341Nice one. My tubelight lit up only this morning :D
Writing the relation as f(x+1) [f(x)+1] = -1
we see that points where f(x) = 0 or f(x) =-1 are points of discontinuity.
So our task is to prove that this happens. In particular we can prove that we can always find a point such that f(x) =0.
Consider any point t \in \mathbb{R}. Let f(t) = k
We consider three cases:
1. k>0
2. k<-1
3.-1<k<0
Case 1: We have f(t+1) = - \frac{1}{k+1}<0
By IMV there exists x_0 \in (t,t+1) such that f(x_0)=0
Case 2: Again the change in sign: f(t+1) = - \frac{1}{k+1} >0 -
ensures x_0 \in (t,t+1) such that f(x_0)=0
Case 3: f(t+1) = - \frac{1}{k+1} <-1 reducing it to Case 2 above so that there exists x_0 \in (t+1,t+2) such that f(x_0)=0
Thus, in all three cases the assumption of continuity in the relevant intervals (which is the basis for applying IMV) leads us to a point of discontinuity in that interval.