Max and Min of Functions without DerivativeDate: 04/22/2003 at 11:36:29 From: Roy Subject: Max and Min of functions without derivative Hello Dr. Math, I was curious to know if there is a general way to find the max and min of cubic functions without using derivatives. Our book does this with the use of graphing calculators, but I was wondering if there is a way to find the critical points without derivatives. Any help is greatly appreciated! Thanks, Roy Delgado Date: 04/22/2003 at 12:36:43 From: Doctor Peterson Subject: Re: Max and Min of functions without derivative Hi, Roy. Yes, there is a way, and it may be very instructive! Look at the graph of a cubic, and recall that if a polynomial has a double root, it will be tangent to the x-axis (at q here): | * | | | * | * * * | | * * * | | * * ------*-------------+------*--------------- p | q | | | | * | | | | | So given a general cubic, if we shift it vertically by the right amount, it will have a double root at one of the turning points. So, given an equation y = ax^3 + bx^2 + cx + d any turning point will be a double root of the equation ax^3 + bx^2 + cx + d - D = 0 for some D, meaning that that equation can be factored as a(x-p)(x-q)^2 = 0 Equating these and expanding, we have a(x-p)(x-q)^2 = ax^3 + bx^2 + cx + d - D a(x-p)(x^2 - 2qx + q^2) = ax^3 + bx^2 + cx + d - D a(x^3 - (p+2q)x^2 + q(2p+q)x - pq^2) = ax^3 + bx^2 + cx + d - D ax^3 - a(p+2q)x^2 + aq(2p+q)x - apq^2 = ax^3 + bx^2 + cx + d - D Since these must be equal for ALL x, we can equate coefficients: a = a -a(p+2q) = b aq(2p+q) = c -apq^2 = d-D We have three equations (ignoring the first, which is simply the reason I multiplied my factored form by a in the first place) in three unknowns, p, q, and D, so we can solve to find the double root q, which is the location of the turning point. To do this, we'll eliminate p by solving the second equation above for p: p = -(b/a + 2q) and putting this into the third equation: aq(-2(b/a + 2q) + q) = c This simplifies to -2bq - 3aq^2 = c 3aq^2 + 2bq + c = 0 (Note that this is the derivative of the cubic we are working with. The rest of the work is just what we would do if we were using calculus, but with different reasoning.) Now we solve this for q using the quadratic formula: -2b +- sqrt(4b^2 - 12ac) -b +- sqrt(b^2 - 3ac) q = ------------------------ = --------------------- 6a 3a This gives both turning points, since there are two ways to make a double root. Our last equation gives the value of D, the y-coordinate of the turning point: D = apq^2 + d = -a(b/a + 2q)q^2 + d = -2aq^3 - bq^2 + d = (aq^3 + bq^2 + cq + d) - (3aq^2 + 2bq + c)q = aq^3 + bq^2 + cq + d (since 3aq^2 + 2bq + c = 0), as we would expect given that x = q; so we don't really have to carry out this step. This is the sort of work that had to be done before calculus was invented. In many cases, calculus is really just a shortcut for algebra. Thanks for asking the question, because I'd never considered approaching it without calculus before! If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 04/22/2003 at 12:47:08 From: Roy Subject: Thank you (Max and Min of functions without derivative) Thanks for the quick response, Dr. Peterson. That is just what I was looking for! I think my students will definitely benefit from seeing this approach and appreciate it once they get into calculus. Roy Delgado |
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