Hey,

It's not differentiable at 3^0.5 because:

3^0.5 = x means that

f(x) = |[(3^0.5)^2]-3|

[3^0.5]^2 becomes 3^(0.5*2) = 3

Which means that f(x) = 0

If we plot the normal x^2 - 3 we find that at y = 0 f(x) crosses the x axis and becomes negative. If we take the absolute value of this we have to flip the graph at this point which means there is a corner at the point x=3

The limit of the derivative approaching from the left does not equal the limit of the derivative approaching from the right and hence it does not exist.

Here is a graph of the absolute value function.