{VERSION 3 0 "IBM INTEL NT" "3.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" 0 21 "" 0 1 0 0 0 1 0 0 0 0 2 0 0 0 0 }{CSTYLE "" -1 256 " " 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 259 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 261 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 262 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 263 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 265 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 266 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 267 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 269 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 270 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 271 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 272 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 273 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 274 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 275 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 276 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 277 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 278 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 279 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 280 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 281 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 282 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 283 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 284 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 285 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 286 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 287 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 288 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 289 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 290 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 291 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 292 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 293 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 294 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 295 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 296 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 297 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 298 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 299 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 300 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 301 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 302 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 303 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 304 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 305 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 306 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 307 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 308 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 309 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 310 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 } {PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 65 "Lab 7 Maple Activities for September 21: (Lesson 4 - Continuity)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 87 "1. In lecture, we used the \+ Intermediate Value Theorem (IVT) to show that the equation " } {XPPEDIT 18 0 "x^2 = sqrt(x+1);" "6#/*$%\"xG\"\"#-%%sqrtG6#,&F%\"\"\" \"\"\"F+" }{TEXT -1 90 " has a solution on the interval (1, 2). Recal l the work we did in order to apply the IVT:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "x^2 = sqrt(x+1);" "6#/*$%\"xG \"\"#-%%sqrtG6#,&F%\"\"\"\"\"\"F+" }{TEXT -1 15 " implies that " } {XPPEDIT 18 0 "x^2-sqrt(x+1) = 0;" "6#/,&*$%\"xG\"\"#\"\"\"-%%sqrtG6#, &F&F(\"\"\"F(!\"\"\"\"!" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 7 " Let f(" }{XPPEDIT 18 0 "x;" "6#%\"xG " }{TEXT -1 4 ") = " }{XPPEDIT 18 0 "x^2-sqrt(x+1);" "6#,&*$%\"xG\"\"# \"\"\"-%%sqrtG6#,&F%F'\"\"\"F'!\"\"" }{TEXT -1 83 ". The function f \+ is continuous on the closed interval [1, 2] since the function " } {TEXT 269 1 "y" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "x^2;" "6#*$%\"xG\"\" #" }{TEXT -1 55 " is continuous on the interval [1, 2] and the functio n " }{TEXT 270 1 "y" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "sqrt(x+1);" "6# -%%sqrtG6#,&%\"xG\"\"\"\"\"\"F(" }{TEXT -1 33 " is continuous on [1, 2 ]. Since " }{TEXT 271 1 "f" }{TEXT -1 93 " is the difference of two f unctions which are continuous on the closed interval [1, 2], then " } {TEXT 272 1 "f" }{TEXT -1 89 " is continuous on this interval. We wil l apply the IVT using the closed interval [1, 2]." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 12 " f(1) = 1 - " }{XPPEDIT 18 0 "sqrt(2);" "6#-%%sqrtG6#\"\"#" }{TEXT -1 5 " < 0" }{TEXT 256 0 " " }}{PARA 0 "" 0 "" {TEXT -1 12 " f(2) = 4 - " }{XPPEDIT 18 0 "sqrt(3) ;" "6#-%%sqrtG6#\"\"$" }{TEXT -1 5 " > 0" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 40 "Thus, by the IVT, there exists a n umber " }{XPPEDIT 18 0 "c;" "6#%\"cG" }{TEXT -1 39 " in the open inter val (1, 2) such that " }{TEXT 257 1 " " }{TEXT -1 2 "f(" }{XPPEDIT 18 0 "c;" "6#%\"cG" }{TEXT -1 6 ") = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 9 "Since f(" }{TEXT 298 1 "x" }{TEXT -1 4 ") = " }{XPPEDIT 18 0 "x^2-sqrt(x+1)" "6#,&*$%\"xG\"\"#\"\"\"-%%sqrtG6 #,&F%F'\"\"\"F'!\"\"" }{TEXT -1 10 ", then f(" }{TEXT 299 1 "c" } {TEXT -1 4 ") = " }{XPPEDIT 18 0 "c^2-sqrt(c+1);" "6#,&*$%\"cG\"\"#\" \"\"-%%sqrtG6#,&F%F'\"\"\"F'!\"\"" }{TEXT -1 12 ". Thus, f(" } {XPPEDIT 18 0 "c;" "6#%\"cG" }{TEXT -1 20 ") = 0 implies that " } {XPPEDIT 18 0 "c^2-sqrt(c+1) = 0;" "6#/,&*$%\"cG\"\"#\"\"\"-%%sqrtG6#, &F&F(\"\"\"F(!\"\"\"\"!" }{TEXT -1 21 ". This implies that " } {XPPEDIT 18 0 "c;" "6#%\"cG" }{TEXT -1 39 " is a solution (root) of th e equation " }{XPPEDIT 18 0 "x^2-sqrt(x+1) = 0;" "6#/,&*$%\"xG\"\"#\" \"\"-%%sqrtG6#,&F&F(\"\"\"F(!\"\"\"\"!" }{TEXT -1 61 " , which is also a solution (root) of the original equation " }{XPPEDIT 18 0 "x^2 = s qrt(x+1)" "6#/*$%\"xG\"\"#-%%sqrtG6#,&F%\"\"\"\"\"\"F+" }{TEXT -1 2 " \+ ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 14 "Also , since f(" }{XPPEDIT 18 0 "c;" "6#%\"cG" }{TEXT -1 12 ") = 0, then " }{TEXT 300 1 "c" }{TEXT -1 8 " is the " }{TEXT 301 1 "x" }{TEXT -1 19 "-coordinate of the " }{TEXT 302 1 "x" }{TEXT -1 24 "-intercept of the graph " }{TEXT 303 1 "y" }{TEXT -1 6 " = f(" }{TEXT 304 1 "x" } {TEXT -1 4 ") = " }{XPPEDIT 18 0 "x^2-sqrt(x+1)" "6#,&*$%\"xG\"\"#\"\" \"-%%sqrtG6#,&F%F'\"\"\"F'!\"\"" }{TEXT -1 45 " . Let's use Maple t o graph the function " }{TEXT 305 1 "y" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "x^2-sqrt(x+1)" "6#,&*$%\"xG\"\"#\"\"\"-%%sqrtG6#,&F%F'\"\"\"F'! \"\"" }{TEXT -1 93 " and find an approximation to this solution (root ), which is accurate to six decimal places." }}{PARA 0 "" 0 "" {TEXT -1 25 "Recall, an approximation " }{TEXT 273 1 "a" }{TEXT -1 17 " to t he solution " }{TEXT 274 1 "c" }{TEXT -1 31 " of an equation is accura te to " }{TEXT 275 1 "n" }{TEXT -1 20 " decimal places if " } {XPPEDIT 18 0 "abs(a-c) < .5*10^(-n);" "6#2-%$absG6#,&%\"aG\"\"\"%\"cG !\"\"*&$\"\"&!\"\"F))\"#5,$%\"nGF+F)" }{TEXT -1 57 ". That is, the ab solute value of the difference between " }{TEXT 276 1 "c" }{TEXT -1 20 ", the solution, and " }{TEXT 277 1 "a" }{TEXT -1 87 ", the approxi mation to this solution, is less than 0.5 times 10 raised to the negat ive " }{TEXT 278 1 "n" }{TEXT -1 60 "th power. But how can you determ ine this if you don't have " }{TEXT 279 1 "c" }{TEXT -1 63 ", the solu tion to the equation. Well, find two approximations " }{TEXT 280 1 "a " }{TEXT -1 6 "1 and " }{TEXT 281 1 "a" }{TEXT -1 18 "2 to the solutio n " }{TEXT 282 1 "c" }{TEXT -1 12 " such that " }{TEXT 283 1 "a" } {TEXT -1 4 "1 < " }{TEXT 284 1 "c" }{TEXT -1 3 " < " }{TEXT 285 1 "a" }{TEXT -1 8 "2 and " }{XPPEDIT 18 0 "abs(a1-a2) < .5*10^(-n);" "6#2- %$absG6#,&%#a1G\"\"\"%#a2G!\"\"*&$\"\"&!\"\"F))\"#5,$%\"nGF+F)" } {TEXT -1 59 ". That is, find two approximations such that the solutio n " }{TEXT 286 1 "c" }{TEXT -1 165 " is between these two approximatio ns and that the absolute value of the difference between these two app roximations is less than 0.5 times 10 raised to the negative " }{TEXT 287 1 "n" }{TEXT -1 27 "th power. Then use either " }{TEXT 288 1 "a" }{TEXT -1 5 "1 or " }{TEXT 289 1 "a" }{TEXT -1 40 "2 as your approxima tion to the solution " }{TEXT 290 1 "c" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 65 "Thus, for this probl em, to find an approximation to the solution " }{TEXT 297 1 "c" } {TEXT -1 18 " of the equation " }{XPPEDIT 18 0 "x^2-sqrt(x+1) = 0;" " 6#/,&*$%\"xG\"\"#\"\"\"-%%sqrtG6#,&F&F(\"\"\"F(!\"\"\"\"!" }{TEXT -1 83 " which is accurate to six decimal places, we will need to find tw o approximations " }{TEXT 291 1 "a" }{TEXT -1 6 "1 and " }{TEXT 292 1 "a" }{TEXT -1 18 "2 to the solution " }{TEXT 293 1 "c" }{TEXT -1 12 " \+ such that " }{TEXT 294 1 "a" }{TEXT -1 4 "1 < " }{TEXT 295 1 "c" } {TEXT -1 3 " < " }{TEXT 296 1 "a" }{TEXT -1 8 "2 and " }{XPPEDIT 18 0 "abs(a1-a2) < .5e-6;" "6#2-%$absG6#,&%#a1G\"\"\"%#a2G!\"\"$\"\"&!\"( " }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "f := x -> x^2 - sqrt(x+1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "f(x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "plot(f(x), x=1..2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 428 "Note: since we know that there is a solution t o the equation between 1 and 2 by the IVT, then we only need to graph \+ the function f on the interval (1, 2). From the graph, it appears t hat the solution (root) is between 1.2 and 1.3. At this point, if we \+ pick any number in the interval [1.2, 1.3] for the solution (root), th e error for our solution is at most 1.3 - 1.2 = 0.1. Now, let's graph f on the interval [1.2, 1.3]. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "plot(f(x), x=1.2..1.3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 336 "From this graph, it appe ars that the solution (root) is between 1.22 and 1.23. At this point, if we pick any number in the interval [1.22, 1.23] for the solution ( root), the error for our solution is at most 1.23 - 1.22 = 0.01 and ou r solution is correct to at least one decimal place. Now, let's graph f on the interval [1.22, 1.23]." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "plot(f(x), x=1.22..1.23);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 342 "From this graph, it appe ars that the solution (root) is between 1.22 and 1.222. At this point , if we pick any number in the interval [1.22, 1.222] for the solution (root), the error for our solution is at most 1.222 - 1.22 = 0.002 an d our solution is correct to at least two decimal places. Now, let's \+ graph f on the interval [1.22, 1.222]." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "plot(f(x), x=1.22..1.222);" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 358 "From this graph, it ap pears that the solution (root) is between 1.2206 and 1.2208. At this \+ point, if we pick any number in the interval [1.2206, 1.2208] for the \+ solution (root), the error for our solution is at most 1.2208 - 1.2206 = 0.0002 and our solution is correct to at least three decimal places . Now, let's graph f on the interval [1.2206, 1.2208]." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "plot(f(x), x=1.2206..1.2208);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 811 "F rom this graph, it appears that the solution (root) is between 1.22074 and 1.22076. At this point, if we pick any number in the interval [1 .22074, 1.22076] for the solution (root), the error for our solution i s at most 1.22076 - 1.22074 = 0.00002 and our solution is correct to a t least four decimal places. We can continue this process of \"zeroin g in\" on the solution (root). However, we can also keep applying the IVT by evaluating the function f at the following numbers starting at 1.220740 and increasing the digit in the millionths place by one unti l we reach 1.220750. We will see that our solution (root) is somewher e between these two numbers. We want to identify the first place wher e the value of the function changes sign. This could be from positive to negative or from negative to positive." }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 142 "f(1.220740); f(1.220741); f(1.220742); f(1.220743) ; f(1.220744); f(1.220745); f(1.220746); f(1.220747); f(1.220748); f(1 .220749); f(1.220750);" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 586 "Note: f(1.220744) < 0 an d f(1.220745) > 0. The IVT tells us that the solution (root) is somew here between 1.220744 and 1.220745. At this point, if we pick any num ber in the interval [1.220744, 1.220745] for the solution (root), the \+ error in our solution is at most 1.220745 - 1.220744 = 0.000001 and ou r solution is correct to at least five decimal places. Again, we will apply the IVT. However, let's \"speed up\" the process by evaluating the function f at 1.2207440, 1.2207445 (the midpoint), and 1.2207450. Again, we want to identify where the value of the function changes s ign." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "f(1.2207440); f(1.2 207445); f(1.2207450);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 271 "Note: f(1.2207440) < 0 and f(1.2207445) \+ > 0. Thus, by the IVT, the root is between 1.2207440 and 1.2207445. \+ Now, evaluate the function f at the following numbers starting at 1.22 07440 and increasing the digit in the ten-millionths place by one unti l we reach 1.2207445." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 83 "f( 1.2207440); f(1.2207441); f(1.2207442); f(1.2207443); f(1.2207444); f( 1.2207445);" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 352 "Note: f(1.2207440) < 0 and f(1.2207441) \+ > 0. Thus, by the IVT, the solution (root) is between 1.2207440 and 1 .2207441. At this point, if we pick any number in the interval [1.220 7440, 1.2207441] for the solution (root), the error in our solution is at most 1.2207441 - 1.2207440 = 0.0000001 and our solution is correct to at least six decimal places." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 177 " You can continue to apply the IVT in or der to find the root correct to as many decimal places as you want. A t this point, we can say that the solution (root) of the equation " } {XPPEDIT 18 0 "x^2-sqrt(x+1) = 0;" "6#/,&*$%\"xG\"\"#\"\"\"-%%sqrtG6#, &F&F(\"\"\"F(!\"\"\"\"!" }{TEXT -1 106 " in the interval (1, 2) is ap proximately 1.2207440 or 1.2207441, which is accurate to six decimal p laces." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 87 "Let's see how well we did with our approximation by asking Maple to s olve the equation " }{XPPEDIT 18 0 "x^2-sqrt(x+1) = 0;" "6#/,&*$%\"xG \"\"#\"\"\"-%%sqrtG6#,&F&F(\"\"\"F(!\"\"\"\"!" }{TEXT -1 32 " , which \+ is also the equation f(" }{TEXT 310 1 "x" }{TEXT -1 8 ") = 0. " }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "solve(f(x)=0); evalf(%);" }} }{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 152 " Our approximation of 1.2207440 or 1.2207441, obtained by the IVT, isn' t too bad. Note: the equation also has a solution (root) in the inter val (-1, 0)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }} }{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 74 "2 . Let's consider Problem 37 on page113. For what value of the consta nt " }{TEXT 268 1 "c" }{TEXT -1 35 " is the function f continuous on ( -" }{XPPEDIT 18 0 "infinity;" "6#%)infinityG" }{TEXT -1 2 ", " } {XPPEDIT 18 0 "infinity;" "6#%)infinityG" }{TEXT -1 2 ")?" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 2 "f(" }{TEXT 258 1 "x " }{TEXT -1 4 ") = " }{XPPEDIT 18 0 "PIECEWISE([c*x+1, x <= 3],[c*x^2- 1, 3 < x]);" "6#-%*PIECEWISEG6$7$,&*&%\"cG\"\"\"%\"xGF*F*\"\"\"F*1F+\" \"$7$,&*&F)F**$F+\"\"#F*F*\"\"\"!\"\"2\"\"$F+" }{TEXT -1 1 " " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "f := x -> piecewise(x<=3, c*x + 1, x>3, c*x^2 - 1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "f(x);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 32 "Note: the piece of the fu nction " }{TEXT 259 1 "y" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "c*x+1;" "6 #,&*&%\"cG\"\"\"%\"xGF&F&\"\"\"F&" }{TEXT -1 70 " is the equation of a line (polynomial). Hence, it is continuous for " }{TEXT 262 1 "x" } {TEXT -1 33 " < 3. The piece of the function " }{TEXT 260 1 "y" } {TEXT -1 3 " = " }{XPPEDIT 18 0 "c*x^2-1;" "6#,&*&%\"cG\"\"\"*$%\"xG\" \"#F&F&\"\"\"!\"\"" }{TEXT -1 74 " is the equation of a parabola (poly nomial). Hence, it is continuous for " }{TEXT 261 1 "x" }{TEXT -1 48 " > 3. Now, we need to check for continuity at " }{TEXT 263 1 "x" } {TEXT -1 116 " = 3. In order for the function to be continuous at thi s number, the point, where the graph of the line stops when " }{TEXT 264 1 "x" }{TEXT -1 78 " = 3, is where the graph of the parabola must \+ start. Thus, we must have that " }{XPPEDIT 18 0 "3*c+1 = 9*c-1;" "6#/ ,&*&\"\"$\"\"\"%\"cGF'F'\"\"\"F',&*&\"\"*F'F(F'F'\"\"\"!\"\"" }{TEXT -1 43 ". We can use Maple to solve this equation." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "solve(3*c+1 = 9*c-1);" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 30 "We can have Maple \+ replace the " }{TEXT 265 1 "c" }{TEXT -1 54 " in our function by this \+ solution of 1/3 by using the " }{TEXT 306 5 "subs " }{TEXT -1 27 "comm and and then using the " }{TEXT 307 7 "unapply" }{TEXT -1 39 " command to have f as a function again:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "subs( c=1/3, f(x) ); f := unapply(%,x);" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 5 "f(x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "plot(f(x), x);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18 "Practice Problems:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 78 "1. Use the Intermediate Value The orem to find one real root to the equation " }{XPPEDIT 18 0 "x^3-3*x+ 1 = 0;" "6#/,(*$%\"xG\"\"$\"\"\"*&\"\"$F(F&F(!\"\"\"\"\"F(\"\"!" } {TEXT -1 27 " to five decimal places. " }{MPLTEXT 0 21 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 78 "2. Use the Intermediate Value Theorem to find one real root to th e equation " }{XPPEDIT 18 0 "x^3-x^2+x-10 = 0;" "6#/,**$%\"xG\"\"$\" \"\"*$F&\"\"#!\"\"F&F(\"#5F+\"\"!" }{TEXT -1 25 " to six decimal pla ces." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 38 "3. For wha t value(s) of the constant " }{TEXT 267 1 "c" }{TEXT -1 35 " is the fu nction f continuous on (-" }{XPPEDIT 18 0 "infinity;" "6#%)infinityG" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "infinity;" "6#%)infinityG" }{TEXT -1 2 ")?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 8 " \+ f(" }{TEXT 266 1 "x" }{TEXT -1 5 ") = " }{XPPEDIT 18 0 "PIECEWIS E([1/2*x^2+c^2, x < 1],[x/c+5, 1 <= x]);" "6#-%*PIECEWISEG6$7$,&*(\"\" \"\"\"\"\"\"#!\"\"%\"xG\"\"#F**$%\"cG\"\"#F*2F-\"\"\"7$,&*&F-F*F0F,F* \"\"&F*1\"\"\"F-" }{TEXT -1 3 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 38 "3. For what value(s) of the constant " }{TEXT 308 1 "c" }{TEXT -1 35 " is the function g continuo us on (-" }{XPPEDIT 18 0 "infinity;" "6#%)infinityG" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "infinity;" "6#%)infinityG" }{TEXT -1 2 ")?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 8 " g(" }{TEXT 309 1 "x" }{TEXT -1 3 ") =" }{XPPEDIT 18 0 "PIECEWISE([x^2-c^2, x < 4] ,[c*x+20, 4 <= x]);" "6#-%*PIECEWISEG6$7$,&*$%\"xG\"\"#\"\"\"*$%\"cG\" \"#!\"\"2F)\"\"%7$,&*&F-F+F)F+F+\"#?F+1\"\"%F)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "0 0 0 " 65 }{VIEWOPTS 1 1 0 1 1 1803 }