Let H(x)=f(x)−3g(x) H(−1)=H(0)=H(2)=3 Applying Rolle's Theorem in the interval [−1,0] H'(x)=f′(x)−3g′(x)=0 for atleast one c∊(−1,0). As H"(x) never vanishes in the interval ⇒Exactly one c∊(−1,0) for which H'(x)=0 Similarly apply Rolle's Theorem in the interval [0,2]. ⇒H'(x)=0has exactly one solution in (0,2)