Show that the equation x3-14x+c=0 has at most one solution in the interval -2,2. such that f'(r)=0.Now f'(r)=. Since risin(a,b), which is contained in-2,2,we have |r|<2,sor2<4.It follows that 3r2-143*4-14=-20. This contradicts f'(r)=0,so the given equation cannot have two real solutions in-2,2. Hence, it has at most one real solution in-2,2.