{VERSION 5 0 "IBM INTEL NT" "5.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "" -1 256 "" 0 1 0 128 0 1 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 257 "" 0 1 0 128 0 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 128 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 128 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 0 1 0 128 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 261 "" 0 1 0 128 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 262 "" 0 1 0 128 128 1 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 263 "" 0 1 0 128 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 0 1 128 0 128 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 265 "" 0 1 128 0 128 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 266 "" 0 1 128 0 128 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 267 "" 0 1 128 0 128 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 0 1 128 0 128 1 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 269 "" 0 1 128 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 270 "" 0 1 128 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 271 "" 0 1 128 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 272 "" 0 1 128 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 273 "" 0 1 0 128 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 274 "" 0 1 128 128 128 1 0 0 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 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 1" -1 3 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 4 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 2" 3 4 1 {CSTYLE "" -1 -1 "" 1 14 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 8 2 0 0 0 0 0 0 -1 0 }} {SECT 0 {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 257 49 "High School Modul es > Algebra by Gregory A. Moore" }}{PARA 3 "" 0 "" {TEXT -1 4 " " }{TEXT 256 31 "Solving Non-Linear Inequalities" }}{PARA 0 "" 0 "" {TEXT -1 57 "\nSolving all types of basic absolute value inequalities. \n" }}{PARA 0 "" 0 "" {TEXT 258 153 "[Directions : Execute the Code Re source section first. Although there will be no output immediately, th ese definitions are used later in this worksheet.]" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 5 "Code " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "restart: with(plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1044 "PolyInequal := proc(expr)\n lo cal k,j,s,t,a,b,minS,maxS,L;\n for k from 1 to nops(expr) do\n \+ s||k := solve( op(k, expr) = 0); od:\n minS := s||1: maxS := s|| 1:\n for k from 1 to nops(expr) do\n if( s||k < minS) then m inS := s||k fi:\n if( s||k > maxS) then minS := s||k fi:\n o d:\n L := (maxS - minS): a:= floor( minS - L/4): b:= ceil( maxS + L /4):\n\n for k from 1 to nops(expr) do\n printf(\"\\n \"); \n for j from a to s||k do printf(\"|----\") od;\n for j from s||k to b do printf(\"|++++\"); od;\n printf(\"|\\t\\t \\t\\t%A\\n\",(op(k,expr)));\n od;\n\n for j from a-1 to b+1 do \+ \n if( j < 0 ) then printf(\"\\t%d \", j); else printf(\"\\t %d \", j); fi: od;\n for k from 1 to 1 do\n printf(\"\\n \") ;\n for j from a-1 to b do \n subs(x = j +.5, expr ) : t := round(%/abs(%)):\n if ( t > 0 ) then printf(\"|++++ \"); else printf(\"|----\") fi; od:\n printf(\"| %A\", \+ expr);\n od; \n print(` `); print(solve( expr > 0)); print(` `);\nend proc:\n\n\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1029 " QuoInequal := proc(top, bottom )\n local k,j,s,t,a,b,minS,maxS,L,ex pr;\n expr := top*bottom;\n for k from 1 to nops(expr) do\n \+ s||k := solve( op(k, expr) = 0); od:\n minS := s||1: maxS := s|| 1:\n for k from 1 to nops(expr) do\n if( s||k < minS) then m inS := s||k fi:\n if( s||k > maxS) then minS := s||k fi:\n o d:\n L := (maxS - minS): a:= floor( minS - L/4): b:= ceil( maxS + L /4):\n\n for k from 1 to nops(expr) do\n printf(\"\\n \"); \n for j from a to s||k do printf(\"|----\") od;\n for j from s||k to b do printf(\"|++++\"); od;\n printf(\"|\\t\\t \\t\\t%A\\n\",(op(k,expr)));\n od;\n\n for j from a-1 to b+1 do \+ \n if( j < 0 ) then printf(\"\\t%d \", j); else printf(\"\\t %d \", j); fi: od;\n for k from 1 to 1 do\n printf(\"\\n \") ;\n for j from a-1 to b do \n subs(x = j +.5, expr ) : t := round(%/abs(%)):\n if ( t > 0 ) then printf(\"|++++ \"); else printf(\"|----\") fi; od:\n printf(\"| %A\", \+ expr);\n od; \n \nend proc:" }}}{PARA 3 "" 0 "" {TEXT -1 0 " " }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 40 " 1. Linear Inequalities \+ " }{TEXT 259 15 " (x - a) > 0" }}{PARA 0 "" 0 "" {TEXT -1 102 "\nIt might be a good idea to review the basic concept of solvi ng a simple linear equality of this type." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "(x-3) > 0; solve(%, x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "(x+7) > 0; solve(% ,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "(x-11) >= 0; solv e(%,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "(x-4) <= 0; so lve(%,x);" }}}{PARA 0 "" 0 "" {TEXT -1 123 "\n\nIts slightly more comp licated when we're not just dealing with |x| but the absolute value of some linear expression of x." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "abs(x - 3) = 10; x = \{solve(%,x)\};" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 32 " 2. Product of T erms " }{TEXT 260 36 " (x - a)(x + b)(x - c) > 0 " }}{PARA 0 "" 0 "" {TEXT -1 89 " \nA polynomial inequality with two o r more factors requires a little more work to solve. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "(x-4)*(x+ 3) > 0;" }}}{PARA 0 "" 0 "" {TEXT -1 72 "\nWe can get a little idea of what is happening by looking at some values" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "for k from -7 to 9 do \n `x = `,k,` Value = `,( k-4)*(k+3); od;" }}}{PARA 0 "" 0 "" {TEXT -1 198 "We can see that the \+ signs are positive then negative and then positive again. There is int eraction between the signs of the two terms.\n\nThe way to solve these is to create a sign graph. \n " }{TEXT 268 1 " " }{TEXT 264 1 "1" }{TEXT 269 66 ". Factor the expression (if its not already f actored)\n " }{TEXT 265 1 "2" }{TEXT 270 90 ". Draw a graph of the sign values for each factor above a common number line\n \+ " }{TEXT 266 1 "3" }{TEXT 271 68 ". Multiply the signs down to \+ get a resulting sign graph\n " }{TEXT 267 1 "4" }{TEXT 272 168 ". \"Read\" the sign graph - look at the original inequality and c heck whether it is >, >=, <, or <= 0, l\n then look fo r the appropriate part of the graph\n" }{TEXT -1 50 "\nLet's create a \+ sign graph for the problem above.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "PolyInequal( (x-4)*(x+3)); " }}}{PARA 0 "" 0 "" {TEXT -1 235 "\nLet's look at some other examples. Even if there are m ore factors, we can use the same method. The only difference will be t hat we have additional rows - so that we have a sign graph for each fa ctor. These examples have three factors." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "(x-3)*(x+1)*(x+5) > 0;\nPolyInequal( (x-3)*(x+1)*(x+5 )); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "(x-7)*(x+1)*(x+2) > 0;\nPolyInequal( (x-7)*(x+1)*(x+2) ); " }}}{PARA 0 "" 0 "" {TEXT -1 41 "\nThese examples have four or more terms.\n" }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 68 "(x-3)*(x+7)*(x+1)*(x+2) > 0;\nPolyInequal( (x- 3)*(x+4)*(x+1)*(x+2) );" }}}{PARA 0 "" 0 "" {TEXT -1 1 "\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 88 "(x-4)*(x-3)*(x-1)*(x+1)*(x+2)*(x) > 0;\nPolyInequal( (x-4)*(x-3)*(x-1)*(x+1)*(x+2)*(x) );" }}}{PARA 0 "" 0 "" {TEXT -1 1 "\n" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 21 " 3. Quotient of Terms" }{TEXT 261 43 " \+ (x - a) / (x - c) > 0" }}{PARA 0 "" 0 "" {TEXT -1 40 "\nN ow lets look at quotient inequalities." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "(x-4)/(x+3) >= 0;" }}} {PARA 0 "" 0 "" {TEXT -1 138 "\nLet's make a list of values so we can \+ compare what happens to the product of these two terms compared to the quotient of these two terms." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 223 "for k from -5 to 8 do \n if((k+3) = 0) \n then print(`x = `,k,` Product = `,(k-4)*(k+3), ` Quotient = undefined`): \n else print(`x = `,k,` Product = `,(k-4)*(k+3), ` Quotient = `, evalf((k- 4)/(k+3),5)); \n\nfi; od;" }}}{PARA 0 "" 0 "" {TEXT -1 426 "\nWhat do \+ we see? There are a couple of things :\n 1. When the product is zero, the quotient is either 0 or undefined. In particular, it's unde fined when the quotient is zero.\n 2. Although the product and \+ quotient take different values, they always take the SAME SIGN.\n\nThu s we can use the same graphing method as above - with one minor change . We need to toss out any value which would make the denominator zero. \n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "QuoInequal((x-4),(x+3) );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "solve( (x-4)/(x+3) \+ >= 0 );" }}}{PARA 0 "" 0 "" {TEXT -1 120 "\nNote that the endpoint of \+ -3 is OPEN rather than closed, while the endpoint of 4 is CLOSED!\n\nO ther examples : \n " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 102 "Ineq := ((x-1)*(x-2))/((x-3)*(x-4)) > 0;\nQuoInequal((x-1)*(x-2), (x-3)*(x-4));\nprint(` `);solve(Ineq );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 94 "Ineq := ((x-6)*(x+3)) / (x-5) > 0;\nQuoInequal( (x- 6)*(x+3), (x-5));\nprint(` `);solve(Ineq );" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 44 " 4. More Complicated Q uotient Inequalities " }{TEXT 273 39 "(x + a) / (x + b) > (x + c) / \+ (x + d) " }}{PARA 0 "" 0 "" {TEXT -1 247 "\nThere are other problems i nvolving quotients which appear to be much more confusing. However, if you simply move all of the terms to one side of the equation, and com bine the rational terms, the problem will be converted to a more famil iar form.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "(x-3)/(x+4) > 1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "(x-3)/(x+4)- 1 > 0 \+ ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "0 < simplify(rhs(%)); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "0 > 1/(x+4);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "solve(%);" }}}{PARA 0 "" 0 " " {TEXT -1 26 "\n\nHere is another example." }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 26 "(x-3)/(x+4) > (x+6)/(x-1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "lhs(%) - rhs(%) < 0;" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 21 "simplify(lhs(%)) < 0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "solve( 7*(2*x+3)/(x-1)/(x+4) < 0 );" }}}{PARA 0 "" 0 "" {TEXT -1 1 "\n" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 43 " 5. Graphing Product Inequalities " } {TEXT 262 32 " (x - a)(x + b)(x - c) > 0" }}{PARA 0 "" 0 "" {TEXT -1 92 "\nYou can visualize an absolute value equation by looking at the intersection of two graphs.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 317 "expr := (x-3)*(x+2): % > 0; \na := -5: b := 5:\ns := solve(expr = 0, x): a1 := min( s): b1 := max(s):\nplots[display]( \+ \n plot( expr, x = a..b, thickness = 2),\n plots[polygonplot]( [[a,-.5],[a1,-.5],[a1,.5],[a,.5]], color = green),\n plots[polygonp lot]( [[b1,-.5],[b,-.5],[b,.5],[b1,.5]], color = green)\n);\n \n\n" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 241 "expr := (x-3)*(x+2): % < 0 ; \na := -5: b := 5:\ns := solve(expr = 0, x): a1 := min( s): b1 := \+ max(s):\nplots[display]( \n plot( expr, x = a..b, thickness = 2), \n plots[polygonplot]( [[a1,-.5],[b1,-.5],[b1,.5],[a1,.5]], color \+ = green)\n);" }}}{PARA 0 "" 0 "" {TEXT -1 51 "\nHere are instances of \+ the more complicated cases. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 313 "expr := (5-x)*(x+2): % < 0; \na := -4: b := 7:\ns := solve(expr = 0, x): a1 := min( s): b1 := max(s):\nplots[display]( \n plot( e xpr, x = a..b, thickness = 2),\n plots[polygonplot]( [[a,-.5],[a1, -.5],[a1,.5],[a,.5]], color = green),\n plots[polygonplot]( [[b1,-.5 ],[b,-.5],[b,.5],[b1,.5]], color = green)\n);" }}}{PARA 0 "" 0 "" {TEXT -1 85 "\n\nNotice the only parts of the solution here are where \+ the graph is BELOW the x axis." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 324 "expr := (x+4)*(x+2)*(x-3): % < 0; \na := -5: b := 3.5:\ns := so lve(expr < 0, x); a1 := -4: a2:= -2: a3:= 3:\nplots[display]( \n \+ plot( expr, x = a..b, thickness = 2),\n plots[polygonplot]( [[a, -.5],[a1,-.5],[a1,.5],[a,.5]], color = green),\n plots[polygonplot ]( [[a2,-.5],[a3,-.5],[a3,.5],[a2,.5]], color = green)\n);" }}}{PARA 0 "" 0 "" {TEXT -1 1 "\n" }}{PARA 0 "" 0 "" {TEXT -1 117 "\nWhen the a bsolute value is LESS than the constant, then we are looking for where the V is BELOW the horizontal line." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 320 "expr := (x+4)*(x+2)*(x-3): % < 0; \na := -5: b := 4: \ns := solve(expr < 0, x); a1 :=-4: a2:= -2: a3:= 3:\nplots[display] ( \n plot( expr, x = a..b, thickness = 2),\n plots[polygonplot ]( [[a,-.5],[a1,-.5],[a1,.5],[a,.5]], color = green),\n plots[poly gonplot]( [[a2,-.5],[a3,-.5],[a3,.5],[a2,.5]], color = green)\n);" }}} {PARA 0 "" 0 "" {TEXT -1 120 "\nWhen the absolute value is GREATER tha n the constant, then we are looking for where the V is ABOVE the horiz ontal line." }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 50 " 6. Graphing Quot ient Inequalities " }{TEXT 263 23 "(x - a) / (x - c) > 0" }}{PARA 0 "" 0 "" {TEXT -1 96 "\nYou can visualize a quotient too . In this case, we're looking for where the graph is positive.\n" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 321 "expr := (x+4)/(x-3): % > 0; solve(%);\na := -9: b := 11: a1 := -4: a2:= 3:\nplots[display ]( \n plot( expr, x = a..b, y = -4..7,thickness = 2, discont = tru e),\n plots[polygonplot]( [[a,-.5],[a1,-.5],[a1,.5],[a,.5]], color = green),\n plots[polygonplot]( [[a2,-.5],[b,-.5],[b,.5],[a2,.5]] , color = green)\n);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 247 "ex pr := (x+4)/(x-3): % < 0; solve(%);\na := -9: b := 11: a1 := - 4: a2:= 3:\nplots[display]( \n plot( expr, x = a..b, y = -4..7,thi ckness = 2, discont = true),\n plots[polygonplot]( [[a1,-.5],[a2,- .5],[a2,.5],[a1,.5]], color = green)\n);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 434 "expr := ((x+4)*(x-1))/((x-3)*(x+2)): % > 0; solve(% );\na := -9: b := 11: a1 := -4: a2 := -2: a3 := 1: a4 := 3:\nplo ts[display]( \n plot( expr, x = a..b, y = -4..7,thickness = 2, dis cont = true),\n plots[polygonplot]( [[a,-.5],[a1,-.5],[a1,.5],[a,. 5]], color = green),\n plots[polygonplot]( [[a2,-.5],[a3,-.5],[a3, .5],[a2,.5]], color = green),\n plots[polygonplot]( [[a4,-.5],[b,- .5],[b,.5],[a4,.5]], color = green)\n);" }}}}{PARA 0 "" 0 "" {TEXT 274 36 "\n \251 2002 Waterloo Maple Inc " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "0 1" 19 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 } {PAGENUMBERS 0 1 2 33 1 1 }