Here is a piecewise polynomial function: f(x) = x^2 + 1 if x <= 1 f(x) = 2x if x > 1 I need to prove that this function is differentiable at x = 1? It's a parabola that turns into a line. It doesn't have any gaps or corners. The limit of f(x) as x approaches 1 is 2, and the limit of f'(x) as x approaches 1 is 2. This was a problem on a test, but my calculus teacher took points off because she says that the function is not differentiable at x = 1. Thanks in advance!
Does [itex]\lim _{h \to 0}\frac{f(1+h) - 2}{h}[/itex] exist? I.e. do the following two exist and are they equal? [tex]\lim _{h \to 0^+}\frac{f(1+h) - 2}{h}\mbox{, and }\lim _{h \to 0^-}\frac{f(1+h) - 2}{h}[/tex]
Yes if we look at the limit of the two different slopes as they approach this point they will converge at 1 giving slope 2. However, if you were to graph the function at this point and or checked for continuity you would see a problem. Lim fx from the right does not equal Lim fx from the left heading toward point 1. lim x^2 evaluated at 1 = 1 and lim 2x evaluated at 1 = 2. This tells you that there would be a big jump at this point and would not be continious as the function output would not agree from the two sides. Even though you should use lim [f(x+h) - F(x)]/h with h->0 to check that a function will be differentiable, you must realize this will only tell us if the two slopes will congerve to the same number at the given point. However, we could have a big jump in a graph in respect to height and still have the same slope for that point. The jump implies disscontinious and thus we fail to meet differentiability. We need the same slope and continuity to imply differentiability, as differentiability a implies the function has same slope and height at the given point in question. If you were to search google you would find that alot of people neglect to mention this important point for piecewise functions. Hope that helps, Kyle Rupps
Yes sorry. I thought the first function was just x^2 and not x^2 + 1. In the x^2 case they would not be continious but would satisfy converging slopes. In the x^2 + 1 case both the slopes converge and the point is continious. My bad, I will try to keep my glasses on rather than off in the future. Looks like that teacher is wrong after all :) Kyle Rupps