. Differential and integral calculus, an introductory course for colleges and engineering schools. through which pass twobranches of a curve is termed a double point. At a double pointthe curve has two tangents, one to each branch, and consequentlyy has two values at such a point. If three branches, four branches,... of a curve pass througha point, that point is called atriple point, quadruple point,. . , and at such a point yfhas three, four, . . generic name for doublepoint, triple point, ... is mul-tiple point. It is not our pur-pose to present here a generalmethod for finding th
. Differential and integral calculus, an introductory course for colleges and engineering schools. through which pass twobranches of a curve is termed a double point. At a double pointthe curve has two tangents, one to each branch, and consequentlyy has two values at such a point. If three branches, four branches,... of a curve pass througha point, that point is called atriple point, quadruple point,. . , and at such a point yfhas three, four, . . generic name for doublepoint, triple point, ... is mul-tiple point. It is not our pur-pose to present here a generalmethod for finding the mul-tiple points of a curve from itsequation. That subject be-longs to a more advancedcourse in mathematics. Sometimes, however, the multiple pointspresent themselves very simply, as is the case in the followingexamples and in some of the exercises. 86. 1. y2 = x2 {x + 3). (a) The curve is symmetrical as to OX but not as to OY. It has noreal points at the left of the line x = — 3. (b) Solving for y, we have y = ±x(x+ 3) by differentiation 3 x+2 „.„ , 3 x + 4. Quadruple Point 2 (3+3)* y = ±s 40c + 3)t 115 116 DIFFERENTIAL CALCULUS §86 When y = 0, x = —4, but y is imaginary and there is no flex. Wheny = oo, x = — 3, but y does not change sign at x = — 3, and there is noflex. Hence the curve has no flexes at all. (c) Consider now the branch for which the radical is +, y = x(x + 3)*, y = 3 x+2 y 3 z+4 2(x+3)2 4(x+3)f When x is j ]M, 2/ is [ M j which means that this branch lies wholly in the first and third quadrants, y is + for all admissible values of x (values greater than —3), and there-fore the branch is convex throughoutits whole extent. When y = 0, x= —2, y = — 2, andy is +. Hence (—2, —2) is a mini-mum point. When y = oo, x = — 3,but y does not change sign here and(—3,0) is neither a maximum norminimum point. But the branch3, 0) and the tangent there is perpendicular to-3. This branch passes through the origin with
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