Let's analyze the functions \( f(x) = \sqrt{x} + 1 \) and \( g(x) = \sqrt[3]{x} + 1 \).
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Domain:
- The function \( f(x) = \sqrt{x} + 1 \) requires \( x \geq 0 \) because the square root function is defined only for non-negative values. Therefore, the domain of \( f(x) \) is \( [0, \infty) \).
- The function \( g(x) = \sqrt[3]{x} + 1 \) can take any real number as input, so its domain is \( (-\infty, \infty) \).
This means the first statement "Both functions have the same domain, \( x > 0 \)" is false; \( f(x) \) has a domain of \( [0, \infty) \) while \( g(x) \) has a domain of \( (-\infty, \infty) \).
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Increasing behavior:
- The derivative of \( f(x) = \sqrt{x} + 1 \) is \( f'(x) = \frac{1}{2\sqrt{x}} \), which is positive for \( x > 0 \). Therefore, \( f(x) \) is increasing on its domain.
- The derivative of \( g(x) = \sqrt[3]{x} + 1 \) is \( g'(x) = \frac{1}{3\sqrt[3]{x^2}} \). This is also positive for all \( x \) (including negative values). Therefore, \( g(x) \) is increasing on its domain.
This means the statement "Both functions are increasing on their domain." is true.
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Points:
- For \( f(x) \):
- \( f(-1) = \text{undefined} \)
- \( f(0) = 1 \)
- For \( g(x) \):
- \( g(-1) = 0 \)
- \( g(0) = 1 \)
This means the statement "Both functions go through the points (-1,0) and (0,1)" is false; only \( g(x) \) goes through (-1, 0).
- For \( f(x) \):
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Range:
- The range of \( f(x) \) is \( [1, \infty) \) because as \( x \) approaches 0, \( f(x) \) approaches 1, and it increases without bound.
- The range of \( g(x) \) is \( (-\infty, \infty) \) because it can take any real value.
This means the statement "Both functions have the same range, (-∞,∞)" is false.
Conclusion
The correct answer is that "Both functions are increasing on their domain."