Question
Compare the given graphs of f(x)=2x−−√ and g(x)=2x−−√3 and determine which of the following is true. (1 point) Responses When x>1 , the function g(x)=2x−−√3 is greater than f(x)=2x−−√ . When x greater than 1 , the function g left parenthesis x right parenthesis equals 2 root index 3 Start Root x End Root is greater than f left parenthesis x right parenthesis equals 2 Start Root x End Root . The graphs have different domains. The graphs have different domains. They are both decreasing on their domains. They are both decreasing on their domains. Both graphs go through (−1,−1) , (0,0) , and (1,1) .
Answers
GPT-4o mini
To compare the functions \( f(x) = 2\sqrt{x} \) and \( g(x) = 2\sqrt[3]{x} \), let's analyze both functions.
1. **Domain**:
- The function \( f(x) = 2\sqrt{x} \) has a domain of \( x \geq 0 \) since the square root function is defined only for non-negative values.
- The function \( g(x) = 2\sqrt[3]{x} \) has a domain of all real numbers \( (-\infty, \infty) \) because the cube root function is defined for all real numbers.
Therefore, the statement "The graphs have different domains" is **true**.
2. **Behavior of the functions**:
- For \( f(x) = 2\sqrt{x} \):
- As \( x \) increases, \( f(x) \) increases. It is an increasing function for \( x \geq 0 \).
- For \( g(x) = 2\sqrt[3]{x} \):
- This function also increases as \( x \) increases. It is an increasing function for all \( x \).
So, the statement "They are both decreasing on their domains" is **false**.
3. **Checking specific points**:
- For \( x = -1 \):
- \( f(-1) \) is not defined, so it cannot pass through (-1, -1).
- For \( x = 0 \):
- \( f(0) = 0 \) and \( g(0) = 0\). So, both functions pass through (0, 0).
- For \( x = 1 \):
- \( f(1) = 2\sqrt{1} = 2 \) and \( g(1) = 2\sqrt[3]{1} = 2\). Both pass through (1, 1).
The functions do not pass through (-1, -1), so "Both graphs go through (-1, -1), (0, 0), and (1, 1)" is **false**.
4. **Comparison for \( x > 1 \)**:
- Let's compare the two functions when \( x > 1 \):
- For both functions: since \( \sqrt{x} > \sqrt[3]{x} \) for \( x > 1\), it follows that \( 2\sqrt{x} > 2\sqrt[3]{x} \).
- Hence, when \( x > 1 \), \( g(x) = 2\sqrt[3]{x} < f(x) = 2\sqrt{x} \).
Thus, the statement "When \( x > 1 \), the function \( g(x) = 2\sqrt[3]{x} \) is greater than \( f(x) = 2\sqrt{x} \)" is **false**.
### Final Answer:
The correct response is **"The graphs have different domains."**
1. **Domain**:
- The function \( f(x) = 2\sqrt{x} \) has a domain of \( x \geq 0 \) since the square root function is defined only for non-negative values.
- The function \( g(x) = 2\sqrt[3]{x} \) has a domain of all real numbers \( (-\infty, \infty) \) because the cube root function is defined for all real numbers.
Therefore, the statement "The graphs have different domains" is **true**.
2. **Behavior of the functions**:
- For \( f(x) = 2\sqrt{x} \):
- As \( x \) increases, \( f(x) \) increases. It is an increasing function for \( x \geq 0 \).
- For \( g(x) = 2\sqrt[3]{x} \):
- This function also increases as \( x \) increases. It is an increasing function for all \( x \).
So, the statement "They are both decreasing on their domains" is **false**.
3. **Checking specific points**:
- For \( x = -1 \):
- \( f(-1) \) is not defined, so it cannot pass through (-1, -1).
- For \( x = 0 \):
- \( f(0) = 0 \) and \( g(0) = 0\). So, both functions pass through (0, 0).
- For \( x = 1 \):
- \( f(1) = 2\sqrt{1} = 2 \) and \( g(1) = 2\sqrt[3]{1} = 2\). Both pass through (1, 1).
The functions do not pass through (-1, -1), so "Both graphs go through (-1, -1), (0, 0), and (1, 1)" is **false**.
4. **Comparison for \( x > 1 \)**:
- Let's compare the two functions when \( x > 1 \):
- For both functions: since \( \sqrt{x} > \sqrt[3]{x} \) for \( x > 1\), it follows that \( 2\sqrt{x} > 2\sqrt[3]{x} \).
- Hence, when \( x > 1 \), \( g(x) = 2\sqrt[3]{x} < f(x) = 2\sqrt{x} \).
Thus, the statement "When \( x > 1 \), the function \( g(x) = 2\sqrt[3]{x} \) is greater than \( f(x) = 2\sqrt{x} \)" is **false**.
### Final Answer:
The correct response is **"The graphs have different domains."**