increasing, the maximum and minimum values of y will occur alternately. This will easily be seen when we consider that whenever the sign of f'(x) changes from + to, y is a maximum, and, whenever it changes from to +, a minimum; and that a change from to can be succeeded only by a change from to +, and conversely. - Modified method of finding Maxima and Minima. 73. Suppose that f(x) has a maximum value when x = x, and that none of the derived functions = ƒ'(x), ƒ"(x), ƒ''''(x), become infinite when x = xo. Then, since f'(x) decreases from, through 0, to, as x passes from x-h, through x, to xh, it appears, by the Lemma of Art. (69), that ƒ ̈(x) must have the sign - for this range of values of x. If ƒ"(x) =−0, when xx, the symbol - 0 being used to denote zero regarded as a limiting state of negative magnitude, then, when x = x,, it is evident that f(x) has a maximum value: from this it follows that, since f(x) now occupies the place of f(x), ƒ"(x) will change sign from +, through 0, to -, and ƒ"(x) will have the sign, as x ranges from xh to x + h. If ƒ" (x) = − 0, when x = x。, then, f(x) now occupying the place of ƒ (x), we see that f(x) will change sign from +, through 0, to -, and f(x) will have the sign as x ranges from x h to xo + h. We may proceed with this reasoning from step to step until we arrive at a derived function of an even order which does not vanish when x = x. Our final conclusion is evidently that, for a maximum value of f(x), we must have ƒ'(x) = 0, and that, of the differential coefficients of f(x), the first which, for a corresponding value of x, does not vanish, must be of an even order, and must be negative. By precisely the same form of reasoning, mutatis mutandis, we may see that, for a minimum value of f(x), the sufficient and necessary conditions are that f'(x) = 0, and that of the derived functions ƒ"(x), ƒ""(x),.... the first which, for a corresponding value of x, does not vanish, shall be of an even order and shall be positive. Hence, to find the maxima and minima of a function f(x), we must equate its first differential coefficient to zero, and thence obtain corresponding values of x: we must then keep differentiating the function until, for each of these values of x, we arrive at a differential coefficient which does not vanish: if, for any one of these values of x, this final differential coefficient is of an even order, the corresponding value of f(x) will be a maximum or a minimum accordingly as the final differential coefficient is negative or positive. If the final differential coefficient is of an odd order, the corresponding value of ƒ(x) will be neither a maximum nor a minimum. (x) being an essentially positive factor. Take then, instead Hence x = + 1 makes y a maximum, and x = the roots of this equation are 1, 2, 3. Now Thus we see that for the values 1, 2, 3, of x, y is respectively a minimum, a maximum, a minimum. Hence x = 0 gives for y a minimum value 4. where (x) = (x − a)3, a quantity essentially positive. Instead, therefore, of dy or f'(x), we may take dx If therefore c be a positive quantity, x = a makes y a minimum; if c be a negative quantity, x = a makes y a maximum. From this and the preceding example it appears that, although either f(x), or its derived functions of sufficiently high orders, may become infinite for a value of x which makes ƒ(x) a maximum or a minimum, yet, if we replace f'(x) by an appropriate function (x) which has always the same sign as ƒ'(x), we may often apply with advantage the rule of Art. (73) for finding such a value of x. Abbreviation of Operation. 74. Suppose that, for a certain function f(x) = y, u being a factor which vanishes when x = a, while v remains and that all the differential coefficients of y of lower orders than the rth will be equal to zero. This consideration enables us to ascertain whether a particular value a of x, which makes dy dx = 0, corresponds to a maximum or a minimum value of y, without being driven to the necessity of obtaining the general expression Ex. 1. Suppose that y is such a function of x that dy dx = (x − 1)(x − 2) (x − 3) (x − 4), and suppose that we desire to know whether x dy dx = 0, corresponds to a maximum or to a minimum value of y. We have, if x = 1, day = (x − 2) (x − 3) (x − 4) = (-) (-) (-) = -: which shews that x = 1 corresponds to a maximum value of y. d'y which shews that x = 1 corresponds to a maximum value of y. Maxima and Minima of implicit Functions of a single Variable. 75. In the preceding articles we have investigated the method of finding the maxima and minima of an explicit function of a single variable. We proceed now to the consideration of those cases in which the function is involved implicitly with its variable. Let y be a function of x by virtue of an equation |