00010001010008main.ACT0001020013differentiation.EAC010000007160Éø‹^tÓ×Ú[Differentiation and [its Applications\AuthoráखMa. Louise Antonette De Las Penas,Phd Associate Professor Ateneo de Manila UniversityQuezon City, Philippines e-mail:mlp@mathsci.math.admu.edu.ph ÿ[ Objective:[In this activity, we [study the consequences[of the Mean Value [Theorem and Rolle's [ Theorem on [Differentiable Functions [from an analytical,[graphical and geometric[perspective. [(See notes below for [more on the Mean Value[Theorem and Rolle's [Theorem)[[ Example 1:[1.Let f(í¸)=í¸3+í¸2-í¸. [Verify that the mean [value theorem holds for[the given function on [[-2,1].[ Solution:[a)Using a graphical [approach,let us [determine the slope of [the secant line [containing the points [(-2,f(-2)),(1,f(1)).[Drag f(í¸) to the ["Graph Window" below[First,tap Analysis-GSolve[-YCal to calculate the [y-values corresponding [to í¸=-2 and í¸=1. [We obtain f(-2)=-2 and[f(1)=1.[To draw the secant line,[tap Analysis-Sketch-Line.[Then press 1 and enter[the points (-2,-2),[(1,1). [The equation of the [secant line is y=í¸.\Graph WindowîÌÊÀGraph2D, †Graph3D@ †LISTSYSL4N†Modify €$ †STATCALC ¤N†STATSYS ¬\N†Sequence, †Sheet4| †Sheet3D°| †SolveEq,†SolveLwr0 †SolveUpr< †StupFLG1H(†StupListpD †StupPict´$ †ViewWindØ †äô †ä †ä †ä †ä$ †ä0 †ä< †äH †äT †ä` †äl †äx †ä„ †ä †äœ †ä¨ †ä ´ †ä!À †ä"Ì †ä#Ø †ä$ä †ä%ð †ä&ü †ä' †ä( †ä) †ä*, †ä+8 †ä,D †ä-P †ä.\ †ä0h †ä1t †ä2€ †ä3Œ †ä4˜ †ä5¤ †äE° †äF¼ †äHÈ †äIÔ †äJà †äKì †äLø †äM †äN †äO †äP( †äQ4 †äR@ †äSL †äTX †ä]d †ä^h †ä_l †ä`p †äat †äbx †ä”| †ä•ˆ †äÍ” †äΠ †äЬ †systemä]listsystemä^]=systemä_ä^systemä`ä_systemäaä`systemäb~‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ‰Ñ aseq_histbNewFolde systemä]listsystemä^]=system‰Ñsystemä]listsystemä^]=system‰Ñsystemä]listsystemä^]=system‰Ñsystemä]listsystemä^]=system‰Ñsystemä]listsystemä^]=system‰Ñsystemä]listsystemä^]=system‰Ñsystemä]listsystemä^]=system‰Ñsystemä]listsystemä^]=system‰Ñsystemä]listsystemä^]=system‰Ñsystemä]listsystemä^]=system‰Ñ  0@PSheet1”øÀšŒ^Sheet2”øÀšŒ^Sheet3”øÀšŒ^Sheet4”øÀšŒ^Sheet5”øÀšŒ^SheetSheet3D 0@PSheet1蚌^Sheet2蚌^Sheet3蚌^Sheet4蚌^Sheet5蚌^SheetSheet3D i™ ™ø‹^tÓ× $ `P`P`]`]] ™$gS$gS ˜]`]]](1…0qy`]#Y‡uYƒ] ˜@ ™ 6Y˜`W‰G6„! ˜O[[b) Next, we use a [geometric approach to [verify the existence of [a number in the interval[(-2,1) that satisfies the[conclusion of the Mean[Value Theorem for the [secant line obtained in a)[We draw f(í¸) in the [Geometry Window and [construct the tangent [line to the curve.To [animate the tangent line,[Click Edit-Animate-[ Go(Once).[From the animation, we[conjecture that the [tangent line assumes a[slope of approximately 1[twice in the whole cycle.[To verify this,we [generate the table of [slopes for the tangent[line.[Tap point on the curve,[select table;then tap [the tangent line,slope [and select table.\Geometry WindowîÎÎ`I™V`8`I™V`8vrALCuxy5Ä@  ÅH!ƒ C@Ð6–°WP`Hd% ™ÈÀ  x^3+x^2-xWP``W‰G6„! ™.Ë@! 3ÊÀ!  ÉÀ# @Ð6–° V‚xCiY™eRW#p ™U g` .É@#  .È@    1ÄH  9 [[Based on the table of [slopes generated, we [conclude that at [approximately í¸=-1.2 [and í¸=.53, the slope [of the tangent lines is [approximately 1.[[x values slope[-2 7.000125002 -1.842105263 5.495958033 -1.684210526 4.141375563 -1.526315789 2.936377547 -1.368421053 1.8809641 -1.210526316 0.97513501 -1.052631579 0.21889051 -0.8947368421 -0.387769529 -0.7368421053 -0.8448450857 -0.5789473684 -1.152336147 -0.4210526316 -1.310242726 -0.2631578947 -1.318554294 -0.1052631579 -1.177268214 0.05263157895 -0.8863976443 0.2105263158 -0.4459425882 0.3684210526 0.1440969547 0.5263157895 0.8837209872 0.6842105263 1.7729295 0.8421052632 2.8117225221 4.000099977[[c)Analytically,we can [calculate the existence[of these numbers by [solving for í¸ in the [equation f'(x)=1.[We obtain the result [í¸=-1.2, í¸=.54 in the [calculation row:R solve(diff(í¸3+í¸2 -í¸)=1,í¸)Rx=- 1.215250437,x= 0.5485837704[Note that there is a [slight difference in the[values obtained due to[numerical approximations.[d)To visualize the [conclusion of the Mean[Value Theorem, we draw[f(í¸), the tangent lines[at í¸=-1.2, í¸=.54,[together with the [secant line in the "Graph[Window".[Tap Analysis-Sketch-[Tangent; enter 1,[í¸=-1.2,í¸=.54 to draw[the tangent lines.[ [ Example 2. [Find an interval [íg,íh] on[which [f(í¸)=í¸4+í¸3-í¸2+í¸-2[satisfies the conditions [of Rolle's Theorem.[ Solution:[a)Graphical and Tabular [ Approach:[Click "2 Graph Window" [below and tap Graph [Button.[Tap Analysis-GSolve-[Root to determine the [roots -2 and 1 of f(í¸).[Thus, the interval we [desire is [-2,1].[Now, to determine the[point satisfying Rolle's[Theorem, we click the[Table Button to generate[the table of slopes.[We conjecture from the[table that f'(x)=0 [somewhere in the interval[ (-1.3,-1.25).[To confirm this [information, we drag [f'(í¸) to "2 Graph [Window" and determine its[root by tapping Analysis-[ GSolve-Root.[We obtain x~-1.29.[Note that f'(x) is [obtained from the [calculation row as [follows:Rdiff(í¸4+í¸3-í¸2 +í¸-2,í¸)R4î’x3+3î’x2-2î’x+1\2 Graph WindowÌÊ€Graph2DT †Graph3Dh †LISTSYSt4N†Modify ¨$ †STATCALC ÌN†STATSYS Ô\N†Sequence0, †Sheet\| †Sheet3DØ| †SolveEqT†SolveLwrX †SolveUprd †StupFLG1p(†StupList˜D †StupPictÜ$ †ViewWind †y1$H†ä@ †äL †äX †äd †äp †ä| †äˆ †ä” †ä  †ä¬ †ä¸ †äÄ †äÐ †äÜ †äè †äô †ä  †ä! †ä" †ä#$ †ä$0 †ä%< †ä&H †ä'T †ä(` †ä)l †ä*x †ä+„ †ä, †ä-œ †ä.¨ †ä/´ t †ä0( †ä14 †ä2@ †ä3L †ä4X †ä5d †äEp †äF| †äHˆ †äI” †äJ  †äK¬ †äL¸ †äMÄ †äNÐ †äOÜ †äPè †äQô †äR †äS †äT †ä]$ †ä^( †ä_, †ä`0 †äa4 †äb8 †ä”< †ä•H †äÍT †äÎ` †äÐl †systemä]listsystemä^]=systemä_ä^systemä`ä_systemäaä`systemäb~ÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕ aseq_histbNewFolde systemä]listsystemä^]=systemÕsystemä]listsystemä^]=systemÕsystemä]listsystemä^]=systemÕsystemä]listsystemä^]=systemÕsystemä]listsystemä^]=systemÕsystemä]listsystemä^]=systemÕsystemä]listsystemä^]=systemÕsystemä]listsystemä^]=systemÕsystemä]listsystemä^]=systemÕsystemä]listsystemä^]=systemÕ  0@PSheet1”øÀšŒ^Sheet2”øÀšŒ^Sheet3”øÀšŒ^Sheet4”øÀšŒ^Sheet5”øÀšŒ^SheetSheet3D 0@PSheet1蚌^Sheet2蚌^Sheet3蚌^Sheet4蚌^Sheet5蚌^SheetSheet3D i™ ™ø‹^tÓ× $  3xx^4+x^3-x^2+x-2`P`P`<`<<$gS$gS ˜<``<<<(1…0qy`<#Y‡uYƒ< ˜dú`dúdú ˜= $0<HT`lx„œ¨´ÀÌØäðü ,8DP\ht€Œ˜¤°¼ÈÔàìø(4@LXdp|ˆ” ¬¸ÄÐÜèô $0<HT`lx„œ¨´ÀÌØäðü ,8DP\ht€Œ˜¤°¼ÈÔàìø(4@LXdp|ˆ” ¬¸ÄÐÜèô $0<HT`lx„œ¨´ÀÌØäðü ,8DP\ht€Œ˜¤°¼ÈÔàìø(4@LXdp|ˆ`P†H`•`6‡PY™3Q``3i`™™™„`…`‰u`YI™‘`€`7D` y™™u`u`y)hu`u•`p` `X `e`E&u`Pe```p$`P9™™Q‡`U`Chu`X€‚`P`%`t™™–ƒ`E`u`˜i™˜gp`@`&$`)Y™—@`5`1hu`t5€Y™0`3 ` v4Y™%`2Bu`u1v ™ `)D`"DU ™`$Chu`9™–G8`i`PY™™€` Fu`wp8``™™™˜xE PY™‰Shu`pW Y™x)`19™ˆ™PY™fFu`AY™T$`G %PY™Ayhu`P ™…bY™))`I€PPY™†u`FThY™d`A`ftPY™’shu`4$™—tY™%`%afPY™p&u`)™™–wY™Y„`9™…fPY™Phu`‰`6IY™@‰`v™‡“5PY™2Bu`bPDSY™$d`Hy™”T‚PY™Shu`54™—‡bY™ `"`XY˜&u`™˜u•t` ™™™“’% ™ ˜•#hu`  cG ™ ™‰`3™™€u ™P ™††u`I™u„@ ™ ™ƒ`R ™P ™y)hu`P™™(‘ ™ ™uI`w™™—u… ™P ™qFu`9I™• ™ ™g` 6U$ ™P ™bhu`™™˜9 ™V%`%4%P ™IFu`G4™˜†‘ ™AD`t9™”DYP ™1“hu`T™™2d ™ i`D f!P ™Bu`‡D™—„ ™ @Y™6y™™P ™66‡PY™’@8 ™$Y™T`f P ™€a‡PY™#p%#•(…„4h"``` ™Q[[b)Analytical Approach:[In the calculation row, [we solve f'(x)=0 to [obtain í¸~-1.29. R solve(diff(í¸4+í¸3-í¸2 +í¸-2)=0)Rx=- 1.288584347,x= 0.2692921734- 0.3485585806î’î,x= 0.2692921734+ 0.3485585806î’î[[Exploratory Exercise:[Verify if the function [f(í¸)=í¸+1í¸-1, [0,2][satisfies the hypotheses[and conclusion of the [Mean Value Theorem on[the given interval.[[ Solution:[a) Graphical and Tabular[ Approach:[Click the "Exploration [Window" below.[Tap the Graph button to[verify that f(í¸) is not[continuous at [0,2].Then [tap the Number Table or[Summary Table button to[verify that f'(1) does [ not exist.[This shows that the [Mean Value Theorem [does not hold for f(í¸) [on the given interval.\Exploration WindowÐÊ„Graph2Dh †Graph3D| †LISTSYSˆ4N†Modify ¼$ †STATCALC àN†STATSYS è\N†SequenceD, †Sheetp| †Sheet3Dì| †SolveEqh†SolveLwrl †SolveUprx †StupFLG1„(†StupList¬D †StupPictð$ †ViewWind †summary0´ †y1ä H†ä †ä †ä †ä( †ä4 †ä@ †äL †äX †äd †äp †ä| †äˆ †ä” †ä  †ä¬ †ä¸ †ä Ä †ä!Ð †ä"Ü †ä#è †ä$ô †ä% †ä& †ä' †ä($ †ä)0 †ä*< †ä+H †ä,T †ä-` †ä.l †ä/x´ †ä0, †ä18 †ä2D †ä3P †ä4\ †ä5h †äEt †äF€ †äHŒ †äI˜ †äJ¤ †äK° †äL¼ †äMÈ †äNÔ †äOà †äPì †äQø †äR †äS †äT †ä]( †ä^, †ä_0 †ä`4 †äa8 †äb< †ä”@ †ä•L †äÍX †äÎd †äÐp †systemä]listsystemä^]=systemä_ä^systemä`ä_systemäaä`systemäb~ÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕÕ aseq_histbNewFolde systemä]listsystemä^]=systemÕsystemä]listsystemä^]=systemÕsystemä]listsystemä^]=systemÕsystemä]listsystemä^]=systemÕsystemä]listsystemä^]=systemÕsystemä]listsystemä^]=systemÕsystemä]listsystemä^]=systemÕsystemä]listsystemä^]=systemÕsystemä]listsystemä^]=systemÕsystemä]listsystemä^]=systemÕ  0@PSheet1”øÀšŒ^Sheet2”øÀšŒ^Sheet3”øÀšŒ^Sheet4”øÀšŒ^Sheet5”øÀšŒ^SheetSheet3D 0@PSheet1蚌^Sheet2蚌^Sheet3蚌^Sheet4蚌^Sheet5蚌^SheetSheet3D i™ ™ø‹^tÓ× $  $(48DHTXdlxˆ”˜¤°¼ÀÌÜèðü ,8DHTdpx„” ¤°¼ÈÌØäxÙððû-7.7 Æp`5`1Ù57.7pfíá(x)ððû-0.02642356982d#V˜ ˜Y-ÙððûUndefined€-Ùððû-0.04455335264ES5&@˜Yfíâ(x)ððû-6.074383868î-3impCƒ†€—Y-ÙððûUndefined€+Ùððû0.013299508252™P‚P˜ f(x) Æððû0.7701149425-3pIBP™ î•ððûUndefined€î–ððû1.298507463)…F0 3x(x+1)/(x-1)`P`P`p`p ™€`€(1…0qy`#Y‡uYƒ ˜IIPI ™  $0<HT`lx„œ¨´ÀÌØäðü ,8`D˜`P ™fffffg`UP —` ™`!x5`P ™`™™™ƒ`€€Ð?ëúÀÄúÀ% ™™™ƒ`P`ufffffgUT™˜3!`™™™˜xE`` ™R[[b)Analytical Approach:[Using a similar solution [given in Example 1a), we[obtain the equation of [the secant line [containing (0,-1) and [(2,3) as y= 2x-1.[Using the calculation row[we check that the [solution to f'(x)=2 is [not real as follows:R solve(diff(í¸+1í¸-1)=2,í¸)Rx=1-î,x=1+î[Moreover, we can verify[in the calculation row [that the tangent line to[graph of f(í¸) at í¸=1[does not exist:[RtanLine(í¸+1í¸-1,í¸,1)R Undefined[This confirms the fact [that the Mean Value [Theorem does not hold[for f(í¸) on [0,2]. [\ NotesîîáàhZMean Value Theorem: Let f be a function that is a)continuous on [a,b] b)differentiable on (a,b) Then there is a number c ík (a,b) such that f'(c)= (f(b)-f(a))/(b-a) Rolle's Theorem: Let f be a function that is a)continuous on [a,b] b)differentiable on (a,b) If f(a)=f(b)=0,then there is at least one number c in (a,b) such that f'(c)=0. ÿ[eActtable1 Ô= $0<HT`lx„œ¨´ÀÌØäðü ,8DP\ht€Œ˜¤°¼ÈÔàìø(4@LXdp|ˆ” ¬¸ÄÐ`•``…`€`u`p`e```U`P`E`@`5`0`%` ````` PY™ Y™PY™Y™PY™Y™PY™Y™PY™Y™PY™Y™PY™Y™PY™Y™PY™Y™Y˜ ˜ ™P ™ ™P ™ ™P ™ ™P ™ ™P ™ ™P ™ ™P ™ ™P ™ ™ P ™