Let's analyze the functions \( f(x) = \sqrt{x} + 1 \) and \( g(x) = \sqrt[3]{x} + 1 \) and evaluate the options provided.
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Domain:
- The domain of \( f(x) = \sqrt{x} + 1 \) is \( x \geq 0 \) (since the square root function is only defined for non-negative \( x \)).
- The domain of \( g(x) = \sqrt[3]{x} + 1 \) is \( (-\infty, \infty) \) (since the cube root function is defined for all real \( x \)).
- Hence, both functions do not have the same domain.
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Range:
- The range of \( f(x) = \sqrt{x} + 1 \) is \( [1, \infty) \) (the minimum value of \( \sqrt{x} \) is 0, so \( \sqrt{x} + 1 \) starts from 1 and goes to \( \infty \)).
- The range of \( g(x) = \sqrt[3]{x} + 1 \) is \( (-\infty, \infty) \) (since \( \sqrt[3]{x} \) can take any real value, adding 1 shifts the entire range but does not restrict it).
- Therefore, both functions do not have the same range.
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Points of Intersection:
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For the point \((-1, 0)\):
- For \( f(-1) \): Not defined (since \(-1 < 0\)).
- For \( g(-1) = \sqrt[3]{-1} + 1 = -1 + 1 = 0\).
- Hence, \((-1, 0)\) is not on \( f(x) \).
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For the point \( (0, 1) \):
- For \( f(0) = \sqrt{0} + 1 = 1\).
- For \( g(0) = \sqrt[3]{0} + 1 = 0 + 1 = 1\).
- Thus, both functions go through the point \( (0, 1) \), but not through the point \( (-1, 0)\).
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Therefore, option a is false.
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Increasing Nature of Functions:
- To determine if both functions are increasing, we can consider their derivatives:
- \( f'(x) = \frac{1}{2\sqrt{x}} \) for \( x > 0\) (positive, therefore \( f(x) \) is increasing).
- \( g'(x) = \frac{1}{3\sqrt[3]{x^2}} \) for all \( x \neq 0\) (positive for all \( x\), thus \( g(x) \) is also increasing).
- Since both functions are increasing (even though \( f(x) \) is defined for \( x \geq 0\) and \( g(x) \) for all \( x\)), option d is true concerning their nature on their respective domains.
- To determine if both functions are increasing, we can consider their derivatives:
Conclusion:
The correct option is d: Both functions are increasing on their domain.
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