367 3.4 Polynomial Functions: Graphs, Applications, and Models Intermediate Value and Boundedness Theorems As Examples 3 and 4 show, one key to graphing a polynomial function is locating its zeros. In the special case where the potential zeros are rational numbers, the zeros are found by the rational zeros theorem. Occasionally, irrational zeros can be found by inspection. For instance, ƒ1x2 = x3 - 2 has the irrational zero 23 2. The next two theorems apply to the zeros of every polynomial function with real coefficients. The first theorem uses the fact that graphs of polynomial functions are continuous curves. The proof requires advanced methods, so it is not given here. Figure 33 illustrates the theorem. x y f(a) f(b) f(c) = 0 b a c y = f(x) Figure 33 Intermediate Value Theorem If ƒ1x2 is a polynomial function with only real coefficients, and if for real numbers a and b the values ƒ1a2 and ƒ1b2 are opposite in sign, then there exists at least one real zero between a and b. This theorem helps identify intervals where zeros of polynomial functions are located. If ƒ1a2 and ƒ1b2 are opposite in sign, then 0 is between ƒ1a2 and ƒ1b2, and so there must be a number c between a and b where ƒ1c2 = 0. CAUTION Be careful when interpreting the intermediate value theorem. If ƒ1a2 and ƒ1b2 are not opposite in sign, it does not necessarily mean that there is no zero between a and b. In Figure 35, ƒ1a2 and ƒ1b2 are both negative, but -3 and -1, which are between a and b, are zeros of ƒ1x2. The intermediate value theorem for polynomials helps limit the search for real zeros to smaller and smaller intervals. In Example 5, we used the theorem to verify that there is a real zero between 2 and 3. To locate the zero more accurately, we can use the theorem repeatedly. (Prior to modern-day methods involving calculators and computers, this was done by hand.) EXAMPLE 5 Locating a Zero Use synthetic division and a graph to show that ƒ1x2 = x3 - 2x2 - x + 1 has a real zero between 2 and 3. S Now Try Exercise 47. −3.1 −4.7 3.1 4.7 Figure 34 Figure 35 x y a f(b) f(b) < 0 f(a) < 0 f(a) b y = f(x) 0 –3 –1 GRAPHING CALCULATOR SOLUTION The graphing calculator screen in Figure 34 indicates that this zero is approximately 2.2469796. (Notice that there are two other zeros as well.) ALGEBRAIC SOLUTION Use synthetic division to find ƒ122 and ƒ132. 2)1 -2 -1 1 2 0 -2 1 0 -1 -1 = ƒ122 3)1 -2 -1 1 3 3 6 1 1 2 7 = ƒ132 Because ƒ122 is negative and ƒ132 is positive, by the intermediate value theorem there must be a real zero between 2 and 3.
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