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- Ch4 sec 4.6 application of differentiation james stewart pdf online

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SECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS |||| 315 68. Show that the curve has two slant asymptotes: and . Use this fact to help sketch the curve. 69. Show that the lines and are slant asymptotes of the hyperbola . 70. Let . Show that This shows that the graph of approaches the graph of , and we say that the curve is asymptotic to the parabola . Use this fact to help sketch the graph of . 71. Discuss the asymptotic behavior of in the same manner as in Exercise 70. Then use your results to help sketch the graph of . 72. Use the asymptotic behavior of to sketch its graph without going through the curve-sketching procedure of this section. f H20849xH20850 H33527 cos x H11001 1H20862x 2 f f H20849xH20850 H33527 H20849x 4 H11001 1H20850H20862x fy H33527 x 2 y H33527 f H20849xH20850 y H33527 x 2 f lim xlH11006H11009 H20851 f H20849xH20850 H11002 x 2 H20852 H33527 0 f H20849xH20850 H33527 H20849x 3 H11001 1H20850H20862x H20849x 2 H20862a 2 H20850 H11002 H20849y 2 H20862b 2 H20850 H33527 1 y H33527H11002H20849bH20862aH20850xy H33527 H20849bH20862aH20850x y H33527H11002x H11002 2y H33527 x H11001 2 y H33527sx 2 H11001 4x 57?60 Find an equation of the slant asymptote. Do not sketch the curve. 58. 59. 60. 61?66 Use the guidelines of this section to sketch the curve. In guideline D find an equation of the slant asymptote. 61. 62. 63. 64. 65. 66. 67. Show that the curve has two slant asymptotes: and . Use this fact to help sketch the curve. y H33527 x H11002 H9266H208622y H33527 x H11001 H9266H208622 y H33527 x H11002 tan H110021 x y H33527 H20849x H11001 1H20850 3 H20849x H11002 1H20850 2 y H33527 2x 3 H11001 x 2 H11001 1 x 2 H11001 1 y H33527 e x H11002 xxy H33527 x 2 H11001 4 y H33527 x 2 H11001 12 x H11002 2 y H33527 H110022x 2 H11001 5x H11002 1 2x H11002 1 y H33527 5x 4 H11001 x 2 H11001 x x 3 H11002 x 2 H11001 2 y H33527 4x 3 H11002 2x 2 H11001 5 2x 2 H11001 x H11002 3 y H33527 2x 3 H11001 x 2 H11001 x H11001 3 x 2 H11001 2x y H33527 x 2 H11001 1 x H11001 1 57. GRAPHING WITH CALCULUS AND CALCULATORS The method we used to sketch curves in the preceding section was a culmination of much of our study of differential calculus. The graph was the final object that we produced. In this section our point of view is completely different. Here we start with a graph produced by a graphing calculator or computer and then we refine it. We use calculus to make sure that we reveal all the important aspects of the curve. And with the use of graphing devices we can tackle curves that would be far too complicated to consider without technology. The theme is the interaction between calculus and calculators. EXAMPLE 1 Graph the polynomial . Use the graphs of and to estimate all maximum and minimum points and intervals of concavity. SOLUTION If we specify a domain but not a range, many graphing devices will deduce a suitable range from the values computed. Figure 1 shows the plot from one such device if we specify that . Although this viewing rectangle is useful for showing that the asymptotic behavior (or end behavior) is the same as for , it is obviously hiding some finer detail. So we change to the viewing rectangle by shown in Figure 2. From this graph it appears that there is an absolute minimum value of about 3 when (by using the cursor) and is decreasing on and increas- ing on . Also, there appears to be a horizontal tangent at the origin and inflec- tion points when and when is somewhere between and . Now let?s try to confirm these impressions using calculus. We differentiate and get f H11033H20849xH20850 H33527 60x 4 H11001 60x 3 H11001 18x H11002 4 fH11032H20849xH20850 H33527 12x 5 H11001 15x 4 H11001 9x 2 H11002 4x H110021H110022xx H33527 0 H20849H110021.62, H11009H20850 H20849H11002H11009, H110021.62H20850fx H11015 H110021.62 H1100215.3 H20851H1100250, 100H20852H20851H110023, 2H20852 y H33527 2x 6 H110025 H33355 x H33355 5 f H11033 fH11032f H20849xH20850 H33527 2x 6 H11001 3x 5 H11001 3x 3 H11002 2x 2 4.6 41,000 _1000 _5 5 y=? FIGURE 1 100 _50 _3 2 y=? FIGURE 2 N If you have not already read Section 1.4, you should do so now. In particular, it explains how to avoid some of the pitfalls of graphing devices by choosing appropriate viewing rectangles. When we graph in Figure 3 we see that changes from negative to positive when ; this confirms (by the First Derivative Test) the minimum value that we found earlier. But, perhaps to our surprise, we also notice that changes from posi- tive to negative when and from negative to positive when . This means that has a local maximum at 0 and a local minimum when , but these were hidden in Figure 2. Indeed, if we now zoom in toward the origin in Figure 4, we see what we missed before: a local maximum value of 0 when and a local minimum value of about when . What about concavity and inflection points? From Figures 2 and 4 there appear to be inflection points when is a little to the left of and when is a little to the right of 0. But it?s difficult to determine inflection points from the graph of , so we graph the sec- ond derivative in Figure 5. We see that changes from positive to negative when and from negative to positive when . So, correct to two decimal places, is concave upward on and and concave downward on . The inflection points are and . We have discovered that no single graph reveals all the important features of this polynomial. But Figures 2 and 4, when taken together, do provide an accurate picture. M EXAMPLE 2 Draw the graph of the function in a viewing rectangle that contains all the important features of the function. Estimate the maximum and minimum values and the intervals of concavity. Then use calculus to find these quantities exactly. SOLUTION Figure 6, produced by a computer with automatic scaling, is a disaster. Some graphing calculators use by as the default viewing rectangle, so let?s try it. We get the graph shown in Figure 7; it?s a major improvement. The -axis appears to be a vertical asymptote and indeed it is because Figure 7 also allows us to estimate the -intercepts: about and . The exact val- ues are obtained by using the quadratic formula to solve the equation ; we get . To get a better look at horizontal asymptotes, we change to the viewing rectangle by in Figure 8. It appears that is the horizontal asymptote and this is easily confirmed: lim xlH11006H11009 x 2 H11001 7x H11001 3 x 2 H33527 lim xlH11006H11009 H20873 1 H11001 7 x H11001 3 x 2 H20874 H33527 1 y H33527 1H20851H110025, 10H20852H20851H1100220, 20H20852 3 H11003 10!* _5 5 y=? FIGURE 6 10 _10 _10 10 y=? FIGURE 7 10 _5 _20 20 y=? y=1 FIGURE 8 x H33527 (H110027 H11006s37 )H208622 x 2 H11001 7x H11001 3 H33527 0 H110026.5H110020.5x lim xl0 x 2 H11001 7x H11001 3 x 2 H33527H11009 y H20851H1100210, 10H20852H20851H1100210, 10H20852 f H20849xH20850 H33527 x 2 H11001 7x H11001 3 x 2 V H208490.19, H110020.05H20850H20849H110021.23, H1100210.18H20850H20849H110021.23, 0.19H20850 H208490.19, H11009H20850H20849H11002H11009, H110021.23H20850f x H11015 0.19x H11015 H110021.23 f H11033f H11033 f xH110021x x H11015 0.35H110020.1 x H33527 0 x H11015 0.35f x H11015 0.35x H33527 0 fH11032H20849xH20850 x H11015 H110021.62 fH11032H20849xH20850fH11032 316 |||| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 10 _30 _3 2 y=f·(x) FIGURE 5 20 _5 _3 2 y=fª(x) FIGURE 3 1 _1 _1 1 y=? FIGURE 4 To estimate the minimum value we zoom in to the viewing rectangle by in Figure 9. The cursor indicates that the absolute minimum value is about when , and we see that the function decreases on and and increases on . The exact values are obtained by differentiating: This shows that when and when and when . The exact minimum value is . Figure 9 also shows that an inflection point occurs somewhere between and . We could estimate it much more accurately using the graph of the second deriv- ative, but in this case it?s just as easy to find exact values. Since we see that when . So is concave upward on and and concave downward on . The inflection point is . The analysis using the first two derivatives shows that Figures 7 and 8 display all the major aspects of the curve. M EXAMPLE 3 Graph the function . SOLUTION Drawing on our experience with a rational function in Example 2, let?s start by graphing in the viewing rectangle by . From Figure 10 we have the feeling that we are going to have to zoom in to see some finer detail and also zoom out to see the larger picture. But, as a guide to intelligent zooming, let?s first take a close look at the expression for . Because of the factors and in the denominator, we expect and to be the vertical asymptotes. Indeed To find the horizontal asymptotes we divide numerator and denominator by : This shows that , so the -axis is a horizontal asymptote. It is also very useful to consider the behavior of the graph near the -intercepts using an analysis like that in Example 11 in Section 2.6. Since is positive, does not change sign at 0 and so its graph doesn?t cross the -axis at 0. But, because of the factor , the graph does cross the -axis at and has a horizontal tangent there. Putting all this information together, but without using derivatives, we see that the curve has to look something like the one in Figure 11. H110021xH20849x H11001 1H20850 3 x f H20849xH20850x 2 x xf H20849xH20850l0 as xlH11006H11009 x 2 H20849x H11001 1H20850 3 H20849x H11002 2H20850 2 H20849x H11002 4H20850 4 H33527 x 2 x 3 H11554 H20849x H11001 1H20850 3 x 3 H20849x H11002 2H20850 2 x 2 H11554 H20849x H11002 4H20850 4 x 4 H33527 1 x H20873 1 H11001 1 x H20874 3 H20873 1 H11002 2 x H20874 2 H20873 1 H11002 4 x H20874 4 x 6 lim xl4 x 2 H20849x H11001 1H20850 3 H20849x H11002 2H20850 2 H20849x H11002 4H20850 4 H33527H11009andlim xl2 x 2 H20849x H11001 1H20850 3 H20849x H11002 2H20850 2 H20849x H11002 4H20850 4 H33527H11009 x H33527 4x H33527 2 H20849x H11002 4H20850 4 H20849x H11002 2H20850 2 f H20849xH20850 H20851H1100210, 10H20852H20851H1100210, 10H20852f f H20849xH20850 H33527 x 2 H20849x H11001 1H20850 3 H20849x H11002 2H20850 2 H20849x H11002 4H20850 4 V (H11002 9 7 , H11002 71 27 )(H11002H11009, H11002 9 7 )H208490, H11009H20850 (H11002 9 7 , 0)fH20849x HS33527 0H20850x H11022 H11002 9 7 f H11033H20849xH20850 H11022 0 f H11033H20849xH20850 H33527 14 x 3 H11001 18 x 4 H33527 2(7x H11001 9H20850 x 4 x H33527H110022 x H33527H110021 f (H11002 6 7 ) H33527H11002 37 12 H11015 H110023.08x H11022 0 x H11021 H11002 6 7 fH11032H20849xH20850 H11021 0H11002 6 7 H11021 x H11021 0fH11032H20849xH20850 H11022 0 fH11032H20849xH20850 H33527H11002 7 x 2 H11002 6 x 3 H33527H11002 7x H11001 6 x 3 H20849H110020.9, 0H20850 H208490, H11009H20850H20849H11002H11009, H110020.9H20850x H11015 H110020.9 H110023.1H20851H110024, 2H20852 H20851H110023, 0H20852 SECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS |||| 317 2 _4 _3 0 y=? FIGURE 9 10 _10 _10 10 y=? FIGURE 10 FIGURE 11 x y 1 2 3_1 4 Now that we know what to look for, we zoom in (several times) to produce the graphs in Figures 12 and 13 and zoom out (several times) to get Figure 14. We can read from these graphs that the absolute minimum is about and occurs when . There is also a local maximum when and a local minimum when . These graphs also show three inflection points near and and two between and . To estimate the inflection points closely we would need to graph , but to compute by hand is an unreasonable chore. If you have a computer algebra system, then it?s easy to do (see Exercise 15). We have seen that, for this particular function, three graphs (Figures 12, 13, and 14) are necessary to convey all the useful information. The only way to display all these features of the function on a single graph is to draw it by hand. Despite the exaggera- tions and distortions, Figure 11 does manage to summarize the essential nature of the function. M EXAMPLE 4 Graph the function . For , estimate all maximum and minimum values, intervals of increase and decrease, and inflection points correct to one decimal place. SOLUTION We first note that is periodic with period . Also, is odd and for all . So the choice of a viewing rectangle is not a problem for this function: We start with by . (See Figure 15.) It appears that there are three local maximum values and two local minimum values in that window. To confirm this and locate them more accurately, we calculate that and graph both and in Figure 16.fH11032f fH11032H20849xH20850 H33527 cosH20849x H11001 sin 2xH20850 H11554 H208491 H11001 2 cos 2xH20850 1.1 _1.1 0 FIGURE 15 1.2 _1.2 0pi pi y=? y=fª(x) FIGURE 16 H20851H110021.1, 1.1H20852H208510, H9266H20852 x H11341 f H20849xH20850 H11341 H33355 1f2H9266f 0 H33355 x H33355 H9266f H20849xH20850 H33527 sinH20849x H11001 sin 2xH20850 f H11033f H11033 0H110021H110021H110025, H1100235,x H11015 2.5H11015211 x H11015 H110020.3H110150.00002x H11015 H1100220 H110020.02 0.05 _0.05 _100 1 y=? FIGURE 12 0.0001 _0.0001 _1.5 0.5 y=? FIGURE 13 500 _10 _1 10 y=? FIGURE 14 318 |||| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION N The family of functions where is a constant, occurs in applications to frequency modulation (FM) synthesis. A sine wave is modulated by a wave with a different frequency . The case where is studied in Example 4. Exercise 25 explores another special case. c H33527 2H20849sin cxH20850 c f H20849xH20850 H33527 sinH20849x H11001 sin cxH20850 Using zoom-in and the First Derivative Test, we find the following values to one deci- mal place. The second derivative is Graphing both and in Figure 17, we obtain the following approximate values: Having checked that Figure 15 does indeed represent accurately for , we can state that the extended graph in Figure 18 represents accurately for . M Our final example is concerned with families of functions. As discussed in Section 1.4, this means that the functions in the family are related to each other by a formula that con- tains one or more arbitrary constants. Each value of the constant gives rise to a member of the family and the idea is to see how the graph of the function changes as the constant changes. EXAMPLE 5 How does the graph of vary as varies? SOLUTION The graphs in Figures 19 and 20 (the special cases and ) show two very different-looking curves. Before drawing any more graphs, let?s see what mem- bers of this family have in common. Since for any value of , they all have the -axis as a horizontal asymptote. A vertical asymp- tote will occur when . Solving this quadratic equation, we get . When , there is no vertical asymptote (as in Figure 19). When , the graph has a single vertical asymptote because When , there are two vertical asymptotes: (as in Figure 20). Now we compute the derivative: This shows that when (if ), when , and x H11021 H110021fH11032H20849xH20850 H11022 0c HS33527 1x H33527H110021fH11032H20849xH20850 H33527 0 fH11032H20849xH20850 H33527H11002 2x H11001 2 H20849x 2 H11001 2x H11001 cH20850 2 x H33527H110021 H11006s1 H11002 c c H11021 1 lim xlH110021 1 x 2 H11001 2x H11001 1 H33527 lim xlH110021 1 H20849x H11001 1H20850 2 H33527H11009 x H33527H110021c H33527 1 c H11022 1x H33527H110021 H11006s1 H11002 c x 2 H11001 2x H11001 c H33527 0 xc lim xlH11006H11009 1 x 2 H11001 2x H11001 c H33527 0 c H33527H110022c H33527 2 cf H20849xH20850 H33527 1H20862H20849x 2 H11001 2x H11001 cH20850V H110022H9266 H33355 x H33355 2H9266 f 0 H33355 x H33355 H9266f Inflection points: H208490, 0H20850, H208490.8, 0.97H20850, H208491.3, 0.97H20850, H208491.8, 0.97H20850, H208492.3, 0.97H20850 Concave downward on: H208490, 0.8H20850, H208491.3, 1.8H20850, H208492.3, H9266H20850 Concave upward on: H208490.8, 1.3H20850, H208491.8, 2.3H20850 f H11033f f H11033H20849xH20850 H33527H11002H208491 H11001 2 cos 2xH20850 2 sinH20849x H11001 sin 2xH20850 H11002 4 sin 2x cosH20849x H11001 sin 2xH20850 Local minimum values: f H208491.0H20850 H11015 0.94, f H208492.1H20850 H11015 0.94 Local maximum values: f H208490.6H20850 H11015 1, f H208491.6H20850 H11015 1, f H208492.5H20850 H11015 1 Intervals of decrease: H208490.6, 1.0H20850, H208491.6, 2.1H20850, H208492.5, H9266H20850 Intervals of increase: H208490, 0.6H20850, H208491.0, 1.6H20850, H208492.1, 2.5H20850 SECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS |||| 319 FIGURE 19 c=2 y= 1 ?+2x+2 2 _2 _5 4 1.2 _1.2 0 pi f f · FIGURE 17 1.2 _1.2 _2pi 2pi FIGURE 18 FIGURE 20 c=_2 2 _2 _5 4 y= 1 ?+2x-2 when . For , this means that increases on and decreases on . For , there is an absolute maximum value . For , is a local maximum value and the intervals of increase and decrease are interrupted at the vertical asymptotes. Figure 21 is a ?slide show? displaying five members of the family, all graphed in the viewing rectangle by . As predicted, is the value at which a transi- tion takes place from two vertical asymptotes to one, and then to none. As increases from , we see that the maximum point becomes lower; this is explained by the fact that as . As decreases from , the vertical asymptotes become more widely separated because the distance between them is , which becomes large as . Again, the maximum point approaches the -axis because as . There is clearly no inflection point when . For we calculate that and deduce that inflection points occur when . So the inflection points become more spread out as increases and this seems plausible from the last two parts of Figure 21. M c x H33527H110021 H11006s3H20849c H11002 1H20850H208623 f H11033H20849xH20850 H33527 2H208493x 2 H11001 6x H11001 4 H11002 cH20850 H20849x 2 H11001 2x H11001 cH20850 3 c H11022 1c H33355 1 c=3c=2c=1c=0c=_1 FIGURE 21 The family of functions ?=1/(?+2x+c) c lH11002H11009 1H20862H20849c H11002 1H20850 l 0xc lH11002H11009 2s1 H11002 c 1cc lH110091H20862H20849c H11002 1H20850 l 0 1 c c H33527 1H20851H110022, 2H20852H20851H110025, 4H20852 fH20849H110021H20850 H33527 1H20862H20849c H11002 1H20850c H11021 1fH20849H110021H20850 H33527 1H20862H20849c H11002 1H20850 c H11022 1H20849H110021, H11009H20850 H20849H11002H11009, H110021H20850fc H33356 1x H11022 H110021fH11032H20849xH20850 H11021 0 320 |||| CHAPTER 4 APPLICATIONS OF DIFFERENTIATION See an animation of Figure 21 in Visual 4.6. TEC 9?10 Produce graphs of that reveal all the important aspects of the curve. Estimate the intervals of increase and decrease and intervals of concavity, and use calculus to find these intervals exactly. 9. 10. 11?12 (a) Graph the function. (b) Use l?Hospital?s Rule to explain the behavior as . (c) Estimate the minimum value and intervals of concavity. Then use calculus to find the exact values. 12. f H20849xH20850 H33527 xe 1H20862x f H20849xH20850 H33527 x 2 ln x11. x l 0 f H20849xH20850 H33527 1 x 8 H11002 2 H11003 10 8 x 4 f H20849xH20850 H33527 1 H11001 1 x H11001 8 x 2 H11001 1 x 3 f1?8 Produce graphs of that reveal all the important aspects of the curve. In particular, you should use graphs of and to esti- mate the intervals of increase and decrease, extreme values, inter- vals of concavity, and inflection points. 1. 2. 3. 4. 5. 6. 7. , 8. f H20849xH20850 H33527 e x x 2 H11002 9 H110024 H33355 x H33355 4f H20849xH20850 H33527 x 2 H11002 4x H11001 7 cos x f H20849xH20850 H33527 tan x H11001 5 cos x f H20849xH20850 H33527 x x 3 H11002 x 2 H11002 4x H11001 1 f H20849xH20850 H33527 x 2 H11002 1 40x 3 H11001 x H11001 1 f H20849xH20850 H33527 x 6 H11002 10x 5 H11002 400x 4 H11001 2500x 3 f H20849xH20850 H33527 x 6 H11002 15x 5 H11001 75x 4 H11002 125x 3 H11002 x f H20849xH20850 H33527 4x 4 H11002 32x 3 H11001 89x 2 H11002 95x H11001 29 f H11033fH11032 f ; EXERCISES4.6 the same time. Find all the maximum and minimum values and inflection points. Then graph in the viewing rectangle by and comment on symmetry. 26?33 Describe how the graph of varies as varies. Graph several members of the family to illustrate the trends that you dis- cover. In particular, you should investigate how maximum and minimum points and inflection points move when changes. You should also identify any transitional values of at which the basic shape of the curve changes. 26. 27. 28. 29. 32. 33. The family of functions , where , , and are positive numbers and , has been used to model the concentration of a drug injected into the blood- stream at time . Graph several members of this family. What do they have in common? For fixed values of and , discover graphically what happens as increases. Then use calculus to prove what you have discovered. 35. Investigate the family of curves given by , where is a real number. Start by computing the limits as . Identify any transitional values of where the basic shape changes. What happens to the maximum or minimum points and inflection points as changes? Illustrate by graphing sev- eral members of the family. 36. Investigate the family of curves given by the equation . Start by determining the transitional value of at which the number of inflection points changes. Then graph several members of the family to see what shapes are possible. There is another transitional value of at which the number of critical numbers changes. Try to discover it graphically. Then prove what you have discovered. 37. (a) Investigate the family of polynomials given by the equa- tion . For what values of does the curve have minimum points? (b) Show that the minimum and maximum points of every curve in the family lie on the parabola . Illus- trate by graphing this parabola and several members of the family. 38. (a) Investigate the family of polynomials given by the equa- tion . For what values of does the curve have maximum and minimum points? (b) Show that the minimum and maximum points of every curve in the family lie on the curve . Illustrate by graphing this curve and several members of the family. y H33527 x H11002 x 3 cf H20849xH20850 H33527 2x 3 H11001 cx 2 H11001 2x y H33527 1 H11002 x 2 cf H20849xH20850 H33527 cx 4 H11002 2x 2 H11001 1 c c f H20849xH20850 H33527 x 4 H11001 cx 2 H11001 x c c x lH11006H11009c f H20849xH20850 H33527 xe H11002cx b aC t H33527 0 b H11022 aC baf H20849tH20850 H33527 CH20849e H11002at H11002 e H11002bt H2085034. f H20849xH20850 H33527 cx H11001 sin xf H20849xH20850 H33527 1 H208491 H11002 x 2 H20850 2 H11001 cx 2 f H20849xH20850 H33527 cx 1 H11001 c 2 x 2 31.f H20849xH20850 H33527 lnH20849x 2 H11001 cH2085030. f H20849xH20850 H33527 e H11002cH20862x 2 f H20849xH20850 H33527 xs c 2 H11002 x 2 f H20849xH20850 H33527 x 4 H11001 cx 2 f H20849xH20850 H33527 x 3 H11001 cx c c cf H20851H110021.2, 1.2H20852H20851H110022H9266, 2H9266H20852 f 13?14 Sketch the graph by hand using asymptotes and intercepts, but not derivatives. Then use your sketch as a guide to producing graphs (with a graphing device) that display the major features of the curve. Use these graphs to estimate the maximum and mini- mum values. 14. 15. If is the function considered in Example 3, use a computer algebra system to calculate and then graph it to confirm that all the maximum and minimum values are as given in the example. Calculate and use it to estimate the intervals of concavity and inflection points. 16. If is the function of Exercise 14, find and and use their graphs to estimate the intervals of increase and decrease and concavity of . 17?22 Use a computer algebra system to graph and to find and . Use graphs of these derivatives to estimate the intervals of increase and decrease, extreme values, intervals of concavity, and inflection points of . 17. 18. 19. , 20. 21. 22. 23?24 (a) Graph the function. (b) Explain the shape of the graph by computing the limit as or as . (c) Estimate the maximum and minimum values and then use calculus to find the exact values. (d) Use a graph of to estimate the x-coordinates of the inflec- tion points. 24. 25. In Example 4 we considered a member of the family of func- tions that occur in FM synthesis. Here we investigate the function with . Start by graphing in the viewing rectangle by . How many local maximum points do you see? The graph has more than are visible to the naked eye. To discover the hidden maximum and minimum points you will need to examine the graph of very carefully. In fact, it helps to look at the graph of at f H11033fH11032 H20851H110021.2, 1.2H20852H208510, H9266H20852 fc H33527 3 f H20849xH20850 H33527 sinH20849x H11001 sin cxH20850 f H20849xH20850 H33527 H20849sin xH20850 sin x f H20849xH20850 H33527 x 1H20862x 23. f H11033 x lH11009x l 0 H11001 CAS f H20849xH20850 H33527 1 1 H11001 e tan x f H20849xH20850 H33527 1 H11002 e 1H20862x 1 H11001 e 1H20862x f H20849xH20850 H33527 H20849x 2 H11002 1H20850e arctan x x H33355 20f H20849xH20850 H33527sx H11001 5 sin x f H20849xH20850 H33527 x 2H208623 1 H11001 x H11001 x 4 f H20849xH20850 H33527 sx x 2 H11001 x H11001 1 f f H11033 fH11032fCAS f f H11033fH11032fCAS f H11033 fH11032 fCAS f H20849xH20850 H33527 H208492x H11001 3H20850 2 H20849x H11002 2H20850 5 x 3 H20849x H11002 5H20850 2 f H20849xH20850 H33527 H20849x H11001 4H20850H20849x H11002 3H20850 2 x 4 H20849x H11002 1H20850 13. SECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS |||| 321 James Stewart Stewart - Calculus - Early Transcedentals 6e calculus

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