importance of graphs for our present work is not, therefore, to determine the exact values of the roots, but to make clearer some principles which are to follow. Graphic Representation of Real and Equal, and Imaginary Roots. 47. The graphs corresponding to the three functions. The graph corresponding to (1) cuts the x-axis in two points, two and four units to the right of the origin. Observe that 2 and 4 are the roots of the equation 2-6x+8 = 0. In equation (1), when x = 3, y=-1, and the corresponding point on the graph is one unit below the x-axis. Now observe that the second member of (2) is obtained from the second member of (1) by adding unity to the latter. Therefore the ordinates in the graph of (2) will be one greater than the corresponding ordinates in the graph of (1), as shown in the figure. This graph is tangent to the x-axis at the point 3, 0, and could have been obtained by sliding the graph of (1) vertically upward one unit. In so doing, the two points where (1) cuts the x-axis would have moved toward each other and coincided in the point where (2) touches the x-axis. The root 3 is a multiple root of the equation 2-6x + 9 = 0. Finally, the second member of (3) is obtained from the second member of (1) by adding two units to the latter. The graph of (3), therefore, could have been obtained by sliding the graph of (1) vertically upward two units, as shown in the figure. This graph does not intersect the x-axis, and therefore indicates the presence of imaginary roots; in fact, the roots of the equation a2-6x+10= 0 are conjugate imaginaries. EXERCISES VIII. --- 1. Locate the points (1, 3); (−1, 3); (2, −2); (−4, −1); (0, −2); (3, 0); (0, 4); (− 3, 0); (0, 0); (4, − 2); (3, 2); (− 5, − 3); (− 5, 4). Draw the graphs of the following equations: Draw, to one set of axes, the graphs of the equations in each of the following examples, and determine their point of intersection: Draw the graph of each of the following functions, and locate each irrational root between two consecutive integers : the term an- can be made to exceed numerically the sum of all the terms that follow it, by taking x large enough, and to exceed the sum of all the terms that precede it, by taking small enough. By taking a large enough, we can make each term in the denominator of (2), and hence the denominator itself, as small as we please. When the denominator is made smaller than an-k, then m>1, numerically. Therefore, from (1), An-kX* > An-k+1Xk−1 + ... + ɑn-12 + ɑn, numerically. In like manner the second part of the principle can be proved. 49. Since a"∞, as xo, it follows from the preceding article that f(x) = ∞, as x = ∞. If we take a large enough to make the term a exceed numerically the sum of all the terms that follow it, f(x) will be positive when a" is positive, and negative when a" is negative. 3x-5x+7x2-19 x 17+00, Ex. 1. when Ex. 2. 29 x2-6x+5+∞, when æ∞, ∞, when æ— — ∞. Principle of Continuity. 50. It was assumed in Art. 38 that the curve is continuous between any two points definitely located; that is, that a point, which we may assume to be describing the curve, nowhere makes a jump. An instance of an interruption of continuity is shown in Fig. 16. Here the point which is describing the curve on arriving at the point P, whose abscissa is a, say, jumps to the point Q. We assume that Q is a finite distance from P, while the abscissa of Q is greater than a by less than any assigned number, however small. Now, let R be a point on the curve, whose abscissa is a +h, h being an infinitesimal. Then, since ordinates represent values of the function, we have FIG. 16. R AP = ƒ (a), and BR = ƒ (a + h). Evidently, f(a+h) - f(a) repre sents the change in the value of the function corresponding to the change h in the value of x. Since BR · AP, = ƒ (a + h) — ƒ (a), remains finite as h = 0, corresponding to an infinitesimal change in the value of x, there is a finite change in the value of the function. On the other hand, let the curve be continuous at a point a, as in Fig. 17. Then f(a+h) − f(a), = BR-AP, = CR. Ash 0, that is, as the point B approaches A, it is evident that CR approaches 0. That is, when the function is continuous at x = a, corresponding to an infinitesimal YI FIG. 17. change in the value of x, there is an infinitesimal change in the value of f(x). We are thus led to the following definition of continuity: The function, f (x), is continuous at x = a, if f(a + h) f(a) = 0, as h=0. 51. A rational integral fraction of x is continuous for all finite values of x. h2 '' (a) + *** h h2 Therefore f'(a) + = ƒ'' (a) + = ƒ""' (a) + ··· = ƒ'(a), and Consequently, 2 B h2 ... h[ƒ'(a) + ', ƒ'"' (a) + ', ƒ'"' (a) + --] ± 0. f(a + h) f(a) = 0 as h = 0, and f(x) is continuous at x = a. But a may be any finite value of x. Therefore f(x) is continuous for all finite values of x. 52. It follows from the principle of continuity that as f(x) changes from f(a) to f(b), it must do so by infinitesimal changes, and therefore pass through every intermediate value. 53. If f(a) and f(b) have opposite signs, at least one real root of ƒ (x) = 0 lies between a and b. By the principle of continuity, a number cannot pass from a positive value to a negative value, or vice versa, without passing through 0. Since f(x), which is continuous in passing from ƒ(a) to ƒ(b), changes its sign, it must pass through 0. This value of x between a and b, for which ƒ(x) becomes 0, is therefore a root of ƒ (x) = 0. In connection with this principle attention is called to the graphs of the functions in Exx. 1-3, Art. 46. Thus, in Ex. 3, ƒ(2) and ƒ(3) have opposite signs, and the graph in Fig. 14 cuts the x-axis once between 2 and 3. In general, if ƒ(a) and f(b) have opposite signs, the corresponding points on the graph are on opposite sides of the x-axis. Therefore the graph must cross the x-axis at least once between a and b. 54. The following principles are derived from the preceding: (i.) Every equation of an odd degree has at least one real root whose sign is opposite to that of its last term. For, f(0)= an; and, by Art. 49, ƒ(+ ∞) = + ∞, ƒ (− x ) = − ∞. If an be positive, ƒ(0) and ƒ(—) have opposite signs, and there is a negative root between 0 and -∞. If an be negative, f(0) and ƒ(+ ∞) have opposite signs, and there is a positive root between 0 and +∞. (ii.) Every equation of an even degree, whose last term is negative, has at least two real roots, one positive and one negative. For, ƒ (0), an, is negative, and ƒ(+∞) = +∞, and ƒ ( − ∞) = + ∞. Since ƒ(0) and ƒ(+∞) have opposite signs, there is a positive root between 0 and +∞. For a similar reason, there is a negative root between 0 and 00. 55. The following is a more general enunciation of the principle given in Art. 53: If f(a) and f(b) have opposite signs, an odd number of real roots lies between a and b. If f(a) and f(b) have like signs, either no real root, or an even number of real roots, lies between a and b. |