If there was a hole in the line at (2,3) and there is another point at (2,1), then would the graph be differentiable at that point and why?

Question

Aurora Alvarado · Accepted Answer

If I understand the description correctly, the answer is no and the reason is below.

The short answer is that the function you have described is not continuous at #2#. It is a theorem that if #f# is differentiable at #c#, then #f# is continuous at #c#. Therefore non-continuous implies non-differentiable. Longer answer "A hole in the line at #(2,3)#" indicates to me that #lim_(xrarr2)f(x) = 3#. The point at #(2,1)# implies that #f(2)=1# Now #f'(2) = lim_(xrarr2)(f(x)-f(2))/(x-2)# # = lim_(xrarr2)(f(x)-1)/(x-2)# This limit has the form #(3-1)/(2-2) = 2/0# which entails that the limit does not exist. Since the derivative is the limit, the derivative also does not exist.

Christopher Allers · Answer

undefined No, the graph would not be differentiable at the point (2,3) if there is a hole at that point. Differentiability requires that the function is continuous and has a defined slope at the point in question. A hole indicates a discontinuity in the function, which means the function is not continuous at that point. Therefore, it would not be differentiable at the point (2,3).