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# Class Note for MATH 125 with Professor Rychlik at UA

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This 2 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Arizona taught by a professor in Fall. Since its upload, it has received 18 views.

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Date Created: 02/06/15
A cubic function with two critical points BY MAREK RYCHLIK Lecture of November 3 2008 Example 1 Let us consider the function yfzzg7127312 Find its critical points in ection points and intervals of monotonicity Determine how many roots the function will have Solution First we nd the critical points by solving the equation y 312721730 This is a quadratic equation A refresher from algebra the equation a12bx c0 has 2 1 or zero solutions depending on the sign of the discriminant A b2 7 4 a c If A gt 0 then there are two roots given by formulas 7bA 11 2a 7b73 12 2a For our case A 7 22 7 4 3 7 3 40 Hence the equation has two roots m 7 072025922 11 12 m 1887425887 MM T Thus these are the two critical points Moreover y gt 0 for z lt x1 and for z gt I and y lt 0 for 1 between 1 and I all because the coe icient at x2 is positive The graph of y which is a parabola crosses the z axis twice and has a global minimum at z g the in ection point of y f I The in ection point is determined from the equation y 61720 1 which gives 13 E 11 In summary the function is strictly increasing there are no critical points and there is one in ection point at z The in ection point is at the minimum point of the rst derivative Thus the function changes most slowly at the in ection point Let us discuss the number of roots that this function has a There can be only one root on each of the intervals 7 00 11 11 13 and 13 12 because on each of them the function is monotonic either monotonically increasing or decreasing b If the critical value at the second critical point u is negative then there will be three roots because the function changes sign on each of the intervals in a On the other hand if the value u is positive then there is no change of sign on the second and third interval and thus there is only one root However u is hard to evaluate by hand exactly so we compute fl 7 1 Since fzg lt fl we have also shown that fzg lt 0 Thus there are three rootsiThus we evaluate f Below is a graph obtained with a free program called GNUplot We picked our window care fully to contain both critical points and the in ection point By a haphazzard pick of the window we may be less successful in confirming the properties of the function that we deter mined GNUplot plot x 1522 fxx3 x2 3x2 fX 35 15 15 1 05 0 05 1 15 2 GNUplot

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