EXAMPLE 10 Show that there is a root of the equation 4x3 − 7x2 + 3x − 2 = 0 between 1 and 2. SOLUTION Let f(x) = 4x3 − 7x2 + 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) = 4 − 7 + 3 − 2 = −2 < 0 and f(2) = 32 − 28 + 6 − 2 = 8 > 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 4x3 − 7x2 + 3x − 2 = 0 has at least one root c in the open interval (1,2) . In fact, we can locate a root more precisely by using the Intermediate Value Theorem again. Since f(1.4) = −0.544 < 0 and f(1.5) = 0.250 > 0 a root must lie between (smaller) and (larger). A calculator gives, by trial and error, f(1.47) = −0.010208 < 0 and f(1.48) = 0.074368 > 0. So a root lies in the open interval (1.47,148)