To compare the graphs of \( f(x) = 2\sqrt{x} \) and \( g(x) = 2\sqrt[3]{x} \), let's analyze each statement.
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They are both decreasing on their domains.
- For \( f(x) = 2\sqrt{x} \), the function is increasing on its domain \( x \geq 0 \).
- For \( g(x) = 2\sqrt[3]{x} \), the function is also increasing for all real \( x \).
- Therefore, this statement is false.
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Both graphs go through (−1,−1), (0,0), and (1,1).
- For \( f(x) = 2\sqrt{x} \), it is only defined for \( x \geq 0 \) (so it does not go through \((-1, -1)\)).
- For \( g(x) = 2\sqrt[3]{x} \), \( g(-1) = 2\sqrt[3]{-1} = 2(-1) = -2 \) (so it does not go through \((-1, -1)\)).
- Therefore, this statement is false.
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When \( x > 1 \), the function \( g(x) = 2\sqrt[3]{x} \) is greater than \( f(x) = 2\sqrt{x} \).
- For \( x > 1 \):
- \( f(x) = 2\sqrt{x} \) grows faster than \( g(x) = 2\sqrt[3]{x} \).
- Specifically, if you calculate values, e.g., for \( x = 2 \):
- \( f(2) = 2\sqrt{2} \approx 2.828 \),
- \( g(2) = 2\sqrt[3]{2} \approx 1.587 \).
- Thus, \( f(x) > g(x) \) when \( x > 1 \).
- Therefore, this statement is false.
- For \( x > 1 \):
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The graphs have different domains.
- \( f(x) = 2\sqrt{x} \) has the domain \( [0, \infty) \).
- \( g(x) = 2\sqrt[3]{x} \) has the domain \( (-\infty, \infty) \).
- Since their domains are indeed different, this statement is true.
In conclusion, the true statement is:
The graphs have different domains.