MT1710: WEEK 3 - 2023/24 Section 1.3: Roots of Polynomial equations (continuation). Example 1.16 Locate the zeros of the expression 10 Let p(x) = x5 + 5x- 5. change There is one change of sign in the numerical coefficients of p(x), hence, by Descartes' Rule of Signs there is at most one positive root for p(x) = 0. Considering p (-x) = - x5-5x - 5, there is no change of sign in the numerical coefficients of p(-x), hence, by Descartes' Rule of Signs there are no negative roots for p(x) = 0. However p(x) is a quintic polynomial and so has to have at least one real roots as the complex roots of equation p(x) = 0 come in pairs. The following Theorem will help to further locate the zeros of a polynomial and help with deciding how many zeros has a function. Theorem 1.2: The Intermediate Value Theorem (Bolzano's Theorem) Let f be a real-valued and continuous function on a closed interval [a, b] and assume that f(a) and f(b) have different signs, i.e., f (a) f (b) < 0. Then there is at least one point x in the open interval (a, b) such that f(x) = 0. Proof of this theorem uses the concept of the supremum of a set, which those of you taking MT194 will meet later this term. The result is also intuitive given f is a continuous function. Example 1.16 (solution continued) Locate the zeros of the expression x5 + 5x - 5 . Let p(x) = x5 + 5x - 5. We have seen that by using Descartes' Rule of Signs this polynomial has at most one positive zero and that it has no negative zeros. To use the Intermediate Value Theorem we need to evaluate p(x) for several values of x so that we identify two values of x which give opposite signs to p(x). In fact, we have p(1)=+1>0 and p(0) =- 5<0->p(1)p(0)<0 Hence, by the Intermediate Value Theorem, there is at least one zero of p(x) in the interval (0, 1). We have located the consecutive integers between which the zero lies.
So, combining the two pieces of information, the only real zero of p(x) lies in the interval (0, 1). If a smaller interval would be needed, we would repeat the process by halving the interval, evaluating p(x) at x = 1/2. In fact the the only real zero of p(x) would lie then in the interval (1/2, 1) and the process could be repeated multiple times to refine the interval again. Example 1.17 Locate the zeros of the expression x5 + x3 - 12x2 + 3x + 2. See solution in lectures. Section 1.4: Curve Sketching. Index of guidelines to follow when sketching a curve: (a) Symmetry; Check if the function is even or odd. In either case we can concentrate on positive values of x and the negative values can be sketched after. (b) Intersections with the axes; The value x = 0 gives the