Asked by C.W
Compare the graphs of f(x)=x−−√+1
and g(x)=x−−√3+1
. Which one of the following options is true?
(1 point)
Responses
Both functions go through the points (−1,0)
and (0,1)
.
Both functions go through the points left parenthesis negative 1 comma 0 right parenthesis and left parenthesis 0 comma 1 right parenthesis .
Both functions have the same domain, x>0
.
Both functions have the same domain, x greater than 0 .
Both functions are increasing on their domain.
Both functions are increasing on their domain.
Both functions have the same range, (−∞,∞)
.
Both functions have the same range, left parenthesis negative infinity comma infinity right parenthesis .
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All Answers 1
Answered by
GPT-5 mini
AI
Correct answer: Both functions are increasing on their domain.
Reasoning in brief:
- f(x) = √x + 1 has domain x ≥ 0 and range y ≥ 1; g(x) = √(x − 3) + 1 has domain x ≥ 3 and range y ≥ 1. So they do not both pass through (−1,0) or (0,1) (only f passes through (0,1)), they do not have the same domain x>0, and their range is not (−∞,∞).
- The square-root function is increasing, and adding 1 does not change that. For x in their domains f′(x)=1/(2√x) (positive for x>0, and f is increasing on [0,∞) in the usual sense) and g′(x)=1/(2√(x−3))>0 for x>3, so both functions are increasing on their domains.
Reasoning in brief:
- f(x) = √x + 1 has domain x ≥ 0 and range y ≥ 1; g(x) = √(x − 3) + 1 has domain x ≥ 3 and range y ≥ 1. So they do not both pass through (−1,0) or (0,1) (only f passes through (0,1)), they do not have the same domain x>0, and their range is not (−∞,∞).
- The square-root function is increasing, and adding 1 does not change that. For x in their domains f′(x)=1/(2√x) (positive for x>0, and f is increasing on [0,∞) in the usual sense) and g′(x)=1/(2√(x−3))>0 for x>3, so both functions are increasing on their domains.
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