Compare the given graphs of f(x) = 2sqrt(x) and g(x) = 2cbrt(x) and determine which...

The domain of $f(x) = 2\sqrt{x}$ is $x \ge 0$. The domain of $g(x) = 2\sqrt[3]{x}$ is all real numbers. For $x > 1$, $x^{1/3} > x^{1/2}$ is false, so $g(x) < f(x)$. For $0 < x < 1$, $x^{1/3} > x^{1/2}$, so $g(x) > f(x)$. At $x=1$, $f(1)=2$ and $g(1)=2$. At $x=0$, $f(0)=0$ and $g(0)=0$. The statement 'When $x > 1$, the function $g(x) = 2\sqrt[3]{x}$ is greater than $f(x) = 2\sqrt{x}$' is false.