 Clearly, \(f(0) = -1 +r\) and \(f(3) = 1 + r\), whence \(f\) takes the value 0 at some point \(x\) in [0,3]. If such \(x\) could be determined, either \(1 \leq x\) or \(x \leq 2\). Since \(f\) equals \(r\) on \([1,2]\), in the first case \(r \leq 0\) and in the second case \(0\leq r\), contradicting the undecidability of the statement \((r\leq 0 \vee 0 \leq r)\).