355 by tanjawi1993

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                                                                                               ‫ﺣﺴﺎب اﻟﻨﻬﺎﻳﺎت‬

                                                                                          ‫- ﻟﺪﻳﻨﺎ :‬
                                                                                           ‫1 + 3‪x‬‬
                                                                       ‫‪lim f ( x) = lim‬‬
                                                                       ‫∞−→ ‪x‬‬        ‫∞−→ ‪x‬‬     ‫‪x‬‬

                                                                                                   ‫3‪x‬‬
                                                                                        ‫‪= lim‬‬
                                                                                           ‫∞−→ ‪x‬‬   ‫‪x‬‬
                                                                                       ‫∞+ = ‪= lim x‬‬
                                                                                                 ‫2‬
                                                                                           ‫∞−→ ‪x‬‬

                                                                                             ‫ﻟﺪﻳﻨﺎ‬
                                                                                  ‫1+ ‪x + x − 2x‬‬
                                                                                           ‫4‬       ‫3‬   ‫2‬
                                                             ‫‪lim f ( x) = lim‬‬
                                                             ‫∞+→ ‪x‬‬        ‫∞+→ ‪x‬‬         ‫3‪x‬‬
                                                                                    ‫4‪x‬‬
                                                                         ‫‪= lim‬‬
                                                                            ‫∞+→ ‪x‬‬   ‫3‪x‬‬
                                                                          ‫∞+ = ‪= lim x‬‬
                                                                                 ‫∞+→ ‪x‬‬



                                                                                 ‫1 + 3‪x‬‬
                                                              ‫‪lim f ( x) = lim‬‬          ‫ﻟﺪﻳﻨﺎ : ∞− =‬
                                                                 ‫−‬
                                                              ‫0→ ‪x‬‬        ‫0→ ‪x‬‬‫−‬
                                                                                  ‫3‪x‬‬
                                                            ‫ﻷن : 1 = )1 + 3 ‪ lim ( x‬و −0 = ‪lim −x‬‬
                                                                               ‫−‬
                                                             ‫0→ ‪x‬‬               ‫0→ ‪x‬‬
                                                                                                       ‫ﻟﺪﻳﻨﺎ :‬
                                                                        ‫1+ ‪x + x − 2x‬‬
                                                                                ‫4‬      ‫3‬           ‫2‬
                                                          ‫‪lim f ( x) = lim‬‬             ‫∞+ =‬
                                                        ‫0→ ‪x‬‬ ‫+‬
                                                                 ‫0→ ‪x‬‬     ‫+‬
                                                                              ‫3‪x‬‬
                                                   ‫ﻷن 1 + 2 ‪ lim x 4 + x3 − 2 x‬و +0 = 3‪lim x‬‬
                                                      ‫+‬               ‫+‬
                                                   ‫0→ ‪x‬‬                ‫0→ ‪x‬‬



                                                          ‫2 أ ﻟﻨﺒﻴﻦ أن ) ‪ ( Δ‬ﻣﻘﺎرب ﻟﻠﻤﻨﺤﻨﻰ ﺑﺠﻮار ∞+‬
                                                                                            ‫ﻟﺪﻳﻨﺎ :‬
                                                                        ‫⎛‬       ‫⎞ 1 2‬
                                          ‫)1 + ‪lim f ( x) − ( x + 1) = lim ⎜ x + 1 − + 3 ⎟ − ( x‬‬
                                          ‫∞+→ ‪x‬‬                   ‫∞+→ ‪x‬‬
                                                                        ‫⎝‬       ‫⎠ ‪x x‬‬
                                                                                 ‫⎞ 1 2 ⎛‬
                                                                           ‫0 = ⎟ 3 + − ⎜ ‪lim‬‬
                                                                          ‫∞+→ ‪x‬‬
                                                                                 ‫⎠ ‪⎝ x x‬‬

               ‫إذن اﻟﻤﺴﺘﻘﻴﻢ ) ‪ ( Δ‬ذو اﻟﻤﻌﺎدﻟﺔ 1 + ‪ y = x‬هﻮ ﺑﺎﻟﻔﻌﻞ ﻣﻘﺎرب ﻣﺎﺋﻞ ﻟﻠﻤﻨﺤﻨﻰ ) ‪ (ζ‬ﺑﺠﻮار ∞+‬
          ‫ب – دراﺳﺔ وﺿﻊ ) ‪ (ζ‬ﺑﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﺴﺘﻘﻴﻢ ) ‪ ( Δ‬ﻋﻠﻰ اﻟﻤﺠﺎل [∞ + '1] ﻟﻜﻞ ‪ x‬ﻣﻦ [∞ + '1] ﻟﺪﻳﻨﺎ :‬
                                                                                               ‫1 2−‬
                                                                            ‫= )1 + ‪f ( x ) − ( x‬‬   ‫+‬
                                                                                                ‫3‪x x‬‬
                                                                                              ‫2 ‪1 − 2x‬‬
                                                                                            ‫=‬
                                                                                                 ‫3‪x‬‬
        ‫ﻟﻜﻞ ‪ x‬ﻣﻦ [∞+ ,1] ﻓﺈن 0 ≺ )1 + ‪( ∀x ∈ ]1 + ∞[ ) f ( x ) − ( x‬‬
                ‫'‬                                                      ‫3‪x‬‬      ‫وﺑﻤﺎ أن 0 ≺ 2 ‪ 1 − 2 x‬و 0‬
                            ‫)‪(Δ‬‬   ‫وﺑﺎﻟﺘﺎﻟﻲ ﻓﺈﻧﻪ ﻓﻲ اﻟﻤﺠﺎل [∞+ ,1] ﻳﻜﻮن اﻟﻤﻨﺤﻨﻰ ) ‪ (ζ‬ﺗﺤﺖ ﻣﻘﺎرﺑﻪ اﻟﻤﺎﺋﻞ‬
                                                                ‫ج - ﺑﺎﻗﻲ اﻟﻔﺮوع اﻟﻼﻧﻬﺎﺋﻴﺔ ﻟﻠﻤﻨﺤﻨﻰ )‪( C‬‬
                                                                                                   ‫ﻟﺪﻳﻨﺎ‬
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                                                                            ‫‪lim‬‬     ‫∞+ = ) ‪f ( x‬‬
                                                                            ‫∞−→ ‪x‬‬

                                                                      ‫)‪f ( x‬‬          ‫1 + 3‪x‬‬
                                                            ‫‪lim‬‬              ‫‪= lim‬‬           ‫و‬
                                                            ‫∞−→‪x‬‬       ‫‪x‬‬       ‫∞−→‪x‬‬     ‫2‪x‬‬
                                                                                        ‫3‪x‬‬
                                                                              ‫‪= lim‬‬
                                                                                ‫∞−→ ‪x‬‬   ‫2‪x‬‬
                                                                            ‫‪= lim‬‬      ‫∞− = ‪x‬‬
                                                                            ‫∞−→‪x‬‬
                  ‫إذن اﻟﻤﻨﺤﻨﻰ ) ‪ (ζ‬ﻳﻘﺒﻞ ﺑﺠﻮار ∞− ﻓﺮﻋﺎ ﺷﻠﺠﻤﻴﺎ إﺗﺠﺎهﻪ ﻣﺤﻮر اﻷراﺗﻴﺐ .‬
                                                                                              ‫ﻟﺪﻳﻨﺎ‬
                                            ‫+‪= lim‬‬     ‫−‪ lim‬و ∞+ = ) ‪f ( x‬‬          ‫∞− = ) ‪f ( x‬‬
                                              ‫0→ ‪x‬‬                        ‫0→ ‪x‬‬

                                  ‫إذن اﻟﻤﺴﺘﻘﻴﻢ ذو اﻟﻤﻌﺎدﻟﺔ 0 = ‪ x‬هﻮ ﻣﻘﺎرب رأﺳﻲ ﻟﻠﻤﻨﺤﻨﻰ ) ‪(ζ‬‬
                                                                                      ‫3 ﺗﺤﺪﻳﺪ ' ‪f‬‬
                      ‫اﻟﺪاﻟﺔ ‪ f‬ﻗﺎﺑﻠﺔ ﻟﻼﺷﺘﻘﺎق ﻋﻠﻰ آﻞ ﻣﻦ اﻟﻤﺠﺎﻟﻴﻦ [∞+ ,0] و [0 ,∞−] وﻟﺪﻳﻨﺎ :‬
                                                                          ‫ﻟﻜﻞ ‪ x‬ﻣﻦ [∞ + '0] :‬

                          ‫−1+ ‪(x ) = ⎛ x‬‬
                                         ‫⎞ 1 2‬
                  ‫‪f‬‬   ‫'‬
                                 ‫⎜‬         ‫+‬  ‫⎟‬
                                ‫⎝‬        ‫⎠3‪x x‬‬
                                  ‫3 2‬
                            ‫4 − 2 +1 =‬
                                  ‫‪x‬‬    ‫‪x‬‬
                                 ‫3 − 2‪x + 2 x‬‬
                                  ‫4‬
                              ‫=‬
                                     ‫4‪x‬‬
                                  ‫4 − )1 + ‪x 4 + 2 x 2 + 1 − 4 ( x‬‬
                                                           ‫2‬      ‫2‬

                               ‫=‬                      ‫=‬
                                          ‫4‪x‬‬                   ‫4‪x‬‬

                                ‫=‬
                                   ‫)3 + 2 ‪( x2 + 1 − 2 )( x2 + 1 + 2 ) = ( x2 − 1)( x‬‬
                                                ‫4‪x‬‬                        ‫4‪x‬‬
                                                                           ‫ﻟﻜﻞ ‪ x‬ﻣﻦ [0 ' ∞−] :‬
                                                            ‫⎛‬      ‫⎞1‬
                                                ‫⎟ + 2‪f ' ( x ) = ⎜ x‬‬
                                                            ‫⎝‬      ‫⎠‪x‬‬
                                                                 ‫1‬
                                                        ‫2 − ‪= 2x‬‬
                                                                 ‫‪x‬‬
                                                           ‫1 − ‪2x‬‬
                                                               ‫3‬
                                                         ‫=‬
                                                              ‫2‪x‬‬
                                                                     ‫4 – ﺟﺪول ﺗﻐﻴﺮات ‪f‬‬
                                    ‫* ﻋﻠﻰ اﻟﻤﺠﺎل [∞ + '0] إﺷﺎرة ) ‪ f ( x‬هﻲ إﺷﺎرة 1 − ‪x‬‬
                                      ‫2‬                 ‫'‬


                                   ‫* ﻋﻠﻰ اﻟﻤﺠﺎل [0 ' ∞−] إﺷﺎرة ) ‪ f ' ( x‬هﻲ إﺷﺎرة 1 − 3 ‪2 x‬‬
                           ‫ﻓﺈن 0 ≺ 3‪ 2 x‬وﻣﻨﻪ 0 ≺ 1 − 3 ‪ 2 x‬ﻟﻜﻞ ‪ x‬ﻣﻦ [0 '∞−]‬      ‫وﺑﻤﺎ أن 0 ≺ ‪x‬‬
                                                                        ‫إذن ﺟﺪول ﺗﻐﻴﺮات ‪ f‬هﻮ :‬
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                                                                             ‫) ‪(ζ‬‬   ‫5 – أ ﻧﻘﻄﺘﺎ اﻧﻌﻄﺎف اﻟﻤﻨﺤﻨﻰ‬
                                                       ‫)1 + ‪2 ( x + 1) ( x 2 − x‬‬
                      ‫= ) ‪( ∀x ∈ ]−∞, 0[ ) f "( x‬‬          ‫3‪x‬‬
                                                                                                         ‫* ﻟﺪﻳﻨﺎ‬

                                                  ‫)1 + ‪− x − 1 2 ( x − x‬‬
                                                               ‫2‬

                                                ‫=‬        ‫.‬
                                                    ‫‪−x‬‬           ‫2‪x‬‬
                                         ‫0 1 + ‪ x 2 − x‬ﻟﻜﻞ ‪ x‬ﻣﻦ ) 0 ≺ ‪( Δ‬‬                               ‫وﺑﻤﺎ أن‬
                                          ‫ﻓﻲ اﻟﻤﺠﺎل [0 '∞−] هﻲ إﺷﺎرة )1 − ‪( − x‬‬              ‫ﻓﺈن إﺷﺎرة ) ‪f ' ( x‬‬
                                               ‫) 3 − 2 ‪−4 ( x‬‬
                  ‫= ) ‪( ∀x ∈ ]0, +∞[ ) f ( x‬‬
                                    ‫''‬

                                                       ‫5‪x‬‬
                                                                                                        ‫ﻟﺪﻳﻨﺎ :‬


                                          ‫=‬
                                               ‫4‬   ‫(‬         ‫()‬
                                                       ‫3 +‪3−x x‬‬         ‫)‬
                                                            ‫5‪x‬‬
                                                ‫(‬           ‫)‬
                                                       ‫إذن إﺷﺎرة ) ‪ f ' ( x‬ﻓﻲ [∞ + '0] هﻲ إﺷﺎرة ‪3 − x‬‬
                                                                                     ‫'‬




                                                                  ‫‪: f‬‬   ‫''‬
                                                                             ‫) ‪(x‬‬   ‫* اﻟﺠﺪول اﻟﺘﺎﻟﻲ ﻳﻌﻄﻲ إﺷﺎرة‬




                                                   ‫‪ f‬ﺗﻨﻌﺪم ﻣﻊ ﺗﻐﻴﻴﺮ اﻹﺷﺎرة ﻓﻲ آﻞ ﻣﻦ )1− ( و 3‬           ‫''‬
                                                                                                             ‫اﻟﺪاﻟﺔ‬
                                                       ‫إذن ﻟﻠﻤﻨﺤﻨﻰ ) ‪ (ζ‬ﻧﻘﻄﺘﺎ اﻧﻌﻄﺎف هﻤﺎ ) 01− ( ‪ A‬هﻮ‬
                                                                ‫'‬

                                                                    ‫⎛‬     ‫⎞ 3 3+4‬
                                                                  ‫,3 ⎜ ‪B‬‬
                                                                    ‫⎜‬             ‫⎟‬
                                                                                  ‫⎟‬
                                                                    ‫⎝‬       ‫⎠ 3 3‬
                                                                    ‫ب * ﻣﻌﺎدﻟﺔ ﻟﻤﻤﺎس ) ‪ (ζ‬ﻓﻲ ) 01− ( ‪A‬‬
                                                                          ‫'‬

                                                                              ‫3 – ‪y = -3x‬‬
                                                                 ‫⎛‬      ‫⎞ 3 3+4‬
                                                                ‫⎜‪B‬‬
                                                                 ‫⎜‬   ‫,3‬       ‫ﻣﻌﺎدﻟﺔ ﻟﻤﻤﺎس ) ‪ (ζ‬ﻓﻲ ⎟‬
                                                                 ‫⎝‬       ‫⎟ 3 3‬‫⎠‬
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                       4    3 3 −8
                  y=     x+
                       3     3 3
                                 (ζ )   ‫6 – رﺳﻢ اﻟﻤﻨﺤﻨﻰ‬

								
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