Let's analyze the two functions \( f(x) = 2\sqrt{x} \) and \( g(x) = 2\sqrt[3]{x} \) step by step.
-
Graph Points Check:
-
\( f(-1) = 2\sqrt{-1} \) is undefined, so \( f \) does not pass through (-1, -1).
-
\( f(0) = 2\sqrt{0} = 0 \) → point (0, 0) is on the graph.
-
\( f(1) = 2\sqrt{1} = 2 \) → point (1, 1) is not on the graph, as y would be 2.
-
\( g(-1) = 2\sqrt[3]{-1} = 2(-1) = -2 \) → point (-1, -1) is not on the graph.
-
\( g(0) = 2\sqrt[3]{0} = 0 \) → point (0, 0) is on the graph.
-
\( g(1) = 2\sqrt[3]{1} = 2 \) → point (1, 1) is not on the graph, as y would be 2.
Statement A: False since neither function passes through (-1, -1) or (1, 1). Both pass through (0, 0).
-
-
Comparing Functions for \( x > 1 \):
- For values greater than 1, we can find typical values of \( f(x) \) and \( g(x) \):
- \( f(2) = 2\sqrt{2} \approx 2.83 \)
- \( g(2) = 2\sqrt[3]{2} \approx 1.59 \)
- In fact, \( f(x) = 2\sqrt{x} \) grows faster than \( g(x) = 2\sqrt[3]{x} \) for \( x > 1 \).
Statement B: False since \( g(x) \) is less than \( f(x) \) when \( x > 1 \).
- For values greater than 1, we can find typical values of \( f(x) \) and \( g(x) \):
-
Domains:
- The domain of \( f(x) = 2\sqrt{x} \) is \( [0, \infty) \).
- The domain of \( g(x) = 2\sqrt[3]{x} \) is \( (-\infty, \infty) \).
Statement C: True since their domains are different.
-
Monotonicity:
- \( f(x) = 2\sqrt{x} \) is increasing on its domain \( [0, \infty) \).
- \( g(x) = 2\sqrt[3]{x} \) is also increasing on its domain \( (-\infty, \infty) \).
Statement D: False since both functions are increasing, not decreasing.
Conclusion: The correct statement is C. the graphs have different domains.
1 answer