If all the points P1, P.,・・・ P, P ⚫ are connected with all the points found by assigning all possible values to x between 0 and 1⁄2, 1 and 1, and so on, the graph of the curve in question (Fig. 12) is obtained. In accordance with the definitions of minimum and maximum, the quantity y will be a minimum at P, and a maximum at P'; because for values of x less than and greater than 2, the values of y are greater than 4, and for values of x less than and greater than – 2, the values of y are algebraically less than — 4. 471. PROBLEM IX.-When will rational integral fractions of the form, ax+br+c b1x+ci y= Further, Ac Ak + A(x − ); hence, for (1), y = B + Ak + A ( x − k + C/4). The variation of an expression of the form, x has already been studied in Problem VIII where p2 would equal A and xx-k. 472. EXAMPLE. The complete reduction of a problem of this character will be illustrated by the example, The equation may be written in the last form of equation (2) as follows: y = 4+ (x-1)+ 2 X- 1 The least positive value which u can have in order that the values 8 or u= 212, and the least negative of x 1 may be real is u2 value is u = — 21/2; the corresponding values of x are x = 1+12 The former corresponds to a minimum and the and x 1-1/2. = latter to a maximum, because, as x increases from x— — ∞ to ∞ to x = 1−√2, u increases from 212; and, as x increases from x = 1-1/2 to x = 0, u decreases from 21/2 to moreover, as x increases from x = 0 to x = 1, 2 х 1 -3; is negative and u will decrease from 3 to oo. Hence, the ordinate u = QA = −21/2 is a maximum. Therefore the maximum value of y is y =4+u=4 — 2√/2 = 2(2−√2). Similarly, as x increases y = 4+ u = 4+2√2 = 2 (2 + √2) is the minimum value of y. The results of this discussion are given in Figure 13. 473. PROBLEM X.-The maximum and the minimum values in the general case may now be determined when The solution of this problem can be reduced to the solution of the problem of the preceding section. By division it is possible to transform (1) into where A and B involve given coefficients of equation (1). Now the study of the variation of y will depend upon the variation of a the corresponding values of y will be found by adding to the values of z. It is sufficient, therefore, to study the values of z when x varies from ∞ to +∞. When is very large, equation (3) may be written The limit of the value of the numerator, as x becomes very large, is A, while the denominator becomes as large as is desired. Therefore, the fraction becomes very small and begins with the value 0 when x=- only to return to 0 when x = +. It is not clear how the value of z varies in the interval between x = ∞ and x = ∞ ; but this may be discovered by proceeding as follows: put v is the quotient of a trinomial of the second degree by a binomial of the first degree. It has already been learned in Problem IX how to determine the variation of such a fraction. After it has been determined when v is a maximum or a minimum, it is possible to determine readily when z is a maximum or a minimum. Since vz = 1, z and must have the same signs. When the absolute value of v increases, that of z decreases, and conversely. Therefore, a maximum value of z corresponds to a minimum value of v and a minimum of z to a maximum of v. 474. This discussion will now be illustrated by two numerical examples. Let EXAMPLE 1. It may be observed first that the denominator has no real roots and therefore is not zero for any real value of x, Write y in the form, and let now vary from ∞ to +∞; then y begins with the value 2 and returns to the same value and remains finite during this interval, since no real value of x will make the denominator of y zero; therefore, the fraction y will pass through a maximum and a minimum. In order to determine these, it is necessary to apply the general method. On dividing the numerator by the denominator the result is 201 8.x-19 Here, u will be a maximum or a minimum according as 8x-19+is a maximum or a minimum. As in problem VIII let Therefore u will be a maximum or a minimum according as and the corresponding values of x will be 8x-19=1/201; The fraction u passes through a maximum for x, and through a minimum for = 4.147. Therefore z, and consequently x2 y, has a minimum for the value x, and a maximum for x. The corresponding values of z, i. e., 1 z, and z are and OM2 = + V §. Find the maximum and minimum values of and trace the variations of the fraction when The condition that the roots of the equation are real is Y 2 = 13+1145 = 12.52, which are correct within 0.01. |