[tex]f(x) = 3\sqrt{x}[/tex]
[tex]f'(x) = -\frac{3}{2\sqrt{x}}[/tex]
When x = 0, f'(0) becomes undefined. Thus, we know that there is no derivative at x = 0. You could also show it by taking the limits, which is the fundamental of calculus. Show that as x approaches 0 from the negative side is not equal to the value as x approaches 0 from the positive side.
What this tells us is that there is no uniquely distinct point that satisfies the continuity of the curve in order for the graph to have a specific gradient at x = 0; thus, this means there is no gradient at x = 0, and hence, the function is not differentiable at x = 0
