Conditions for Real-valued RootsDate: 04/01/98 at 13:32:19 From: Heather Subject: Calculus Given the equation ax^3 + bx^2 + c = 0, use the first and second derivatives to help find conditions that ensure the equation has 3 distinct real-valued roots. So far I've figured this out.... f'(x) = 3ax^2 + 2bx = x(3ax + 2b) = 3ax + 2b = 0 3ax = -2b -2b x = ---- or 0 3a f''(x) = 6ax + 2b = 0 6ax = -2b -b x = --- 3a Also, a does not equal 0, and b does not equal 0; and f'(x) is a parabola and f''(x) is a line. So where do I go from here? Date: 04/01/98 at 16:41:41 From: Doctor Rob Subject: Re: Calculus f(x) has three real roots if and only if g(x) = f(x)/a, = x^3 + (b/a)*x^2 + c/a, = x^3 + B*x^2 + C, has three real roots. We know that for large enough positive x, g(x) > 0, and for small enough negative x, g(x) < 0. You have found the two critical points x = 0, y = C, and x = -2*B/3, y = (4*B^3 + 27*C)/27. Call these points P and Q, respectively. At these points, g''(0) = 2*B, and g''(-2*b/3) = -2*B. Thus one is a local maximum, the other a local minimum, by the appropriate second derivative test. A root means that the curve y = g(x) crosses the axis, that is, changes sign. Moving from large values of x where g(x) > 0, to smaller values, we must cross the axis, reach a local minimum, cross the axis again, reach a local maximum, then cross the axis a third time, before reaching the very small values of x for which g(x) < 0 is guaranteed. There are two cases, depending on the sign of B. Case 1: B > 0. Then P is the local minimum and Q is the local maximum. Then there must be a zero between P and +infinity, one between P and Q, and one between Q and -infinity. That means that C < 0 and (4*B^3+27*C)/27 > 0. The last condition is equivalent to 4*B^3 + 27*C > 0, or C > -4*B^3/27. Thus B > 0 > C > -4*B^3/27. Case 2: B < 0. Then Q is the local minimum and P is the local maximum. Then there must be a zero between Q and +infinity, one between P and Q, and one between P and -infinity. That means that C > 0 and (4*B^3+27*C)/27 < 0. The last condition is equivalent to 4*B^3 + 27*C < 0, or C < -4*B^3/27. Thus B < 0 < C < -4*B^3/27. In either case, B and C must have opposite signs, and B and 4*B^3 + 27*C must have the same sign. In terms of a, b, and c, b > 0 > c > -4*b^3/27*a^2, or b < 0 < c < -4*b^3/27*a^2. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 04/01/98 at 16:56:24 From: Doctor Anthony Subject: Re: Calculus The method is as follows. First find conditions for the quadratic f'(x) to have real, distinct roots. f'(x) = 3ax^2 + 2bx = 0 so x = 0 and x = -2b/3a Now you require the two turning points to be on opposite sides of the x axis, so (assuming a and b have same sign) we require f(-2b/3a) > 0 and f(0) < 0 so c < 0 If a and b have opposite signs, we require c > 0. a(-2b/3a)^3 + b(-2b/3a)^2 + c > 0 -8b^3/(27a^2) + 4b^3/(9a^2) + c > 0 (4/27)(b^3/a^2) + c > 0 4b^3 + 27a^2*c > 0 If a and b have the same sign, then c < 0 and we require b > 0 and |4b^3| > 27a^2*c If a and b have opposite signs, then c > 0 and 4b^3 + 27a^2*c < 0 So we require b < 0 and |4b^3| > 27a^2*c We can summarize as follows: if a and b same signs then c < 0 and |4b^3| > 27a^2*c if a and b opposite signs then b < 0, C > 0, and |4b^3| > 27a^2*c These would be sufficient conditions for three real roots. There is no need to use the second derivative as the above are sufficient. There is a point of inflection, given by f''(x) = 0, and this occurs at x = -b/(3a), i.e. halfway between the maximum and minimum points. -Doctor Anthony, The Math Forum Check out out web site! http://mathforum.org/dr.math/ |
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