Show that there is a root of the equation 5x3 − 10x2 + 3x − 2 = 0 between 1 and 2. SOLUTION Let f(x) = 5x3 − 10x2 + 3x − 2 = 0. We are looking for a solution of the given equation, that is, a number c between 1 and 2 such that f(c) = . Therefore we take a = , b = , and N = in the Intermediate Value Theorem. We have f(1) = 5 − 10 + 3 − 2 = −4 < 0 and f(2) = 40 − 40 + 6 − 2 = 4 > 0. Thus f(1) < 0 < f(2); that is N = 0 is a number between f(1) and f(2). Now f is continuous since it is a polynomial, so that the Intermediate Value Theorem says there is a number c between and such that f(c) = . In other words, the equation 5x3 − 10x2 + 3x − 2 = 0 has at least one root c in the open interval . In fact, we can locate a root more precisely by using the Intermediate Value Theorem again. Since f(1.7) = −1.235 < 0 and f(1.8) = 0.160 > 0 a root must lie between (smaller) and (larger). A calculator gives, by trial and error, f(1.78) = −0.145240 < 0 and f(1.79) = 0.005695 > 0. Do a root lies in the open interval .