Smoothing a Piece-wise Function Instructor: Christine Breiner View the complete course: ocw.mit.edu/18-... License: Creative Commons BY-NC-SA More information at ocw.mit.edu/terms More courses at ocw.mit.edu
@Simon Hoogendal: The derivatives of left and right are never going to be equal in terms of their actual function definitions. The left side of the piecewise function is a linear function with a constant slope, and the right is a parabola with a variable slope. No matter how you define the left side of the function, this will always be the case since the problem tells us that the function must be linear at x 1 of the derivative function of each must be equal. If they aren't, it means that the slope of the tangent line at x=1 is different in each function. So again, the derivative functions of each part of the piecewise function will never be equal in definition, but they must each give the same output when x=1.
This is only possible line ax+b. b/c exactly at (1,2) there's a tangent line that has to coincide with the line ax+b in order to be smooth, however you can skip this step by just saying f(x) is riemann integrable. As a side note: all the solutions for ax+b, (a,b) is isomorphic to R^2.
In the second part, you impose that the function be *continuously differentiable at x=1, which is technically stronger than differentiability. For some functions, f'(a) might exist even if the limit of f'(x) as x goes to a does not.
If a=x and b = 1, so the second function was y=(x^2)+1, then wouldn't the resulting function have a derivative, because the combination would just create the function y=(x^2)+1 ?
