Answer: No
Explanation:
Apply the derivative
[tex]f(x) = \sqrt{x}\\\\f(x) = x^{1/2}\\\\f'(x) = (1/2)x^{-1/2}\\\\f'(x) = \frac{1}{2x^{1/2}}\\\\f'(x) = \frac{1}{2\sqrt{x}}\\\\[/tex]
Though we run into a problem since f ' (0) is undefined, due to the zero in the denominator.
Therefore, f(x) is not differentiable at x = 0.
The function is continuous, but only on the interval [tex]x \ge 0[/tex] and not over the entire set of real numbers R.
