To analyze the functions \( f(x) = \sqrt{x - 1} \) and \( g(x) = \sqrt[3]{x - 3} \), we need to determine their domains and ranges.
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Domain:
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For \( f(x) = \sqrt{x - 1} \): The expression under the square root must be non-negative, so: \[ x - 1 \geq 0 \implies x \geq 1. \] Thus, the domain of \( f(x) \) is \( [1, \infty) \).
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For \( g(x) = \sqrt[3]{x - 3} \): The cube root function is defined for all real numbers, so the domain of \( g(x) \) is: \[ (-\infty, \infty). \]
Conclusion about domains: The functions do not have the same domain.
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Range:
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For \( f(x) = \sqrt{x - 1} \): As \( x \) varies from 1 to \( \infty \), \( f(x) \) can take values ranging from 0 to \( \infty \). Hence, the range of \( f(x) \) is \( [0, \infty) \).
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For \( g(x) = \sqrt[3]{x - 3} \): As \( x \) varies from \( -\infty \) to \( \infty \), \( g(x) \) can also take all real values, including negative ones. Therefore, the range of \( g(x) \) is \( (-\infty, \infty) \).
Conclusion about ranges: The functions do not have the same range.
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Comparing Values:
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For \( f(x) \) at \( x = 1 \): \[ f(1) = \sqrt{1 - 1} = 0. \]
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For \( g(x) \) at \( x = 1 \): \[ g(1) = \sqrt[3]{1 - 3} = \sqrt[3]{-2} < 0. \]
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As \( x \) increases further (for \( x \geq 1 \)), \( f(x) \) increases from 0 upward, while \( g(x) \) continues to take on values from negative infinity to positive values. This implies \( f(x) > g(x) \) for \( x \geq 1 \).
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Conclusion: The last statement is true; the values of \( f(x) > g(x) \) for all \( x \) such that \( 0 \leq x < \infty \) (noting that \( g(x) \) is not defined for \( x < 3 \) but \( f(x) \) starts from \( x = 1 \)) captures the essence.
Final Evaluation: The correct statement is:
- The values of \( f(x) > g(x) \) for all values of \( x \) such that \( 0 \leq x < \infty\).
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