To compare the functions \( f(x) = \sqrt{x} + 1 \) and \( g(x) = \sqrt[3]{x} + 1 \), let's analyze their properties:
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Domain:
- \( f(x) = \sqrt{x} + 1 \): The square root function is only defined for \( x \geq 0 \). Therefore, the domain of \( f(x) \) is \( x \geq 0 \).
- \( g(x) = \sqrt[3]{x} + 1 \): The cube root function is defined for all real numbers. Therefore, the domain of \( g(x) \) is \( (-\infty, \infty) \).
Conclusion: Both functions do not have the same domain.
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Points:
- For \( f(x) \):
- At \( x = -1 \): \( f(-1) \) is undefined because the square root of a negative number is not a real number.
- At \( x = 0 \): \( f(0) = \sqrt{0} + 1 = 1 \).
- For \( g(x) \):
- At \( x = -1 \): \( g(-1) = \sqrt[3]{-1} + 1 = -1 + 1 = 0 \).
- At \( x = 0 \): \( g(0) = \sqrt[3]{0} + 1 = 0 + 1 = 1 \).
Conclusion: The functions do not go through the points \( (-1, 0) \) and \( (0, 1) \); \( f(x) \) is undefined at \( x = -1 \).
- For \( f(x) \):
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Range:
- \( f(x) = \sqrt{x} + 1 \): The minimum value occurs at \( x = 0 \) is \( f(0) = 1 \). As \( x \) increases, \( f(x) \) increases without bound. Thus, the range is \( [1, \infty) \).
- \( g(x) = \sqrt[3]{x} + 1 \): The cube root function can take on any real value, and adding 1 shifts the entire graph up by 1. Therefore, the range is all real numbers: \( (-\infty, \infty) \).
Conclusion: Both functions do not have the same range.
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Increasing Nature:
- Both \( f(x) = \sqrt{x} + 1 \) and \( g(x) = \sqrt[3]{x} + 1 \) are increasing functions in their respective domains.
- \( f(x) \) is increasing for \( x \geq 0 \), and \( g(x) \) is increasing for all \( x \).
Conclusion: Both functions are increasing on their domains.
After analyzing the provided options, the true statement is:
Both functions are increasing on their domain.