To analyze the two functions \( f(x) = \sqrt{x} + 1 \) and \( g(x) = \sqrt{3^x} + 1 \):
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Domain:
- The domain of \( f(x) = \sqrt{x} + 1 \) is \( x \geq 0 \) because the square root is undefined for negative numbers.
- The domain of \( g(x) = \sqrt{3^x} + 1 \) is all real numbers (\( x \in \mathbb{R} \)), as \( 3^x \) is always non-negative for any real \( x \), making \( \sqrt{3^x} \) defined everywhere.
- Thus, the statement "Both functions have the same domain, \( x > 0 \)" is false.
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Range:
- The range of \( f(x) = \sqrt{x} + 1 \) starts from \( f(0) = 1 \) and goes to \( \infty \), so the range is \( [1, \infty) \).
- The range of \( g(x) = \sqrt{3^x} + 1 \) is similar: \( g(0) = \sqrt{3^0} + 1 = 1 + 1 = 2 \) when \( x = 0 \) and it goes to \( \infty \) as \( x \to \infty \). Therefore, the range is \( [2, \infty) \).
- Both functions do not have the same range, so the statement "Both functions have the same range, \( (-\infty, \infty) \)" is false.
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Increasing Behavior:
- \( f(x) = \sqrt{x} + 1 \) is increasing for \( x \geq 0 \) because the derivative \( f'(x) = \frac{1}{2\sqrt{x}} \) is positive for \( x > 0 \).
- \( g(x) = \sqrt{3^x} + 1 \) is also increasing for all real \( x \), as its derivative \( g'(x) = \frac{\ln(3)}{2\sqrt{3^x}} \) is positive for all \( x \).
- Thus, the statement "Both functions are increasing on their domain" is true.
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Points on the Graph:
- Checking the points:
- For \( f(-1) \): Not defined as \( f(x) = \sqrt{x} + 1 \) requires \( x \geq 0 \).
- For \( f(0) = 1 \), and checking \( g(-1) \): \( g(-1) = \sqrt{3^{-1}} + 1 = \frac{1}{\sqrt{3}} + 1 \neq 0 \).
- Thus, neither function goes through the points \((-1, 0)\) and \((0, 1)\).
- Checking the points:
Based on the above analysis, the only true statement is:
Both functions are increasing on their domain.
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