To analyze the given functions \( f(x) = 2\sqrt{x} \) and \( g(x) = 2\sqrt[3]{x} \), we will compare their properties, including their domains, behavior (increasing/decreasing), and specific points.
-
Domains:
- For \( f(x) = 2\sqrt{x} \): The domain is \( x \geq 0 \) because the square root function is only defined for non-negative numbers.
- For \( g(x) = 2\sqrt[3]{x} \): The domain is all real numbers (\( -\infty < x < \infty \)) because the cube root function is defined for all real numbers.
Conclusion: The graphs have different domains.
-
Behavior (Increasing/Decreasing):
- \( f(x) = 2\sqrt{x} \) is increasing because the square root function increases as \( x \) increases.
- \( g(x) = 2\sqrt[3]{x} \) is also increasing because the cube root function increases as \( x \) increases.
Conclusion: They are both increasing on their domains (not decreasing).
-
Comparison for \( x > 1 \):
- To compare \( g(x) \) and \( f(x) \) for \( x > 1 \):
- For \( f(x) = 2\sqrt{x} \), at \( x = 1 \), \( f(1) = 2 \).
- For \( g(x) = 2\sqrt[3]{x} \), at \( x = 1 \), \( g(1) = 2 \).
- For points where \( x > 1 \):
- As \( x \) increases above 1, \( \sqrt{x} \) increases faster than \( \sqrt[3]{x} \), which means \( f(x) \) will exceed \( g(x) \).
Conclusion: When \( x > 1 \), the function \( g(x) \) is not greater than \( f(x) \).
- To compare \( g(x) \) and \( f(x) \) for \( x > 1 \):
-
Specific Points:
- At \( x = -1 \), \( f(-1) \) is not defined since the square root cannot have a negative input.
- At \( x = 0 \), both \( f(0) = 0 \) and \( g(0) = 0 \).
- At \( x = 1 \), both \( f(1) = 2 \) and \( g(1) = 2 \).
Conclusion: The statement that both graphs go through the points (−1,−1), (0,0), and (1,1) is incorrect.
Based on this analysis, the accurate statement is:
The graphs have different domains.
11 answers