Differentiability, Intervals, Inflection Points of Piecewise FunctionDate: 05/13/98 at 19:47:26 From: Cathy Emineth Subject: AP Calculus: Integral Okay, this is my problem (from an AP exam): Let f be a function defined by f(x) = {2x - x^2 for x < or = 1 {x^2 + kx + p for x > 1 (a) For what values of k and p will f be continuous and differentiable? (b) For the values of k and p found in (a), on what interval or intervals is f increasing? (c) Using the values of k and p found in part (a), find all points of inflection of the graph of f. Support your conclusion. Yuck, right? This is an AP essay question, which means that all work MUST be shown and clearly labeled. For starters, I set x^2 + kx + p = 1. I found that this worked when k = 1, p = -1, or vice versa. Then all values greater than one work in this equation. For part (b), I drew a blank. Help! This is pre-1992, which means no calculators of any sort are allowed. My class has been getting one of these lovelies every week since the beginning of the year. I consider this unfair, because I never planned on taking the AP exam, but AP Calculus is the only course offered in my school for the senior year besides Senior Math. Date: 05/16/98 at 09:07:13 From: Doctor Schwa Subject: Re: AP Calculus: Integral Thanks for letting me know how you started working on the problem. That helps me understand what you're thinking about and lets me write an answer that (I hope!) is more useful to you. For a function to be continuous, as you point out, x^2 + kx + p must equal 1. But at what value of x? The transition between the two "pieces" of the function happens at x = 1. So that's where the values must match. You correctly plugged in x = 1 to find 2x - x^2 = 1, but you also have to plug in x = 1 at the other end, so that the limit as x->1 from below equals the limit as x->1 from above, in order for the function to be continuous. The function also has to be differentiable, so the slope from below must equal the slope from above. That is, the derivative 2 - 2x, which approaches 0 as x->1, must equal 2x + k, which approaches 2 + k as x->1. That's enough information to solve for k and p, and there is only one possibility. (You have two facts, that the function must match up and the slope must match up; and you have two variables, k and p, which is usually enough information to solve the problem completely). Another way to think about this whole thing that doesn't require as much calculus is to think about (2x - x^2) as a downward-opening parabola that has just hit its maximum (vertex) at x = 1. So to continue it with the upward-opening parabola x^2 + kx + p, we have to have its vertex also at the same point (1,1), which gives you another way to find values for k and p. Now we could find where it's increasing by taking derivatives: 2 - 2x when x < 1 2x - 2 when x > 1 and then check where the first derivative is positive. You might notice at x = 1 that the derivative is zero. Does that mean it's not increasing there? It depends on how you define "increasing." There was some interesting talk about this on the ap-calc mailing list recently; let me know if you're interested in talking more about the various possible ways of defining an increasing function. You could also use the graph of these two half-parabolas to figure out when it's increasing (and again, the vertex point (1,1) will be a special case, depending on your definition of "increasing"). For part (c), you can take the second derivative and see if there are any points where it changes sign. Those would be inflection points. I find it convenient to draw a number line to keep track of the intervals where the second derivative is positive or negative. It is a shame that there aren't other choices of courses, such as the new AP Statistics. What is Senior Math like? Is the problem that there's nothing of intermediate level between that and the AP calc? At my school, Gunn HS in Palo Alto, CA, we have AP Stat, Calc AB, and Calc BC, as well as Precalculus, which is probably pretty similar to your "Senior Math." -Doctor Schwa, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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