is the square root of the absolute value of x differentiable at 0? if so what does the tangent look like?

Answer 1

Well the square root of the absolute value is simply the square root of positive values reflected in the y axis (x=0). As it is reflected at the point x=0, any tangent at this point is going to be along the x-axis (horizontal). However the absolute value means the function may not be differentiable here, as the curve is not smooth, the function not continuous. I can't be sure about this bit, but if it works it should be horizontal.

Answer 2

y = √|x|

The limit of the slope of the curve as x→0 is -∞ from the left and +∞ from the right. It is not differentiable at x = 0. The graph comes to a sharp point at x = 0.