To analyze and compare the functions \( f(x) = 2\sqrt{x} \) and \( g(x) = 2\sqrt[3]{x} \):
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Function Analysis:
- \( f(x) = 2\sqrt{x} \): This function is defined for \( x \geq 0 \) since the square root is not defined for negative numbers. It is an increasing function, starting from the origin (0,0) and increasing as \( x \) increases.
- \( g(x) = 2\sqrt[3]{x} \): This function is defined for all \( x \) (both positive and negative), as the cube root is defined for all real numbers. It is also an increasing function.
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Behavior When \( x > 1 \):
- For \( x > 1 \), both functions are positive, but we need to compare their values:
- For \( x = 1 \): \( f(1) = 2\sqrt{1} = 2 \) and \( g(1) = 2\sqrt[3]{1} = 2 \).
- For \( x = 2 \): \( f(2) = 2\sqrt{2} \approx 2.828 \) and \( g(2) = 2\sqrt[3]{2} \approx 2.519 \) (since \( \sqrt[3]{2} \) is less than \( \sqrt{2} \)).
- Thus, for \( x > 1 \), \( f(x) > g(x) \).
- For \( x > 1 \), both functions are positive, but we need to compare their values:
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Monotonicity:
- Both functions are increasing, not decreasing, on their respective domains.
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Domains:
- The domain of \( f(x) \) is \( [0, \infty) \).
- The domain of \( g(x) \) is \( (-\infty, \infty) \).
- Therefore, the domains are different.
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Points of Intersection:
- Both functions do intersect at the point (1, 2) and both pass through the origin (0, 0) as well. However, they do not pass through (-1, -1).
Based on the analysis, the correct statements among the given options are:
- "When \( x > 1 \), the function \( g(x) = 2\sqrt[3]{x} \) is less than \( f(x) = 2\sqrt{x} \)", which is the negation of one of the given responses.
- "They are both decreasing on their domains" is incorrect because both functions are increasing.
- "The graphs have different domains" is correct.
- "Both graphs go through (−1,−1), (0,0), and (1,1)" is incorrect as they do not both pass through (−1,−1).
The correct response that matches our findings is:
- "The graphs have different domains."
1 answer