Therefore the negative roots of f(x) = 0 are the positive roots of ƒ(— x) = 0, and hence follows Descartes's rule for negative roots: The number of negative roots of the equation f(x) = 0 can not be greater than the number of variations of signs in ƒ(— x). one EXAMPLE. The equation + 15 x2 + 7x − 11 = 0 has variation of sign and can not have more than one positive root. Again f(x) = x2 + 15 x2 7x 110 has one variation of sign and therefore f(x) can not have more than one negative root. 818. Determination of the Existence of Imaginary Roots by means of Descartes's Rule. In case the sum of the maximum number of positive and negative roots is less than the degree of the equation we are sure that the equation has imaginary roots. EXAMPLE. The equation x + 15x2 + 7x-11= 0 has, 2817, at most one positive root and,. 2817, at most one negative root. Hence the given equation can not have more than two real roots, and therefore must have at least two imaginary roots. f(x+h)=p(x+h)"+p1(x+h)"~1+p2(x+h)"−2+ . . . . +Pn−1(x+h)+Pn• .... Expanding (x+ h)", (x + h)"−1, . .. by the binomial theorem and arranging the whole result according to the ascending powers of h, we have hn n(n +Pn-1 ... +2pn-2] + [~(x-1) ( − 2) . . . . 21.]. The first line of the expression is evidently fr). The coefficient of h is represented by f'(x), of " by ƒ"(x), and in general the coefficient 2! Hence f'(x), ƒ''(x), ƒ'""'(x), . are called respectively the first, second, and third derivatives of f(x) with respect to x. It follows that f'(x) is obtained from f(x) by multiplying each term in f(x) by the exponent of x in that term and diminishing the exponent of x by 1. EXAMPLE. Find the derivatives of 3 * 2 x3 = - f(c) 3x2 x35x+7 f'(x) 36 x 12 x = - f""(x)=72 x - 12 f(x)=72 x0 f'''''(x) = 0. 820. Equal Roots.-If the equation f(x) = 0 has p roots, equal to a, then f'(x) = 0 has p 1 roots equal to a. If the equation f(r)=0 has p roots equal to a it is divisible p times by x-a or by (-a); let the quotient of f(x)=0 by (x—a)o be F(x); then we have Using the expansion for f(x+h) and F(x+h), 2819, and the expansion of [(x-a)+h] by the binomial theorem, we have h 12 ƒ(x) + ' '; ƒ'(x) +'",ƒƒ''(x) + . . . . . = [(x−a)2+p(.c−a)»-h+ . ...] 1! X 1! 2! × [F(x) + '"; F"'(x)+'" F"(x)] =(x—a)oF(x)+[(x—a)oF′(x)+p(x—a)p-1F(x)]h+terms in h2, h3, . . . Thus we have two integral polynomials which are identically equal for all finite values of h; hence the coefficients of like powers of h are equal, thus ƒ'(x)=(x—a)oF"′(x)+p(x—a)p−1F(x). Therefore, f(r) contains the factors x-a repeated p. 2 times; that is, f'(c) has p-1 roots equal to a. Similarly, if b is a root. of f(x) = 0 repeated r- -1 times, then b is a root of f'(x) = 0 *The notation ƒ'(x), ƒ''(x), ƒ'''(x)..., becomes inconvenient when the number of accents is large, and hence f(x) is used for the coefficient of ! repeated r 2 times, and so on. Therefore, f(x)=0 has or has not equal roots according as f(x) and f'(x) have or have not the common factor, x - a, or some power of x -a. 821. Hence it follows that the equal roots of f(x)=0 are given by finding the greatest common divisor of f(x)=0 and f'(x)=0, and placing it equal to zero, then solving the resulting equation. EXAMPLE. Find the equal and non-equal roots of the equation f(x)=x-5x+5x+9x3-14x2-4x+8= 0. Hence f'(x)=6x-25 x 20 x3 +27 x2 - 28 x 4. The G. C. D. of f(x) and ƒ'(x) is x+1 and (x-2)2; hence (x+1)° and (x-2)3 are factors of f(x). The remaining factor of f(x) is x — Therefore -1 is a double root, 2 is a triple root, and 1 a single root of f(x)=0. -1. 822. Continuity of a Rational Integral Function of x.-If f(x) is a rational integral polynomial in x, and x is made to vary by infinitesimal increments (637) from a to a larger number b, we shall prove that ƒ(x) at the same time varies by infinitesimal increments; in this case it is said that f(x) varies continuously with x. Let c and ch be any two values of x lying between a and b. Hence we have, 1819, (1) ƒ(c+h) — ƒ(c)=h ƒ'(e)+" ƒ"(c) + ƒ""" (c) + . . . h3 3! 2! Since f(x), f'(x), ƒ"(x), . . . are all finite for x = c, the limit of the second member of equation (1) is 0 for h±0 (8668). Therefore, the lim h=0 [f(c+h) − f(c)] = 0. Hence, to infinitesimal changes in x there correspond infinitesimal changes in f(x), and as x changes continuously from a to b, the function f(x) changes continuously from f(a) to f(b). 823. When f'(c) is positive, f(x) increases with x; and when ƒ'(c) is negative, f(x) diminishes as x increases. (n − 1)! ƒ "(c) ]. The limit of the bracket for h± 0 is ƒ'(c) (1668). This theorem may be made more perspicuous by aid of the graphical representation of the following example. 824. Theorem.—Iƒ ƒ(a) and ƒ(b) have contrary signs then the equation f(x) = 0 must have at least one real root situated between a and b. -20 25 Y' For, since f(x) changes continuously from f(a) to f(b), and therefore passes through all intermediary values as a changes continuously from a to b, and since by hypothesis one of the signs of f(a) and f(b) is positive and the other negative, it follows that f(x), for some value of x between a and b, must take the value zero situated between f(a) and ƒ(b). Caution. It does not follow that f(x) = 0 has but one root between a and b; nor does it also follow, in case f(a) and ƒ(b) have the same sign, that f(x) = 0 has no root between a and b. 825. Theorem.-Every equation of an odd degree has at least one real root, whose sign is the opposite of that of the last term. Let the equation be f(x)="+P1x”−1 Then we have Pn +Pn-1+Pn = 0. for x, f(x) is negative, n being odd, [Reciprocal of 2668] [Reciprocal of 1668] If p, is positive then f(x)=0 must have at least one real negative root between co and 0 (2824); if p, is negative then f(x)=0 must have a real positive root between 0 and ∞ (8824). n EXAMPLE. Show that f(x)=x+5 x 20 x Q. E. D. 19 x-2=0 Here has a root between 2 and 3 and a root between 4 and -5. we have f(2)8 = 4) 10 Thus f(x)=0 has a positive root between +2 and +3, whose sign is opposite to that of the last term, -2. 826. Theorem.—If the last term of an equation of an even degree is negative the equation has two real roots, one positive and one negative. Consider the equation in $825, and for n even let Hence there is at least one real root between between 0 and + (824). Q. E. D. ∞ and 0, and one 827. Theorem.-If two numbers, a and b, are substituted for x in f(x), and f(a) and f(b) have contrary signs, then an odd number of roots of f(x) =0 lies between a and b; if f(a) and (b) have the same signs then either no root or an even number of roots of f(x)=0 lies between a and b. We The theorem in 824 is a particular case of this theorem. give a proof for the first part of the theorem and the second part may be established in a similar manner. Let a be less than b, and suppose that of the roots of f(x)=0, a1, a21 a,, and no others, lie between a and b. Let the quotient of f(x) divided by (x-a,) (x-α)... (x-a), be F(x), then we have |