Prove for the function f defined so that f(x)=|x-3|, for each real number x, the real number 1 is not the derivative of f at p=3.
Consider the functions in two situations :
If x<3 --> x-3<0 --> |x-3|=3-x
If x>3 --> x-3>0 --> |x-3|=x-3
the derivative goes to -1 from the left, and goes to 1 from the right.