Question
The domain of \( h(x)=\sqrt{-x}+6 \) is \( (-\infty, 0] \) (Type your answer in interval notation.) The range of \( h(x)=\sqrt{-x}+6 \) is \( \square \). (Type your answer in interval notation.)
Ask by Marshall Cervantes. in the United States
Feb 26,2025
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Answer
The range of \( h(x) = \sqrt{-x} + 6 \) is \( [6, \infty) \).
Solution
Since h(x) = √(-x) + 6, we first ensure the radicand is nonnegative:
-x ≥ 0 → x ≤ 0.
Thus, the domain is (-∞, 0].
For the range, note that √(-x) is always nonnegative. The minimum value occurs when:
√(-x) = 0 → h(x) = 0 + 6 = 6.
As x becomes more negative (approaches -∞), √(-x) increases without bound, leading to h(x) increasing without bound.
Therefore, the range of h(x) is [6, ∞).
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Simplify this solution Bonus Knowledge
The function \( h(x) = \sqrt{-x} + 6 \) has a domain of \( (-\infty, 0] \) because the expression inside the square root must be non-negative, which is true for any \( x \) less than or equal to 0. To find the range, notice that as \( x \) approaches \( -\infty \), \( \sqrt{-x} \) becomes very large, leading \( h(x) \) to tend towards \( \infty \). When \( x = 0 \), \( h(x) = \sqrt{0} + 6 = 6 \). Thus, the range of \( h(x) \) is \( [6, \infty) \).
