \][/tex]
- Solving for [tex]\( x \)[/tex]:
[tex]\[ x \geq -3 \][/tex]
- Since we are only considering [tex]\( y \)[/tex] values that actually satisfy [tex]\( y < \sqrt{x + 3} + 1 \)[/tex], [tex]\( x \)[/tex] must be strictly greater than [tex]\(-3\)[/tex].
- Therefore, the domain is:
[tex]\[ x > -3 \][/tex]
2. Range:
- Consider the smallest value [tex]\( y \)[/tex] can take. This happens when [tex]\( x \)[/tex] is at its lowest permissible value from the domain, which is as [tex]\( x \)[/tex] approaches [tex]\(-3\)[/tex].
- When [tex]\( x = -3 \)[/tex]:
[tex]\[ y = \sqrt{x + 3} + 1 = \sqrt{-3 + 3} + 1 = \sqrt{0} + 1 = 1 \][/tex]
- However, the inequality specifies [tex]\( y \)[/tex] must be less than [tex]\( \sqrt{x + 3} + 1 \)[/tex], so [tex]\( y \)[/tex] can never actually be 1.
- Therefore, [tex]\( y \)[/tex] must be greater than 1 to satisfy the inequality.
- Hence, the range is:
[tex]\[ y > 1 \][/tex]
Combining both parts, we get:
- The domain is [tex]\( x > -3 \)[/tex]
- The range is [tex]\( y > 1 \)[/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{\text{A. } x > -3 \text{ and } y > 1 } \][/tex]
