the first case the dc (remembering the definition) is positive, and in the second case negative. IOI. Again, in order that a function may have a maximum or minimum, it is obvious, from what has been said, that the function must first increase and then diminish, or first diminish and then increase; and therefore in either case the dc must change its sign. 102. In order that any quantity, which is varying continuously may change its sign, it is evident that it must pass through the value 0, from positive to negative, or from negative to positive; and, therefore, in order that there may be a maximum or minimum the de must be equal to 0. In other words, when a function reaches one of its greatest or least values it neither increases nor diminishes, at that instant, and therefore its rate of variation is 0, and therefore dc rate of variation of function = rate of variation of variable 0 rate of variation of variable =0. We have, then, a relation from which we may find the value of the variable which produces this maximum or minimum. 103. Suppose we have an expression or function 8+6x-x2, and we wish to find for what value of the variable x it will be a maximum or a minimum. We know that de must be 0, in order that there may be a maximum or minimum. But dc-6-2x, x=3; and for this value of x the function 8+6x-x2=8+6×3-32 =8+18-9 =17; and this is, therefore, a maximum or a minimum : we have to determine which. Now, if we substitute in the function values a little larger and a little smaller than 3, we shall see whether the values immediately on either side are both greater or both less than 17. If x=1, function = 13; From this we see that for the value 3, the function has a value, which is greater than those immediately on either side of it, and therefore this value of the function, namely 17, is a maximum. 104. We might have arrived at this conclusion equally well by substituting these values, 1, 2, 3, etc., in the de; for since the value of the dc must change sign-i.e., pass through the value 0-we may see, by substituting these values, whether it is passing from positive to negative, in which case the function must be a maximum; or from negative to positive, in which case the function must have attained a minimum value. From this we see that the dc has passed from positive to negative, and therefore the value of the function given by the value 3 of the variable is a maximum. 105. Let us take another example. Suppose the function to be x3-9x2+24x-7; we wish to find what value of the variable makes this a maximum or a minimum, and what is the value of that maximum or minimum. x-3=±1, therefore, x=4 and x=2 are the two solutions. Now let us substitute, as before, in the function, numbers immediately larger and immediately smaller than these, and also in the dc. If x=1, function = 9, and dc=+3; From this we see that when x=2 the function is a maximum; since, firstly, the dc passes from positive to negative; and, secondly, from the values of the function which immediately precede and immediately succeed the value of the function when x=2. Similarly we see that when x=4, the function is a minimum. 106. Now dc is the de of de (see Art. 37), that is, it gives the rate of variation of de1; and, when the function is a maximum, it has been increasing and is about to decrease, and the rate of its variation, which is given by dc1, has been decreasing until the function arrives at the maximum, and then de1 =0. Therefore the dc, must itself have been receiving negative increments, and therefore dc2, which gives its rate of variation, must be negative. Similarly for a minimum, dc, must be a positive. So that we have a third method of testing whether the function be a maximum or minimum, for the particular value of the variable, provided it has a dc2 which does not vanish. In the first case which we considered function=8+6x − x2, which shows that the function has a maximum value. In the second case which we considered function=-9x2+24x-7, dc1 =3x2-18x+24, dc2=6x-18. =2, we have Substituting in de2 the value x= dc2-12-18=-6; and therefore, as before, for the value 2 of the variable the function has a maximum value. Again, substituting the value x=4, we have dc2=24-18=+6; and therefore, as before, for the value 4 of the variable the function is a minimum. 107. We will conclude this part of the subject with one more example. "Divide a straight line into two parts, so that the rectangle contained by the parts may be a maximum.” Let a be the straight line, and x one of the parts, a-x=other part, the rectangle(α-x)x dc1 = a -2x, From the sign of dc, we see that there is a maxi XX. The Tangent to a Curve. 108. Let OPQ be any curve, and P a point on it. Then the ratio of PM: OM will give the position of the point P, and likewise the tangent of the angle which the chord OP makes with OX. Suppose a point to be taken in the curve near to P, viz., P'; then, if P'M' be drawn parallel to PM (which is perpendicular to OX), and PÑ be drawn parallel to OX, the ratio of P'M': OM' gives the position of P'; and if OPQ, instead of being a curve, were a straight line, the ratio P'M' : OM' would be equal to PM: OM, i.e., the straight line PP' would pass through 0, if produced. PP' T IN M M Now if we know the position of P, the position of P' and the direction of the chord PP' are determined by the ratio of P'N: PN, which is also the tangent of the angle which the chord PP' makes with PN or OX. Let P' move up gradually towards P, and eventually coincide with it, then the angie P'PT ultimately vanishes (see Art. 57) and the directions of the arc, chord, and tangent are the same, and are identical with that of the tangent; and the tangent of the angle which the tangent at P makes with PN or OX is represented by the ratio of the very small increase in PM to the very small increase in OM, |