Validity of Piecewise Differentiation for Functions with Discontinuities

[Ted123]

[PLAIN]http://img863.imageshack.us/img863/9868/heaviside.png

 

[HallsofIvy]

Are you sure the function is defined as you say? Yes, as given, f is NOT continuous because it is not defined at x= -1 and at x= 1.

If the function were defined as
f(x)= \left{\begin{array}{cc} -1, & x<-1 \\ x, & -1\le x\le 1 \\ 1, & x> 1\end{array}
then it would be continuous and the first and second derivatives exactly as you say.

 

[Ted123]

Are you sure the function is defined as you say? Yes, as given, f is NOT continuous because it is not defined at x= -1 and at x= 1.

If the function were defined as
f(x)= \left{\begin{array}{cc} -1, & x<-1 \\ x, & -1\le x\le 1 \\ 1, & x> 1\end{array}
then it would be continuous and the first and second derivatives exactly as you say.

The question and the solution is exactly how they appear. So even if the statement in the solutions 'f being continuous...' is false, is the method of piecewise differentiation to obtain f' still valid?