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

1 answer

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.