Question

compare the given graphs of f (x) = 2√x and g (x) = 2∛x and determine which of the following is true.
A. Both graphs go through (-1, -1), (0, 0), and (1, 1).
B. when x > 1, the function g (x) = 2∛x is greater than f (x) = 2√x
C. the graphs have different domains
D. they are both decreasing on their domains

Answer 1

Let's analyze the two functions \( f(x) = 2\sqrt{x} \) and \( g(x) = 2\sqrt[3]{x} \) step by step.

1. **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).

2. **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 \( x > 1 \).

3. **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.

4. **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**.