To compare the functions \( f(x) = \sqrt{4x} \) and \( g(x) = \sqrt[3]{4x} \), we need to analyze their domains and ranges.
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Domain:
- For \( f(x) = \sqrt{4x} \): The expression inside the square root must be non-negative, so \( 4x \geq 0 \). This means \( x \geq 0 \). Therefore, the domain of \( f \) is \( [0, \infty) \).
- For \( g(x) = \sqrt[3]{4x} \): The cube root is defined for all real numbers, so the domain of \( g \) is \( (-\infty, \infty) \).
Conclusion: The functions have different domains.
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Range:
- The range of \( f(x) = \sqrt{4x} \) is \( [0, \infty) \) because the square root function produces non-negative outputs.
- The range of \( g(x) = \sqrt[3]{4x} \) is \( (-\infty, \infty) \) because the cube root can take any real number.
Conclusion: The functions have different ranges.
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Increasing/Decreasing Behavior:
- \( f(x) = \sqrt{4x} \) is an increasing function for \( x \geq 0 \) since its derivative \( f'(x) = \frac{2}{\sqrt{4x}} \) is positive for \( x > 0 \).
- \( g(x) = \sqrt[3]{4x} \) is also an increasing function for all \( x \) since its derivative \( g'(x) = \frac{4}{3(4x)^{2/3}} \) is positive for all \( x \).
Conclusion: Both functions are increasing.
Summary:
- The functions have different domains.
- The functions have different ranges.
- Both functions are increasing.
Given this analysis, the option that describes the functions accurately is:
Option #3: The functions have different domains, they have different ranges, and both functions are increasing.