Question

Is the function $f(x)=\sqrt{x}$ differentiable at x=0 and continuous over R?

Answer 1

Answer:  No

Explanation:

Apply the derivative

$f(x) = \sqrt{x}\\\\f(x) = x^{1/2}\\\\f'(x) = (1/2)x^{-1/2}\\\\f'(x) = \frac{1}{2x^{1/2}}\\\\f'(x) = \frac{1}{2\sqrt{x}}$

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 $x \ge 0$ and not over the entire set of real numbers R.