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

Question

(1 point)
Responses

The graphs have different domains.
The graphs have different domains.

They are both decreasing on their domains.
They are both decreasing on their domains.

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 .

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 .
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Answer 1

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.