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
The graphs have different domains.
The graphs have different domains.
Both graphs go through (−1,−1)
, (0,0)
, and (1,1)
.
Both graphs go through left parenthesis negative 1 comma negative 1 right parenthesis , left parenthesis 0 comma 0 right parenthesis , and left parenthesis 1 comma 1 right parenthesis .
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 .
They are both decreasing on their domains.
and g(x)=2x−−√3
and determine which of the following is true.
(1 point)
Responses
The graphs have different domains.
The graphs have different domains.
Both graphs go through (−1,−1)
, (0,0)
, and (1,1)
.
Both graphs go through left parenthesis negative 1 comma negative 1 right parenthesis , left parenthesis 0 comma 0 right parenthesis , and left parenthesis 1 comma 1 right parenthesis .
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 .
They are both decreasing on their domains.
Answers
GPT-4o mini
To compare the functions \( f(x) = 2\sqrt{x} \) and \( g(x) = 2\sqrt[3]{x} \), we analyze their graphs and properties:
1. **Domain:**
- The function \( f(x) = 2\sqrt{x} \) is defined for \( x \geq 0 \) (since square roots are only defined for non-negative numbers).
- The function \( g(x) = 2\sqrt[3]{x} \) is defined for all real numbers \( x \).
- So, the domains are indeed different: \( f(x) \) has a domain of \( [0, \infty) \) while \( g(x) \) has a domain of \( (-\infty, \infty) \).
2. **Points:**
- For \( f(x) \):
- At \( x = -1 \): \( f(-1) \) is not defined.
- At \( x = 0 \): \( f(0) = 2\sqrt{0} = 0 \).
- At \( x = 1 \): \( f(1) = 2\sqrt{1} = 2 \).
- For \( g(x) \):
- At \( x = -1 \): \( g(-1) = 2\sqrt[3]{-1} = -2 \).
- At \( x = 0 \): \( g(0) = 2\sqrt[3]{0} = 0 \).
- At \( x = 1 \): \( g(1) = 2\sqrt[3]{1} = 2 \).
- So, they both go through \( (0, 0) \) and \( (1, 2) \), and do not go through \( (-1, -1) \).
3. **Behavior for \( x > 1 \):**
- For \( x > 0 \):
- \( f(x) = 2\sqrt{x} \) is an increasing function that grows monotonically.
- \( g(x) = 2\sqrt[3]{x} \) is also an increasing function but increases at a slower rate after \( x = 1 \).
- If we compare both functions for \( x > 1 \), \( f(x) \) will be greater than \( g(x) \) specifically for values of \( x > 1 \) since \( \sqrt{x} > \sqrt[3]{x} \).
4. **Monotonicity:**
- Both functions \( f(x) \) and \( g(x) \) are increasing on their respective domains.
Considering everything:
- **True statements**:
- The graphs have different domains.
- They do **not** both go through \( (-1, -1) \), \( (0, 0) \), and \( (1, 1) \); however, they do go through \( (0, 0) \) and \( (1, 2) \).
- For \( x > 1 \), \( f(x) > g(x) \) which means that statement about \( g(x) \) being greater is false.
- They are not both decreasing; both functions are increasing on their domains.
### Conclusion
The correct answer is:
**The graphs have different domains.**
1. **Domain:**
- The function \( f(x) = 2\sqrt{x} \) is defined for \( x \geq 0 \) (since square roots are only defined for non-negative numbers).
- The function \( g(x) = 2\sqrt[3]{x} \) is defined for all real numbers \( x \).
- So, the domains are indeed different: \( f(x) \) has a domain of \( [0, \infty) \) while \( g(x) \) has a domain of \( (-\infty, \infty) \).
2. **Points:**
- For \( f(x) \):
- At \( x = -1 \): \( f(-1) \) is not defined.
- At \( x = 0 \): \( f(0) = 2\sqrt{0} = 0 \).
- At \( x = 1 \): \( f(1) = 2\sqrt{1} = 2 \).
- For \( g(x) \):
- At \( x = -1 \): \( g(-1) = 2\sqrt[3]{-1} = -2 \).
- At \( x = 0 \): \( g(0) = 2\sqrt[3]{0} = 0 \).
- At \( x = 1 \): \( g(1) = 2\sqrt[3]{1} = 2 \).
- So, they both go through \( (0, 0) \) and \( (1, 2) \), and do not go through \( (-1, -1) \).
3. **Behavior for \( x > 1 \):**
- For \( x > 0 \):
- \( f(x) = 2\sqrt{x} \) is an increasing function that grows monotonically.
- \( g(x) = 2\sqrt[3]{x} \) is also an increasing function but increases at a slower rate after \( x = 1 \).
- If we compare both functions for \( x > 1 \), \( f(x) \) will be greater than \( g(x) \) specifically for values of \( x > 1 \) since \( \sqrt{x} > \sqrt[3]{x} \).
4. **Monotonicity:**
- Both functions \( f(x) \) and \( g(x) \) are increasing on their respective domains.
Considering everything:
- **True statements**:
- The graphs have different domains.
- They do **not** both go through \( (-1, -1) \), \( (0, 0) \), and \( (1, 1) \); however, they do go through \( (0, 0) \) and \( (1, 2) \).
- For \( x > 1 \), \( f(x) > g(x) \) which means that statement about \( g(x) \) being greater is false.
- They are not both decreasing; both functions are increasing on their domains.
### Conclusion
The correct answer is:
**The graphs have different domains.**