Compare the given graphs of f(x)=2x−−√ and g(x)=2x−−√3 and determine which of the following is true.

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

1. **Domains**:
- \( f(x) = 2\sqrt{x} \): The square root function is defined for \( x \geq 0 \). Therefore, the domain of \( f \) is \( [0, \infty) \).
- \( g(x) = 2\sqrt[3]{x} \): The cube root function is defined for all real numbers. Therefore, the domain of \( g \) is \( (-\infty, \infty) \).
- **Conclusion**: The graphs have different domains: \( f(x) \) is only defined for non-negative values, while \( g(x) \) is defined for all real numbers.

2. **Monotonicity**:
- \( f(x) \): As \( x \) increases, \( \sqrt{x} \) increases, so \( f(x) \) is increasing for \( x \geq 0 \).
- \( g(x) \): As \( x \) increases, \( \sqrt[3]{x} \) also increases, meaning \( g(x) \) is increasing for all real numbers.
- **Conclusion**: Neither function is decreasing on its domain.

3. **Comparison when \( x > 1 \)**:
- When \( x > 1 \):
- \( f(x) = 2\sqrt{x} \)
- \( g(x) = 2\sqrt[3]{x} \)
- To compare, we can evaluate both functions at a few points greater than 1.
- For \( x = 2 \):
- \( f(2) = 2\sqrt{2} \approx 2.83 \)
- \( g(2) = 2\sqrt[3]{2} \approx 2 \times 1.26 \approx 2.52 \)
- For \( x = 3 \):
- \( f(3) = 2\sqrt{3} \approx 3.46 \)
- \( g(3) = 2\sqrt[3]{3} \approx 2 \times 1.44 \approx 2.88 \)
- In both cases, \( f(x) > g(x) \) for \( x > 1 \).
- **Conclusion**: \( f(x) \) is greater than \( g(x) \) when \( x > 1 \).

4. **Intersection points**:
- Evaluating \( f \) and \( g \) at specific values:
- \( f(-1) = 2\sqrt{-1} \) is not defined.
- \( f(0) = 0 \) and \( g(0) = 0 \) (both go through (0,0)).
- \( f(1) = 2 \) and \( g(1) = 2 \) (both go through (1,1)).
- **Conclusion**: Only \( (0,0) \) and \( (1,1) \) are common, but both do not pass through \( (-1, -1) \), as \( f(-1) \) is undefined.

Now, summarizing what we found:

- **The graphs have different domains.** (True)
- **They are both decreasing on their domains.** (False)
- **When \( x > 1 \), \( g(x) \) is greater than \( f(x) \).** (False)
- **Both graphs go through \( (-1, -1), (0, 0), (1, 1) \).** (False)

Therefore, the true statement is: **The graphs have different domains.**