Question
Compare the graphs of f(x)=x−−√+1 and g(x)=x−−√3+1 . Which one of the following options is true?
(1 point)
Responses
Both functions are increasing on their domain.
Both functions are increasing on their domain.
Both functions have the same range, (−∞,∞).
Both functions have the same range, left parenthesis negative infinity comma infinity right parenthesis .
Both functions have the same domain, x>0.
Both functions have the same domain, x greater than 0 .
Both functions go through the points (−1,0) and (0,1).
(1 point)
Responses
Both functions are increasing on their domain.
Both functions are increasing on their domain.
Both functions have the same range, (−∞,∞).
Both functions have the same range, left parenthesis negative infinity comma infinity right parenthesis .
Both functions have the same domain, x>0.
Both functions have the same domain, x greater than 0 .
Both functions go through the points (−1,0) and (0,1).
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To analyze the functions \( f(x) = \sqrt{x} + 1 \) and \( g(x) = \sqrt[3]{x} + 1 \), we will investigate their domains, behaviors, and characteristics.
1. **Domains**:
- For \( f(x) = \sqrt{x} + 1 \): The domain is \( x \geq 0 \) since the square root function is defined for non-negative values.
- For \( g(x) = \sqrt[3]{x} + 1 \): The domain is all real numbers, \( x \in (-\infty, \infty) \), because the cube root function is defined for all real values of \( x \).
Therefore, the statement "Both functions have the same domain, \( x > 0 \)" is **false**.
2. **Ranges**:
- For \( f(x) \): The minimum value occurs at \( x = 0 \), giving \( f(0) = 1 \). As \( x \) increases, \( f(x) \) increases without bound; therefore, the range of \( f(x) \) is \( [1, \infty) \).
- For \( g(x) \): As \( x \) moves from negative to positive infinity, \( g(x) \) takes values from \( -\infty \) to \( \infty \) (since it can output negative values when \( x < 0 \)). Thus, the range of \( g(x) \) is \( (-\infty, \infty) \).
Therefore, the statement "Both functions have the same range, \( (-\infty, \infty) \)" is **false**.
3. **Monotonicity**:
- For \( f(x) = \sqrt{x} + 1 \): The derivative \( f'(x) = \frac{1}{2\sqrt{x}} \) for \( x > 0 \) is always positive, indicating that \( f(x) \) is increasing on its domain \( [0, \infty) \).
- For \( g(x) = \sqrt[3]{x} + 1 \): The derivative \( g'(x) = \frac{1}{3\sqrt[3]{x^2}} \) is always positive for all \( x \neq 0 \), suggesting that \( g(x) \) is increasing everywhere on its domain.
Therefore, the statement "Both functions are increasing on their domain" is **true**.
4. **Points of intersection**:
- To check if both functions go through the points \((-1,0)\) and \((0,1)\):
- For \( f(x) \): \( f(0) = \sqrt{0} + 1 = 1 \) (true); \( f(-1) \) is not defined (false).
- For \( g(x) \): \( g(0) = \sqrt[3]{0} + 1 = 1 \) (true); \( g(-1) = \sqrt[3]{-1} + 1 = 0\) (true).
Hence, the statement "Both functions go through the points (−1, 0) and (0, 1)" is **false** because f(-1) is not in the domain of \( f(x) \).
### Conclusion:
The only true statement is:
**Both functions are increasing on their domain.**
1. **Domains**:
- For \( f(x) = \sqrt{x} + 1 \): The domain is \( x \geq 0 \) since the square root function is defined for non-negative values.
- For \( g(x) = \sqrt[3]{x} + 1 \): The domain is all real numbers, \( x \in (-\infty, \infty) \), because the cube root function is defined for all real values of \( x \).
Therefore, the statement "Both functions have the same domain, \( x > 0 \)" is **false**.
2. **Ranges**:
- For \( f(x) \): The minimum value occurs at \( x = 0 \), giving \( f(0) = 1 \). As \( x \) increases, \( f(x) \) increases without bound; therefore, the range of \( f(x) \) is \( [1, \infty) \).
- For \( g(x) \): As \( x \) moves from negative to positive infinity, \( g(x) \) takes values from \( -\infty \) to \( \infty \) (since it can output negative values when \( x < 0 \)). Thus, the range of \( g(x) \) is \( (-\infty, \infty) \).
Therefore, the statement "Both functions have the same range, \( (-\infty, \infty) \)" is **false**.
3. **Monotonicity**:
- For \( f(x) = \sqrt{x} + 1 \): The derivative \( f'(x) = \frac{1}{2\sqrt{x}} \) for \( x > 0 \) is always positive, indicating that \( f(x) \) is increasing on its domain \( [0, \infty) \).
- For \( g(x) = \sqrt[3]{x} + 1 \): The derivative \( g'(x) = \frac{1}{3\sqrt[3]{x^2}} \) is always positive for all \( x \neq 0 \), suggesting that \( g(x) \) is increasing everywhere on its domain.
Therefore, the statement "Both functions are increasing on their domain" is **true**.
4. **Points of intersection**:
- To check if both functions go through the points \((-1,0)\) and \((0,1)\):
- For \( f(x) \): \( f(0) = \sqrt{0} + 1 = 1 \) (true); \( f(-1) \) is not defined (false).
- For \( g(x) \): \( g(0) = \sqrt[3]{0} + 1 = 1 \) (true); \( g(-1) = \sqrt[3]{-1} + 1 = 0\) (true).
Hence, the statement "Both functions go through the points (−1, 0) and (0, 1)" is **false** because f(-1) is not in the domain of \( f(x) \).
### Conclusion:
The only true statement is:
**Both functions are increasing on their domain.**
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