We have in fact demonstrated that f is continuous on the closed interval, even though f is not continuous at either of the values 0 or 5. Since the discontinuous characteristic of the delta function cannot exactly. Thus, the graph represents the function except that it has a hole at x 3. Again, polyfit is sufficient, but you could do it easily enough using a simple matrix solve using one call to backslash. Not all piecewise-defined functions are discontinuous where the function. Put a break after the first two points, now fit the data as a pair of discontinuous line segments. Let us see if f has a discontinuity at x 2.

Since lim x1 f (x) f (1), there is no discontinuity at x 1. Notice that each piece is a polynomial function, so they are continuous by themselves. If someone were to ask, 'For what values of x is f continuous?' then I know that I would answer this, because f is discontinuous at 0 and 5:īut if someone were to say 'Describe the intervals on which f is continuous,' then how can I possibly dispute it if a student answers this? The answer is simple enough, at least in 1-d. Let us examine where f has a discontinuity. Thus, by definition of continuity on a closed interval, f is continuous on the closed interval, since it is continuous on the open interval (0,5), continuous from the right at 0, and continuous from the left at 5. The definition of 'f is continuous on the closed interval ' is that f is continuous on (a,b) and f is continuous from the right at a and f is continuous from the left at b.

The definition of 'f is continuous from the left at b' is: Thus f is continuous from the right at 0. The definition of 'f is continuous from the right at a' is: I know that f is not continuous at 0 or at 5.

Discontinuous piecewise function