Calculo 1, lista de cálculo 1 engenharia química, Exercícios de Cálculo

Baixe Calculo 1, lista de cálculo 1 engenharia química e outras Exercícios em PDF para Cálculo, somente na Docsity! 1 Universidade Federal de Alagoas Instituto de Matemática Banco de Questões Calculo 1 (Resolução) Organizador: Carlos Alberto Santos Barbosa Maceio, Brasil Janeiro de 2015 2 Apresentação O objetivo principal desta resolução do Banco de Questões (extraído das avaliações escritas de turmas de Cálculo Unificado da Universidade Federal de Alagoas - UFAL) é auxiliar no desempenho dos estudantes da disciplina Cálculo 1, que durante o processo de conhecimento começam a inserir-se no aprendizado de modo a melhorar seu desempenho acadêmico, mostrando-lhes uma noção de como resolver questões das provas realizadas, visando sempre a clareza e objetividade na obtenção dos resultados e suas implicações referentes ao quesito. 5 2015.1 ............................................................................................................ 292 11.1 1ª Prova – 11 de Abril de 2015 ........................................................... 292 11.2 2ª Prova – 08 de Maio de 2015........................................................... 299 11.3 2ª Prova – 09 de Maio de 2015........................................................... 306 11.4 3ª Prova – 16 de Outubro de 2015 ..................................................... 315 11.5 3ª Prova – 17 de Outubro de 2015 ..................................................... 324 11.6 4ª Prova – 13 de Novembro de 2015 .................................................. 332 11.7 4ª Prova – 14 de Novembro de 2015 .................................................. 343 11.8 Reavaliação da 1ª Média – 27 de Novembro de 2015 ........................ 353 11.9 Reavaliação da 1ª Média – 28 de Novembro de 2015 ........................ 364 11.10 Reavaliação da 2ª Média – 27 de Novembro de 2015 ...................... 371 11.11 Reavaliação da 2ª Média – 28 de Novembro de 2015 ...................... 380 11.12 Avaliação Final – 03 de Dezembro de 2015 ..................................... 391 2015.2 ............................................................................................................ 412 12.1 1ª Prova – 12 de Fevereiro de 2016 ................................................... 412 12.2 1ª Prova – 13 de Fevereiro de 2016 ................................................... 418 12.3 2ª Prova – 11 de Março de 2016 ........................................................ 427 12.4 2ª Prova – 12 de Março de 2016 ........................................................ 434 12.5 3ª Prova – 08 de Abril de 2016 ........................................................... 440 12.6 3ª Prova – 09 de Abril de 2016 ........................................................... 448 12.7 4ª Prova – 06 de Maio de 2016........................................................... 457 12.8 4ª Prova – 07 de Maio de 2016........................................................... 468 12.9 Reavaliação da 1ª Média – 20 de Maio de 2016 ................................ 477 12.10 Reavaliação da 1ª Média – 21 de Maio de 2016............................... 487 12.11 Reavaliação da 2ª Média – 20 de Maio de 2016............................... 495 12.12 Reavaliação da 2ª Média – 21 de Maio de 2016............................... 504 12.13 Avaliação Final – 27 de Maio de 2016 .............................................. 513 2016.1 ............................................................................................................ 535 13.1 1ª Prova – 22 de Julho de 2016 .......................................................... 535 13.2 1ª Prova – 23 de Julho de 2016 .......................................................... 543 13.3 2ª Prova – 19 de Agosto de 2016 ....................................................... 551 13.4 2ª Prova – 20 de Agosto de 2016 ....................................................... 559 13.5 3ª Prova – 23 de Setembro de 2016 ................................................... 567 13.6 3ª Prova – 24 de Setembro de 2016 ................................................... 577 13.7 4ª Prova – 21 de Outubro de 2016 ..................................................... 585 13.8 4ª Prova – 22 de Outubro de 2016 ..................................................... 591 13.9 Reavaliação da 1ª Média – 27 de Outubro de 2016 ........................... 599 6 13.10 Reavaliação da 1ª Média – 29 de Outubro de 2016 ......................... 607 13.11 Reavaliação da 2ª Média – 27 de Outubro de 2016 ......................... 613 13.12 Reavaliação da 2ª Média – 29 de Outubro de 2016 ......................... 622 13.13 Avaliação Final – 04 de Novembro de 2016 ..................................... 629 2016.2 ............................................................................................................ 650 14.1 1ª Prova – 17 de Fevereiro de 2017 ................................................... 650 14.2 1ª Prova – 18 de Fevereiro de 2017 ................................................... 658 14.3 2ª Prova – 24 de Março de 2017 ........................................................ 666 14.4 2ª Prova – 25 de Março de 2017 ........................................................ 674 14.5 3ª Prova – 28 de Abril de 2017 ........................................................... 682 14.6 3ª Prova – 29 de Abril de 2017 ........................................................... 690 14.7 4ª Prova – 19 de Maio de 2017........................................................... 697 14.8 4ª Prova – 20 de Maio de 2017........................................................... 705 14.9 Reavaliação da 1ª Média – 26 de Maio de 2017 ................................ 713 14.10 Reavaliação da 1ª Média – 27 de Maio de 2017............................... 721 14.11 Reavaliação da 2ª Média – 26 de Maio de 2017............................... 728 14.12 Reavaliação da 2ª Média – 27 de Maio de 2017............................... 737 14.13 Avaliação Final – 02 de Junho de 2017 ............................................ 745 7 Capítulo 1 2005 1.1 1ª Avaliação-21 de Fevereiro de 2005 1. 𝑓(𝑥) = 𝑥3 − 1 5𝑥3 − 20𝑥2 + 15𝑥 𝑓(𝑥) = (𝑥 − 1)(𝑥2 + 𝑥 + 1) 5𝑥(𝑥 − 1)(𝑥 − 3) 𝐸𝑠𝑐𝑟𝑒𝑣𝑒𝑛𝑑𝑜 𝑓(𝑥) 𝑑𝑒𝑠𝑠𝑎 𝑓𝑜𝑟𝑚𝑎, 𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑐𝑜𝑛𝑐𝑙𝑢𝑖𝑟 𝑞𝑢𝑒 𝐷(𝑓) = ℝ − {0, 1, 3}. ∗ 𝐴𝑠𝑠í𝑛𝑡𝑜𝑡𝑎𝑠 𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑖𝑠: −𝐷𝑖𝑧𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑎 𝑟𝑒𝑡𝑎 𝑥 = 𝑎 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑠𝑒 𝑝𝑒𝑙𝑜 𝑚𝑒𝑛𝑜𝑠 𝑢𝑚 𝑑𝑜𝑠 𝑠𝑒𝑔𝑢𝑖𝑛𝑡𝑒𝑠 𝑐𝑎𝑠𝑜𝑠 𝑜𝑐𝑜𝑟𝑟𝑒𝑟: lim 𝑥→𝑎+ 𝑓(𝑥) = ±∞ 𝑜𝑢 lim 𝑥→𝑎− 𝑓(𝑥) = ±∞ 𝑉𝑎𝑚𝑜𝑠 𝑣𝑒𝑟𝑖𝑓𝑖𝑐𝑎𝑟 𝑠𝑒 𝑎𝑠 𝑟𝑒𝑡𝑎𝑠 𝑥 = 0; 𝑥 = 1 𝑒 𝑥 = 3 𝑠ã𝑜 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎𝑠 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑖𝑠. ∗ 𝑂𝑏𝑠: 𝑠𝑒 𝑥 → 0+, 𝑒𝑛𝑡ã𝑜 𝑥 > 0. 𝐿𝑜𝑔𝑜, 5𝑥 > 0. lim 𝑥→0+ 𝑓(𝑥) = lim 𝑥→0+ (𝑥 − 1)⏞ −1 ↑ (𝑥2 + 𝑥 + 1)⏞ 1 ↑ 5𝑥⏟ ↓ 0+ (𝑥 − 1)⏟ ↓ −1 (𝑥 − 3)⏟ ↓ −3 = lim 𝑥→0+ (𝑥 − 1)(𝑥2 + 𝑥 + 1)⏞ −1 ↑ 5𝑥(𝑥 − 1)(𝑥 − 3)⏟ ↓ 0+ = −∞ 𝐿𝑜𝑔𝑜, 𝑎 𝑟𝑒𝑡𝑎 𝑥 = 0 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑎𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑒 𝑓(𝑥). ∗ 𝑂𝑏𝑠: 𝑠𝑒 𝑥 → 1, 𝑒𝑛𝑡ã𝑜 𝑥 ≠ 1. 𝐿𝑜𝑔𝑜, 𝑥 − 1 ≠ 0 𝑒, 𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜, (𝑥 − 1) (𝑥 − 1) = 1, ∀𝑥 ≠ 1. lim 𝑥→1+ 𝑓(𝑥) = lim 𝑥→1+ (𝑥 − 1)(𝑥2 + 𝑥 + 1) 5𝑥(𝑥 − 1)(𝑥 − 3) = lim 𝑥→1+ (𝑥2 + 𝑥 + 1)⏞ 3 ↑ 5𝑥(𝑥 − 3)⏟ ↓ −10 = − 3 10 lim 𝑥→1− 𝑓(𝑥) = lim 𝑥→1− (𝑥 − 1)(𝑥2 + 𝑥 + 1) 5𝑥(𝑥 − 1)(𝑥 − 3) = lim 𝑥→1− (𝑥2 + 𝑥 + 1)⏞ 3 ↑ 5𝑥(𝑥 − 3)⏟ ↓ −10 = − 3 10 𝐿𝑜𝑔𝑜, 𝑎 𝑟𝑒𝑡𝑎 𝑥 = 1 𝑛ã𝑜 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑎𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑒 𝑓(𝑥). ∗ 𝑂𝑏𝑠: 𝑠𝑒 𝑥 → 3+, 𝑒𝑛𝑡ã𝑜 𝑥 > 3. 𝐿𝑜𝑔𝑜, 𝑥 − 3 > 0. lim 𝑥→3+ 𝑓(𝑥) = lim 𝑥→3+ (𝑥 − 1)⏞ 2 ↑ (𝑥2 + 𝑥 + 1)⏞ 13 ↑ 5𝑥⏟ ↓ 15 (𝑥 − 1)⏟ ↓ 2 (𝑥 − 3)⏟ ↓ 0+ = lim 𝑥→0+ (𝑥 − 1)(𝑥2 + 𝑥 + 1)⏞ 26 ↑ 5𝑥(𝑥 − 1)(𝑥 − 3)⏟ ↓ 0+ = +∞ 𝐿𝑜𝑔𝑜, 𝑎 𝑟𝑒𝑡𝑎 𝑥 = 3 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑎𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑒 𝑓(𝑥). 10 ∗ 𝑓(2) = 2.23 − 2 = 16 − 2 = 14. 𝐶𝑜𝑚𝑜 lim 𝑥→2 𝑓(𝑥) = 𝑓(2), 𝑒𝑛𝑡ã𝑜 𝑓(𝑥) é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 𝑥 = 2. 𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑐𝑜𝑛𝑐𝑙𝑢í𝑚𝑜𝑠 𝑞𝑢𝑒 𝑓(𝑥) é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑝𝑎𝑟𝑎 𝑡𝑜𝑑𝑜 𝑥 ∈ ℝ. 𝑏) 𝑇𝑜𝑚𝑒 𝑓(2) = 14, 𝑒 𝑐𝑎𝑙𝑐𝑢𝑙𝑒𝑚𝑜𝑠 𝑓(3). 𝐸𝑛𝑡ã𝑜, 𝑡𝑒𝑚𝑜𝑠: 𝑓(3) = 2.33 − 3 = 54 − 3 = 51. 𝑆𝑒𝑗𝑎 𝑐 ∈ ℝ 𝑡𝑎𝑙 𝑞𝑢𝑒 𝑓(𝑐) = 15. 𝐶𝑜𝑚𝑜 𝑓(𝑥) é 𝑢𝑚𝑎 𝑓𝑢𝑛çã𝑜 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 𝑓𝑒𝑐ℎ𝑎𝑑𝑜 [2, 3] 𝑒 𝑓(2) < 𝑓(𝑐) < 𝑓(3), 𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑔𝑎𝑟𝑎𝑛𝑡𝑖𝑟 𝑝𝑒𝑙𝑜 𝑇𝑒𝑜𝑟𝑒𝑚𝑎 𝑑𝑜 𝑉𝑎𝑙𝑜𝑟 𝐼𝑛𝑡𝑒𝑟𝑚𝑒𝑑𝑖á𝑟𝑖𝑜 𝑞𝑢𝑒 𝑒𝑥𝑖𝑠𝑡𝑒 𝑎𝑙𝑔𝑢𝑚 𝑐 ∈ (2, 3) 𝑡𝑎𝑙 𝑞𝑢𝑒 𝑓(𝑐) = 15. 4. ℎ(𝑥) = 𝑥|𝑥 − 1| |𝑥 − 1| = { 𝑥 − 1, 𝑠𝑒 𝑥 ≥ 1 −(𝑥 − 1), 𝑠𝑒 𝑥 < 1 ℎ(𝑥) = { 𝑥2 − 𝑥, 𝑠𝑒 𝑥 ≥ 1 −𝑥2 + 𝑥, 𝑠𝑒 𝑥 < 1 𝑎) 𝑃𝑟𝑖𝑚𝑒𝑖𝑟𝑜 𝑣𝑎𝑚𝑜𝑠 𝑣𝑒𝑟𝑖𝑓𝑖𝑐𝑎𝑟 𝑠𝑒 ℎ(𝑥) é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 𝑥 = 1. ∗ ℎ(1) = 12 − 1 = 1 − 1 = 0. lim 𝑥→1+ ℎ(𝑥) = lim 𝑥→1+ (𝑥2 − 𝑥) = lim 𝑥→1+ 𝑥2 − lim 𝑥→1+ 𝑥 = 1 − 1 = 0. lim 𝑥→1− ℎ(𝑥) = lim 𝑥→1− (−𝑥2 + 𝑥) = lim 𝑥→1− −𝑥2 + lim 𝑥→1− 𝑥 = −1 + 1 = 0. 𝐶𝑜𝑚𝑜 lim 𝑥→1+ ℎ(𝑥) = lim 𝑥→1− ℎ(𝑥) , 𝑒𝑛𝑡ã𝑜 lim 𝑥→1 ℎ(𝑥) = 0. 𝑇𝑒𝑚𝑜𝑠, 𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑞𝑢𝑒 lim 𝑥→1 ℎ(𝑥) = 𝑓(1). 𝐿𝑜𝑔𝑜, ℎ(𝑥) é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 𝑥 = 1. ℎ+′(1) = lim 𝑥→1+ ℎ(𝑥) − ℎ(1) 𝑥 − 1 = lim 𝑥→1+ 𝑥2 − 𝑥 𝑥 − 1 = lim 𝑥→1+ 𝑥(𝑥 − 1) (𝑥 − 1) = lim 𝑥→1+ 𝑥 = 1. ℎ− ′(1) = lim 𝑥→1− ℎ(𝑥) − ℎ(1) 𝑥 − 1 = lim 𝑥→1− −𝑥2 + 𝑥 𝑥 − 1 = lim 𝑥→1− −𝑥(𝑥 − 1) (𝑥 − 1) = lim 𝑥→1− 𝑥 = −1. ∗ 𝐶𝑜𝑚𝑜 𝑎𝑠 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎𝑠 𝑙𝑎𝑡𝑒𝑟𝑎𝑖𝑠 𝑒𝑚 𝑥 = 1 𝑠ã𝑜 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒𝑠, 𝑜𝑢 𝑠𝑒𝑗𝑎, ℎ+ ′ (1) ≠ ℎ− ′ (1), 𝑒𝑛𝑡ã𝑜 ℎ(𝑥) 𝑛ã𝑜 é 𝑑𝑒𝑟𝑖𝑣á𝑣𝑒𝑙 𝑒𝑚 𝑥 = 1. 𝑏) 𝑉𝑎𝑚𝑜𝑠 𝑣𝑒𝑟𝑖𝑓𝑖𝑐𝑎𝑟 𝑠𝑒 𝑜 𝑝𝑜𝑛𝑡𝑜 (2, 2) 𝑝𝑒𝑟𝑡𝑒𝑛𝑐𝑒 à ℎ(𝑥). ℎ(2) = 22 − 2 = 4 − 2 = 2. 𝐷𝑜 𝐶á𝑙𝑐𝑢𝑙𝑜 1 𝑠𝑎𝑏𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑜 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑑𝑒 𝑢𝑚𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 𝑛𝑢𝑚 𝑝𝑜𝑛𝑡𝑜 é 𝑑𝑎𝑑𝑜 𝑝𝑒𝑙𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑎 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎 𝑑𝑎 𝑓𝑢𝑛çã𝑜 𝑛𝑎𝑞𝑢𝑒𝑙𝑒 𝑝𝑜𝑛𝑡𝑜. 𝐷𝑎𝑑𝑜 𝑢𝑚 𝑝𝑜𝑛𝑡𝑜 (𝑥𝑜 , 𝑦𝑜) 𝑒 (𝑚) 𝑜 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑑𝑒 𝑢𝑚𝑎 𝑟𝑒𝑡𝑎 𝑡𝑒𝑚𝑜𝑠: 𝑦 − 𝑦𝑜 = 𝑚(𝑥 − 𝑥𝑜) 11 𝑂𝑛𝑑𝑒,𝑚 = ℎ′(2). 𝐿𝑜𝑔𝑜, 𝑦 − 2 = ℎ′(2)(𝑥 − 2) ℎ′(2) = lim 𝑥→2 ℎ(𝑥) − ℎ(2) 𝑥 − 2 = lim 𝑥→2 𝑥2 − 𝑥 − 2 𝑥 − 2 = lim 𝑥→2 (𝑥 + 1)(𝑥 − 2) (𝑥 − 2) = lim 𝑥→2 (𝑥 + 1) = 3. 𝐿𝑜𝑔𝑜, 𝑎 𝑒𝑞𝑢𝑎çã𝑜 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 𝑛𝑜 𝑝𝑜𝑛𝑡𝑜 (2, 2) é: 𝑦 − 2 = 3(𝑥 − 2) 𝑦 − 2 = 3𝑥 − 6 𝑦 = 3𝑥 − 4 12 Capítulo 2 2007 2.1 1ª Prova-15 de Setembro de 2007 1. 𝑓(𝑥) = |𝑥 − 3| 𝑓(𝑥) = { 𝑥 − 3, 𝑠𝑒 𝑥 ≥ 3 −(𝑥 − 3), 𝑠𝑒 𝑥 < 3 𝑎) 𝐷𝑖𝑧𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑢𝑚𝑎 𝑓𝑢𝑛çã𝑜 𝑓(𝑥) é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑛𝑜 𝑝𝑜𝑛𝑡𝑜 𝑥 = 𝑎 𝑠𝑒, 𝑒 𝑠𝑜𝑚𝑒𝑛𝑡𝑒 𝑠𝑒, 1) 𝑓(𝑎) 𝑒𝑥𝑖𝑠𝑡𝑒; 2) lim 𝑥→𝑎 𝑓(𝑥) 𝑒𝑥𝑖𝑠𝑡𝑒; 3) 𝑓(𝑎) = lim 𝑥→𝑎 𝑓(𝑥). 𝑃𝑟𝑜𝑐𝑒𝑑𝑒𝑛𝑑𝑜 𝑒𝑠𝑠𝑎 𝑐𝑜𝑛𝑑𝑖çã𝑜 𝑝𝑎𝑟𝑎 𝑥 = 3 𝑡𝑒𝑚𝑜𝑠: 1) 𝑓(3) = |3 − 3| = |0| = 0. 2) lim 𝑥→3 𝑓(𝑥) . 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑛𝑑𝑜 𝑜𝑠 𝑙𝑖𝑚𝑖𝑡𝑒𝑠 𝑙𝑎𝑡𝑒𝑟𝑎𝑖𝑠, 𝑡𝑒𝑚𝑜𝑠: lim 𝑥→3+ 𝑓(𝑥) = lim 𝑥→3+ (𝑥 − 3) = lim 𝑥→3+ 𝑥 − lim 𝑥→3+ 3 = 3 − 3 = 0. lim 𝑥→3− 𝑓(𝑥) = lim 𝑥→3+ (−𝑥 + 3) = lim 𝑥→3+ −𝑥 + lim 𝑥→3+ 3 = −3 + 3 = 0. 𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑐𝑜𝑚𝑜 lim 𝑥→3+ 𝑓(𝑥) = lim 𝑥→3− 𝑓(𝑥) 𝑒𝑛𝑡ã𝑜 lim 𝑥→3 𝑓(𝑥) = 0. 3)𝑓(3) = lim 𝑥→3 𝑓(𝑥) = 0. ∗ 𝐿𝑜𝑔𝑜, 𝑓 é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 𝑥 = 3. ∗ 𝐴𝑛𝑎𝑙𝑖𝑠𝑎𝑛𝑑𝑜 𝑎 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑏𝑖𝑙𝑖𝑑𝑎𝑑𝑒 𝑑𝑒 𝑓 𝑒𝑚 𝑥 = 3 𝑡𝑒𝑚𝑜𝑠: 𝑓′ + (3) = lim 𝑥→3+ 𝑓(𝑥) − 𝑓(3) 𝑥 − 3 = lim 𝑥→3+ 𝑥 − 3 − 0 𝑥 − 3 = lim 𝑥→3+ (𝑥 − 3) (𝑥 − 3) = 1. 𝑓′ − (3) = lim 𝑥→3− 𝑓(𝑥) − 𝑓(3) 𝑥 − 3 = lim 𝑥→3− −(𝑥 − 3) − 0 𝑥 − 3 = lim 𝑥→3− − (𝑥 − 3) (𝑥 − 3) = −1. 𝐶𝑜𝑚𝑜 𝑓′ + (3) ≠ 𝑓′ − (3) 𝑡𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑓 𝑛ã𝑜 é 𝑑𝑒𝑟𝑖𝑣á𝑣𝑒𝑙 𝑒𝑚 𝑥 = 3. 𝑏) 𝑥² = 𝑥³ + 2 → 𝑥³ − 𝑥2 + 2 = 0 𝑆𝑒𝑗𝑎 𝑓(𝑥) = 𝑥3 − 𝑥2 + 2. 𝐶𝑎𝑙𝑐𝑢𝑙𝑒𝑚𝑜𝑠 𝑓(−2) 𝑒 𝑓(0). 𝑓(−2) = (−2)3 − (−2)2 + 2 = −8 − 4 + 2 = −10. 𝑓(0) = 03 − 02 + 2 = 0 − 0 + 2 = 2. 𝑓(𝑐) = 0. 𝐶𝑜𝑚𝑜 𝑓 é 𝑢𝑚𝑎 𝑓𝑢𝑛çã𝑜 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑎𝑙 𝑒, 𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 𝑓𝑒𝑐ℎ𝑎𝑑𝑜 [−2,0], 𝑒 𝑓(−2) < 𝑓(𝑐) < 𝑓(0), 𝑝𝑒𝑙𝑜 𝑇𝑒𝑜𝑟𝑒𝑚𝑎 𝑑𝑜 𝑉𝑎𝑙𝑜𝑟 𝐼𝑛𝑡𝑒𝑟𝑚𝑒𝑑𝑖á𝑟𝑖𝑜 𝑒𝑥𝑖𝑠𝑡𝑒 𝑎𝑙𝑔𝑢𝑚 𝑐 ∈ (−2, 0) 𝑡𝑎𝑙 𝑞𝑢𝑒 𝑓(𝑐) = 0. 𝐷𝑒 𝑚𝑜𝑑𝑜 𝑞𝑢𝑒 𝑐 é 𝑢𝑚 𝑛ú𝑚𝑒𝑟𝑜 𝑐𝑢𝑗𝑜 𝑞𝑢𝑎𝑑𝑟𝑎𝑑𝑜 é 𝑖𝑔𝑢𝑎𝑙 𝑎𝑜 𝑠𝑒𝑢 𝑐𝑢𝑏𝑜 𝑠𝑜𝑚𝑎𝑑𝑜 𝑐𝑜𝑚 2. 15 𝑓𝑟𝑎çã𝑜 𝑝𝑜𝑟 |𝑥|. 𝑁𝑜𝑡𝑒 𝑞𝑢𝑒, 𝑠𝑒 𝑥 → +∞ 𝑒𝑛𝑡ã𝑜 |𝑥| = 𝑥 𝑒 𝑠𝑒 𝑥 → −∞, |𝑥| = −𝑥. 𝐸𝑚 𝑎𝑚𝑏𝑜𝑠, 𝑣𝑎𝑙𝑒 𝑎 𝑛𝑜𝑡𝑎çã𝑜 |𝑥| = √𝑥2. 4. 𝑎) lim 𝑥→8 √7 + √𝑥 3 − 3 𝑥 − 8 = lim 𝑥→8 (√7 + √𝑥 3 − 3) 𝑥 − 8 . (√7 + √𝑥 3 + 3) (√7 + √𝑥 3 + 3) = lim 𝑥→8 √𝑥 3 − 2 (𝑥 − 8) (√7 + √𝑥 3 + 3) = lim 𝑥→8 (√𝑥 3 − 2) (√𝑥 3 − 2)(√𝑥2 3 + 2√𝑥 3 + 4) (√7 + √𝑥 3 + 3) = lim 𝑥→8 1 (√𝑥2 3 + 2√𝑥 3 + 4) (√7 + √𝑥 3 + 3) = 1 (4 + 4 + 4)(3 + 3) = 1 (12)(9) = 1 72 . 𝑏)𝑓(𝑥) = 3 + ⟦𝑥⟧ 2 . 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑟 𝑜𝑠 𝑙𝑖𝑚𝑖𝑡𝑒𝑠 𝑙𝑎𝑡𝑒𝑟𝑎𝑖𝑠 𝑛𝑜 𝑝𝑜𝑛𝑡𝑜 𝑒𝑚 𝑞𝑢𝑒 𝑥 = 3. lim 𝑥→3+ 𝑓(𝑥) = lim 𝑥→3+ (3 + ⟦𝑥⟧ 2 ) = lim 𝑥→3+ 3 + lim 𝑥→3+ ⟦𝑥⟧ 2 = 3 + 3 2 = 9 2 ∗ 𝑂𝑏𝑠: 𝑠𝑒 𝑥 → 3+, 𝑒𝑛𝑡ã𝑜 𝑥 > 3. 𝐿𝑜𝑔𝑜, ⟦𝑥⟧ = 3 lim 𝑥→3− 𝑓(𝑥) = lim 𝑥→3− (3 + ⟦𝑥⟧ 2 ) = lim 𝑥→3− 3 + lim 𝑥→3− ⟦𝑥⟧ 2 = 3 + 2 2 = 4 ∗ 𝑂𝑏𝑠: 𝑠𝑒 𝑥 → 3−, 𝑒𝑛𝑡ã𝑜 𝑥 < 3. 𝐿𝑜𝑔𝑜, ⟦𝑥⟧ = 2. 5. 𝑠 = 256𝑡 − 16𝑡2 𝑎) 𝑣(𝑡) = 𝑑𝑠 𝑑𝑡 = 𝑠′(𝑡) = 256 − 32𝑡 → 𝑣(𝑡) = 256 − 32𝑡 ∴ 𝑣(6) = 256 − 32 × 6 = 256 − 192 = 64 𝑚/𝑠 𝑏) 𝑣(𝑡) = 0 → 256 − 32𝑡 = 0 32𝑡 = 256 𝑡 = 8𝑠 𝑐) 𝑠(8) = 256 × 8 − 32 × 8 𝑠(8) = 2.048 − 256 𝑠(8) = 1.792𝑚 16 2.2 2ª Prova-05 de Outubro de 2007 1. 𝑥2 + 𝑦2 = 4 𝑃𝑜𝑟 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎çã𝑜 𝑖𝑚𝑝𝑙í𝑐𝑖𝑡𝑎 𝑡𝑒𝑚𝑜𝑠: 2𝑥 + 2𝑦. 𝑦′ = 0 𝑦′ = − 𝑥 𝑦 ∗ 𝐸𝑞𝑢𝑎çã𝑜 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑞𝑢𝑒 𝑐𝑜𝑛𝑡é𝑚 𝑜 𝑠𝑒𝑔𝑚𝑒𝑛𝑡𝑜 𝑂𝑃: 𝑦 = 𝑚. 𝑥 𝑂𝑛𝑑𝑒 𝑚 é 𝑜 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑞𝑢𝑒 é 𝑑𝑎𝑑𝑜 𝑐𝑜𝑚𝑜 𝑚 = 𝑡𝑔𝛼 , 𝑠𝑒𝑛𝑑𝑜 𝛼 𝑜 â𝑛𝑔𝑢𝑙𝑜 𝑓𝑜𝑟𝑚𝑎𝑑𝑜 𝑒𝑛𝑡𝑟𝑒 𝑜 𝑠𝑒𝑔𝑚𝑒𝑛𝑡𝑜 𝑂𝑃 𝑒 𝑜 𝑒𝑖𝑥𝑜 𝑑𝑜𝑠 𝑥, 𝑜𝑢 𝑠𝑒𝑗𝑎, 𝛼 = 75°. 𝑦 = 𝑡𝑔75°. 𝑥 𝑡𝑔75° = 𝑡𝑔(30° + 45°) = 𝑡𝑔30° + 𝑡𝑔45° 1 − 𝑡𝑔30°. 𝑡𝑔45° = 1 √3 + 1 1 − 1 √3 = 1 + √3 √3 √3 − 1 √3 = √3 + 1 √3 − 1 ∗ 𝐴 𝑒𝑞𝑢𝑎çã𝑜 𝑑𝑒 𝑢𝑚𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 à 𝑐𝑖𝑟𝑐𝑢𝑛𝑓𝑒𝑟ê𝑛𝑐𝑖𝑎 𝑥2 + 𝑦2 = 4 é 𝑑𝑎𝑑𝑎 𝑝𝑒𝑙𝑎 𝑠𝑒𝑔𝑢𝑖𝑛𝑡𝑒 𝑒𝑥𝑝𝑟𝑒𝑠𝑠ã𝑜: 𝑦 − 𝑦𝑜 = − 𝑥𝑜 𝑦𝑜 (𝑥 − 𝑥𝑜) 𝑦. 𝑦𝑜 − 𝑦𝑜 2 = −𝑥. 𝑥𝑜 + 𝑥𝑜 2 𝑦. 𝑦𝑜 = −𝑥. 𝑥𝑜 + 𝑥𝑜 2 + 𝑦𝑜 2 𝑦. 𝑦𝑜 = −𝑥. 𝑥𝑜 + 4 𝑦 = − 𝑥𝑜 𝑦𝑜 𝑥 + 4 𝑦𝑜 ∗ 𝐶𝑜𝑚𝑜 𝑜 𝑠𝑒𝑔𝑚𝑒𝑛𝑡𝑜 𝑂𝑃 é 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 à 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 𝑛𝑜 𝑝𝑜𝑛𝑡𝑜 𝑃, 𝑒𝑛𝑡ã𝑜 𝑜 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 é 𝑜 𝑖𝑛𝑣𝑒𝑟𝑠𝑜 𝑠𝑖𝑚é𝑡𝑟𝑖𝑐𝑜 𝑑𝑜 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑞𝑢𝑒 𝑐𝑜𝑛𝑡é𝑚 𝑜 𝑠𝑒𝑔𝑚𝑒𝑛𝑡𝑜 𝑂𝑃. − 𝑥𝑜 𝑦𝑜 = − 1 √3 + 1 √3 − 1 = 1 − √3 1 + √3 𝐷𝑒𝑣𝑒𝑚𝑜𝑠 𝑝𝑟𝑜𝑣𝑎𝑟 𝑞𝑢𝑒 1 − √3 1 + √3 = 𝑡𝑔165°. 𝑃𝑜𝑖𝑠, 𝑒𝑠𝑠𝑒 é 𝑜 â𝑛𝑔𝑢𝑙𝑜 𝑓𝑜𝑟𝑚𝑎𝑑𝑜 𝑒𝑛𝑡𝑟𝑒 𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 𝑒 𝑎 𝑑𝑖𝑟𝑒çã𝑜 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑎 𝑑𝑜 𝑒𝑖𝑥𝑜 𝑑𝑜𝑠 𝑥. 𝑡𝑔165° = 𝑡𝑔(120° + 45°) = 𝑡𝑔120° + 𝑡𝑔45° 1 − 𝑡𝑔120°. 𝑡𝑔45° = −√3 + 1 1 + √3 = 1 − √3 1 + √3 ∗ 𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 𝑓𝑜𝑟𝑚𝑎 𝑢𝑚 â𝑛𝑔𝑢𝑙𝑜 𝑑𝑒 15° 𝑐𝑜𝑚 𝑎 𝑑𝑖𝑟𝑒çã𝑜 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑎 𝑑𝑜 𝑒𝑖𝑥𝑜 𝑑𝑜𝑠 𝑥. 2. 𝑎) 3𝑥. 𝑎𝑟𝑐𝑡𝑔(𝑥 + 𝑦) + 107 = 𝑥²𝑦 𝑃𝑜𝑟 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎çã𝑜 𝑖𝑚𝑝𝑙í𝑐𝑖𝑡𝑎 𝑡𝑒𝑚𝑜𝑠: 𝑑 𝑑𝑥 (3𝑥. 𝑎𝑟𝑐𝑡𝑔(𝑥 + 𝑦)) + 𝑑 𝑑𝑥 (107) = 𝑑 𝑑𝑥 (𝑥2𝑦) 3𝑎𝑟𝑐𝑡𝑔(𝑥 + 𝑦) + 3𝑥. 1 + 𝑦′ 1 + (𝑥 + 𝑦)2 + 0 = 2𝑥𝑦 + 𝑥2𝑦′ 17 3𝑎𝑟𝑐𝑡𝑔(𝑥 + 𝑦) + 3𝑥 + 3𝑥𝑦′ 1 + (𝑥 + 𝑦)2 = 2𝑥𝑦 + 𝑥2𝑦′ 3𝑎𝑟𝑐𝑡𝑔(𝑥 + 𝑦)[1 + (𝑥 + 𝑦)2] + 3𝑥 + 3𝑥𝑦′ = 2𝑥𝑦[1 + (𝑥 + 𝑦)2] + 𝑥2𝑦′[1 + (𝑥 + 𝑦)2] 𝑦′[𝑥2(1 + (𝑥 + 𝑦)2) − 3𝑥] = 3𝑎𝑟𝑐𝑡𝑔(𝑥 + 𝑦)[1 + (𝑥 + 𝑦)2] + 3𝑥 − 2𝑥𝑦[1 + (𝑥 + 𝑦)2] 𝑦′ = 3𝑎𝑟𝑐𝑡𝑔(𝑥 + 𝑦)[1 + (𝑥 + 𝑦)2] + 3𝑥 − 2𝑥𝑦[1 + (𝑥 + 𝑦)2] 𝑥2[1 + (𝑥 + 𝑦)2] − 3𝑥 𝑏) 𝑓(𝑥) = log3 5 𝑥2+1 . 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟 𝑓′(𝑥). 𝑆𝑒𝑗𝑎 𝑢 = 𝑥2 + 1 𝑣 = 5𝑢 𝑦 = 𝑓(𝑣) = log3 𝑣 𝑃𝑒𝑙𝑎 𝑟𝑒𝑔𝑟𝑎 𝑑𝑎 𝑐𝑎𝑑𝑒𝑖𝑎 𝑡𝑒𝑚𝑜𝑠: 𝑑𝑦 𝑑𝑥 = 𝑑𝑢 𝑑𝑥 . 𝑑𝑣 𝑑𝑢 . 𝑑𝑦 𝑑𝑣 𝑑𝑦 𝑑𝑥 = (2𝑥). 5𝑢 ln(5) . 1 𝑣. ln(3) 𝑑𝑦 𝑑𝑥 = 2𝑥. 5𝑥 2+1 ln(5) 5𝑥 2+1 ln(3) 𝑓′(𝑥) = 2𝑥. ln(5) ln(3) 3. 𝑓(𝑥) = 2𝑠𝑒𝑛(𝑥) + 𝑠𝑒𝑛2(𝑥) 𝑓′(𝑥) = 2cos(𝑥) + 2. 𝑠𝑒𝑛(𝑥). cos(𝑥) 𝐸𝑛𝑐𝑜𝑛𝑡𝑟𝑎𝑟 𝑜𝑠 𝑝𝑜𝑛𝑡𝑜𝑠 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 𝑑𝑒 [0,2𝜋] 𝑜𝑛𝑑𝑒 𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 𝑎𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑒 𝑓(𝑥) é ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙. (𝐸𝑛𝑐𝑜𝑛𝑡𝑟𝑎𝑟 𝑜𝑠 𝑣𝑎𝑙𝑜𝑟𝑒𝑠 𝑑𝑒 𝑥 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 [0, 2𝜋] 𝑜𝑛𝑑𝑒 𝑓′(𝑥) = 0) 𝑓′(𝑥) = 0 → 2 cos(𝑥) + 2. 𝑠𝑒𝑛(𝑥). cos(𝑥) = 0 2 cos(𝑥) [1 + 𝑠𝑒𝑛(𝑥)] = 0 { 2cos(𝑥) = 0 1 + 𝑠𝑒𝑛(𝑥) = 0 ∗ 𝐷𝑎 𝑝𝑟𝑖𝑚𝑒𝑖𝑟𝑎 𝑠𝑒𝑛𝑡𝑒𝑛ç𝑎 𝑡𝑒𝑚𝑜𝑠: cos(𝑥) = 0 → 𝑥 = 𝜋 2 𝑒 𝑥 = 3𝜋 2 ∗ 𝐷𝑎 𝑠𝑒𝑔𝑢𝑛𝑑𝑎 𝑠𝑒𝑛𝑡𝑒𝑛ç𝑎 𝑡𝑒𝑚𝑜𝑠: 1 + 2𝑠𝑒𝑛(𝑥) = 0 𝑠𝑒𝑛(𝑥) = −1 → 𝑥 = 3𝜋 2 𝐿𝑜𝑔𝑜, 𝑜𝑠 𝑝𝑜𝑛𝑡𝑜𝑠 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 [0, 2𝜋] 𝑜𝑛𝑑𝑒 𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 𝑎𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑒 𝑓(𝑥) é ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙, 𝑠ã𝑜: ( 𝜋 2 , 3) , ( 3𝜋 2 ,−1). ∗ 𝑓 ( 𝜋 2 ) = 2𝑠𝑒𝑛 ( 𝜋 2 ) + 𝑠𝑒𝑛2 ( 𝜋 2 ) = 2 + 12 = 2 + 1 = 3. ( 𝜋 2 , 3) ; ∗ 𝑓 ( 3𝜋 2 ) = 𝑠𝑒𝑛 ( 3𝜋 2 ) + 𝑠𝑒𝑛2 ( 3𝜋 2 ) = −1 + (−1)2 = −1 + 1 = 0. ( 3𝜋 2 , 0) ; 4. 20 2.3 1ª Prova-06 de Outubro de 2007 1. 𝑎) 𝑓(𝑥) = 𝑥. 𝑠𝑒𝑛 ( 1 𝑥 ) − 𝑐𝑜𝑠 ( 1 𝑥 ) . 𝑃𝑟𝑜𝑣𝑎𝑟 𝑞𝑢𝑒 𝑓(𝑥) 𝑝𝑜𝑠𝑠𝑢𝑖 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙. −𝐷𝑖𝑧𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑎 𝑟𝑒𝑡𝑎 𝑦 = 𝑎 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑠𝑒 𝑢𝑚 𝑑𝑜𝑠 𝑐𝑎𝑠𝑜𝑠 𝑎 𝑠𝑒𝑔𝑢𝑖𝑟 𝑜𝑐𝑜𝑟𝑟𝑒𝑟: lim 𝑥→+∞ 𝑓(𝑥) = 𝑎 𝑜𝑢 lim 𝑥→−∞ 𝑓(𝑥) = 𝑎. lim 𝑥→+∞ 𝑓(𝑥) = lim 𝑥→+∞ (𝑥. 𝑠𝑒𝑛 ( 1 𝑥 ) − 𝑐𝑜𝑠 ( 1 𝑥 )) . 𝐹𝑎𝑧𝑒𝑛𝑑𝑜 𝑎 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑖çã𝑜 𝜃 = 1 𝑥 𝑒 𝑎𝑗𝑢𝑠𝑡𝑎𝑛𝑑𝑜 𝑜 𝑙𝑖𝑚𝑖𝑡𝑒, 𝑡𝑒𝑚𝑜𝑠: ∗ 𝑆𝑒 𝑥 → +∞, 𝑒𝑛𝑡ã𝑜 𝜃 → 0. lim 𝜃→0 ( 𝑠𝑒𝑛(𝜃) 𝜃 − 𝑐𝑜𝑠(𝜃)) = lim 𝜃→0 𝑠𝑒𝑛(𝜃) 𝜃 − lim 𝜃→0 cos(𝜃) = 1 − 1 = 0. 𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑎 𝑟𝑒𝑡𝑎 𝑦 = 0 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑎𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑒 𝑓(𝑥). lim 𝑥→−∞ 𝑓(𝑥) = lim 𝑥→−∞ (𝑥. 𝑠𝑒𝑛 ( 1 𝑥 ) − 𝑐𝑜𝑠 ( 1 𝑥 )) . 𝐹𝑎𝑧𝑒𝑛𝑑𝑜 𝑎 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑖çã𝑜 𝜃 = 1 𝑥 𝑒 𝑎𝑗𝑢𝑠𝑡𝑎𝑛𝑑𝑜 𝑜 𝑙𝑖𝑚𝑖𝑡𝑒, 𝑡𝑒𝑚𝑜𝑠: ∗ 𝑆𝑒 𝑥 → −∞, 𝑒𝑛𝑡ã𝑜 𝜃 → 0. lim 𝜃→0 ( 𝑠𝑒𝑛(𝜃) 𝜃 − 𝑐𝑜𝑠(𝜃)) = lim 𝜃→0 𝑠𝑒𝑛(𝜃) 𝜃 − lim 𝜃→0 cos(𝜃) = 1 − 1 = 0. 𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑎 𝑟𝑒𝑡𝑎 𝑦 = 0 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑎𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑒 𝑓(𝑥). 𝑏) 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟 𝑑𝑦 𝑑𝑥 , 𝑜𝑛𝑑𝑒 𝑥. arccos(𝑥 + 𝑦) − 𝜋 = 𝑦2. 𝑃𝑒𝑙𝑎 𝑑𝑒𝑟𝑖𝑣𝑎çã𝑜 𝑖𝑚𝑝𝑙í𝑐𝑖𝑡𝑎 𝑡𝑒𝑚𝑜𝑠: arccos(𝑥 + 𝑦) + 𝑥 ( −(1 + 𝑦′) √1 − (𝑥 + 𝑦)2 ) − 0 = 2𝑦. 𝑦′ arccos(𝑥 + 𝑦) [√1 − (𝑥 + 𝑦)2] − 𝑥 − 𝑥𝑦′ = 2𝑦. 𝑦′ [√1 − (𝑥 + 𝑦)2] 𝑦′ [2𝑦√1 − (𝑥 + 𝑦)2 + 𝑥] = arccos(𝑥 + 𝑦) [√1 − (𝑥 + 𝑦)2] − 𝑥 𝑦′ = arccos(𝑥 + 𝑦) [√1 − (𝑥 + 𝑦)2] − 𝑥 2𝑦√1 − (𝑥 + 𝑦)2 + 𝑥 2. 𝑦 = 2𝑥2 − 1 . 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟 𝑎 𝑒𝑞𝑢𝑎çã𝑜 𝑑𝑎𝑠 𝑟𝑒𝑡𝑎𝑠 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒𝑠 á 𝑐𝑢𝑟𝑣𝑎 𝑞𝑢𝑒 𝑝𝑎𝑠𝑠𝑎𝑚 𝑝𝑒𝑙𝑜 𝑝𝑜𝑛𝑡𝑜 (4, 13). 𝑦′ = 4𝑥 ∗ 𝐴 𝑒𝑞𝑢𝑎çã𝑜 𝑑𝑒 𝑢𝑚𝑎 𝑟𝑒𝑡𝑎 𝑑𝑎𝑑𝑜 𝑢𝑚 𝑝𝑜𝑛𝑡𝑜 𝑝𝑒𝑟𝑡𝑒𝑛𝑐𝑒𝑛𝑡𝑒 à 𝑒𝑙𝑎 𝑒 𝑠𝑒𝑢 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑡𝑒𝑚𝑜𝑠: 𝑦 − 𝑦𝑜 = 𝑚(𝑥 − 𝑥𝑜) 𝑦 − 13 = 4𝑥(𝑥 − 4) 2𝑥2 − 1 − 13 = 4𝑥2 − 16𝑥 21 2𝑥2 − 16𝑥 + 14 = 0 𝑥2 − 8𝑥 + 7 = 0 ∆= 64 − 28 = 36 𝑥 = 8 ± 6 2 → 𝑥′ = 7 𝑒 𝑥′′ = 1 ∗ 𝑃𝑎𝑟𝑎 𝑥 = 1, 𝑡𝑒𝑚𝑜𝑠 𝑦′ = 4 𝐿𝑜𝑔𝑜, 𝑎 𝑒𝑞𝑢𝑎çã𝑜 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 𝑡𝑒𝑚 𝑎 𝑠𝑒𝑔𝑢𝑖𝑛𝑡𝑒 𝑓𝑜𝑟𝑚𝑎: 𝑦 − 13 = 4(𝑥 − 4) 𝑦 − 13 = 4𝑥 − 16 𝑦 = 4𝑥 − 3 ∗ 𝑃𝑎𝑟𝑎 𝑥 = 7, 𝑡𝑒𝑚𝑜𝑠 𝑦′ = 28 𝐿𝑜𝑔𝑜, 𝑎 𝑒𝑞𝑢𝑎çã𝑜 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 𝑡𝑒𝑚 𝑎 𝑠𝑒𝑔𝑢𝑖𝑛𝑡𝑒 𝑓𝑜𝑟𝑚𝑎: 𝑦 − 13 = 28(𝑥 − 4) 𝑦 − 13 = 28𝑥 − 112 𝑦 = 28𝑥 − 99 3. 𝑓(𝑥) = 3𝑥 + |𝑥| 𝑒 𝑔(𝑥) = 3𝑥 4 − |𝑥| 4 ; 𝑃𝑟𝑜𝑣𝑎𝑟 𝑞𝑢𝑒 𝑓′(0) 𝑒 𝑔′(0) 𝑛ã𝑜 𝑒𝑥𝑖𝑠𝑡𝑒𝑚,𝑚𝑎𝑠 𝑞𝑢𝑒 (𝑓 ∘ 𝑔)′(0) 𝑒𝑥𝑖𝑠𝑡𝑒. 𝑓′ + (0) = lim 𝑥→0+ 𝑓(𝑥) − 𝑓(0) 𝑥 − 0 = lim 𝑥→0+ 4𝑥 − 0 𝑥 = lim 𝑥→0+ 4𝑥 𝑥 = 4. 𝑓′ − (0) = lim 𝑥→0− 𝑓(𝑥) − 𝑓(0) 𝑥 − 0 = lim 𝑥→0− 2𝑥 − 0 𝑥 = lim 𝑥→0− 2𝑥 𝑥 = 2. ∗ 𝐶𝑜𝑚𝑜 𝑓′ + (0) ≠ 𝑓′ − (0) , 𝑒𝑛𝑡ã𝑜 𝑓 𝑛ã𝑜 é 𝑑𝑒𝑟𝑖𝑣á𝑣𝑒𝑙 𝑒𝑚 𝑥 = 0. 𝑔′ + (0) = lim 𝑥→0+ 𝑔(𝑥) − 𝑔(0) 𝑥 − 0 = lim 𝑥→0+ 𝑥 2 − 0 𝑥 = lim 𝑥→0+ 𝑥 2 𝑥 = 1 2 . 𝑔′ − (0) = lim 𝑥→0− 𝑔(𝑥) − 𝑔(0) 𝑥 − 0 = lim 𝑥→0− 𝑥 − 0 𝑥 = lim 𝑥→0− 𝑥 𝑥 = 1. ∗ 𝐶𝑜𝑚𝑜 𝑔′ + (0) ≠ 𝑔′ − (0) , 𝑒𝑛𝑡ã𝑜 𝑔 𝑛ã𝑜 é 𝑑𝑒𝑟𝑖𝑣á𝑣𝑒𝑙 𝑒𝑚 𝑥 = 0. −𝑉𝑎𝑚𝑜𝑠 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑟 (𝑓 ∘ 𝑔)(𝑥): 𝑓(𝑔(𝑥)) = 3. 𝑔(𝑥) + |𝑔(𝑥)| = 9𝑥 4 − 3|𝑥| 4 + | 3𝑥 4 − |𝑥| 4 | 𝑓(𝑔(𝑥)) = { 9𝑥 4 − 3𝑥 4 + 3𝑥 4 − 𝑥 4 , 𝑠𝑒 𝑥 ≥ 0 9𝑥 4 + 3𝑥 4 − ( 3𝑥 4 + 𝑥 4 ) , 𝑠𝑒 𝑥 < 0 𝑓(𝑔(𝑥)) = { 2𝑥, 𝑠𝑒 𝑥 ≥ 0 2𝑥, 𝑠𝑒 𝑥 < 0 𝐿𝑜𝑔𝑜, 𝑓(𝑔(𝑥)) = 2𝑥. (𝑓 ∘ 𝑔)′ + (0) = lim 𝑥→0+ (𝑓 ∘ 𝑔)(𝑥) − (𝑓 ∘ 𝑔)(0) 𝑥 − 0 = lim 𝑥→0+ 6𝑥 4 + 2𝑥 4 − 0 𝑥 = lim 𝑥→0+ 2𝑥 𝑥 = 2. (𝑓 ∘ 𝑔)′ − (0) = lim 𝑥→0− (𝑓 ∘ 𝑔)(𝑥) − (𝑓 ∘ 𝑔)(0) 𝑥 − 0 = lim 𝑥→0− 3𝑥 − 𝑥 − 0 𝑥 = lim 𝑥→0− 2𝑥 𝑥 = 2. ∗ 𝐶𝑜𝑚𝑜 (𝑓 ∘ 𝑔)′ + (0) = (𝑓 ∘ 𝑔)′ − (0) , 𝑒𝑛𝑡ã𝑜 𝑓 ∘ 𝑔 é 𝑑𝑒𝑟𝑖𝑣á𝑣𝑒𝑙 𝑒𝑚 𝑥 = 0. 𝐸 (𝑓 ∘ 𝑔)′(0) = 2. 22 4. 𝑎) 𝑦 = 𝑡𝑔(𝑥) − 1 sec(𝑥) . 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟 𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 𝑒𝑚 𝑥 = 0. ∗ 𝑃𝑎𝑟𝑎 𝑥 = 0 𝑡𝑒𝑚𝑜𝑠 𝑦 = 𝑡𝑔(0) − 1 sec(0) = 0 − 1 1 = −1. (0, −1) 𝑦′ = 𝑠𝑒𝑐2(𝑥). sec(𝑥) − [𝑡𝑔(𝑥) − 1] sec(𝑥) . 𝑡𝑔(𝑥) 𝑠𝑒𝑐²(𝑥) 𝑦′ = 𝑠𝑒𝑐3(𝑥) − sec(𝑥) . 𝑡𝑔2(𝑥) + sec(𝑥) . 𝑡𝑔(𝑥) 𝑠𝑒𝑐²(𝑥) 𝑦′ = 𝑠𝑒𝑐2(𝑥) − 𝑡𝑔2(𝑥) + 𝑡𝑔(𝑥) 𝑠𝑒𝑐(𝑥) ; 𝑡𝑔2(𝑥) + 1 = 𝑠𝑒𝑐2(𝑥) → 𝑠𝑒𝑐2(𝑥) − 𝑡𝑔2(𝑥) = 1 𝑦′ = 1 + 𝑡𝑔(𝑥) 𝑠𝑒𝑐(𝑥) → 𝑦′(0) = 1 + 𝑡𝑔(0) 𝑠𝑒𝑐(0) = 1 + 0 1 = 1. ∗ 𝐿𝑜𝑔𝑜, 𝑒𝑞𝑢𝑎çã𝑜 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 𝑒𝑚 𝑥 = 0 é: 𝑦 + 1 = 1(𝑥 − 0) 𝑦 + 1 = 𝑥 𝑦 = 𝑥 − 1 𝑏) 𝑓(𝑥) = 𝑐𝑜𝑠 [𝑡𝑔√cos(𝑥)] . 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟 𝑓′(𝑥). 𝑆𝑒𝑗𝑎 𝑢 = cos(𝑥) ; 𝑣 = √𝑢 ; 𝑧 = 𝑡𝑔(𝑣); 𝑦 = 𝑓(𝑧) = cos(𝑧) 𝑃𝑒𝑙𝑎 𝑅𝑒𝑔𝑟𝑎 𝑑𝑎 𝐶𝑎𝑑𝑒𝑖𝑎 𝑡𝑒𝑚𝑜𝑠: 𝑑𝑦 𝑑𝑥 = 𝑑𝑢 𝑑𝑥 . 𝑑𝑣 𝑑𝑢 . 𝑑𝑧 𝑑𝑣 . 𝑑𝑦 𝑑𝑧 𝑑𝑦 𝑑𝑥 = (−𝑠𝑒𝑛(𝑥)). 1 2√𝑢 . (𝑠𝑒𝑐2(𝑣)). (−𝑠𝑒𝑛(𝑧)) 𝑦′ = 𝑓′(𝑥) = (−𝑠𝑒𝑛(𝑥)). 1 2√cos(𝑥) . (𝑠𝑒𝑐²√cos(𝑥)). (−𝑠𝑒𝑛 (𝑡𝑔√cos(𝑥))) 𝑓′(𝑥) = 𝑠𝑒𝑛(𝑥)𝑠𝑒𝑛 (𝑡𝑔√cos(𝑥)) 𝑠𝑒𝑐²√cos(𝑥) 2√cos(𝑥) 𝑐) 𝑔(𝑥) = √𝑥 + √𝑥 . 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟 𝑔′(𝑥). 𝑆𝑒𝑗𝑎 𝑢 = 𝑥 + √𝑥; 𝑦 = 𝑓(𝑢) = √𝑢 ; 𝑃𝑒𝑙𝑎 𝑅𝑒𝑔𝑟𝑎 𝑑𝑎 𝐶𝑎𝑑𝑒𝑖𝑎 𝑡𝑒𝑚𝑜𝑠: 𝑑𝑦 𝑑𝑥 = 𝑑𝑢 𝑑𝑥 . 𝑑𝑦 𝑑𝑢 𝑑𝑦 𝑑𝑥 = (1 + 1 2√𝑥 ) ( 1 2√𝑢 ) 𝑑𝑦 𝑑𝑥 = (1 + 1 2√𝑥 )( 1 2√𝑥 + √𝑥 ) 5. 𝑎) 𝑓(𝑥) = 5𝑎𝑟𝑐𝑡𝑔(𝑠𝑒𝑛 (𝑥2)). 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟 𝑓′(𝑥). 𝑆𝑒𝑗𝑎 𝑢 = 𝑥2; 𝑣 = 𝑠𝑒𝑛(𝑢); 𝑧 = 𝑎𝑟𝑐𝑡𝑔(𝑣); 𝑦 = 𝑓(𝑧) = 5𝑧 . 𝑃𝑒𝑙𝑎 𝑅𝑒𝑔𝑟𝑎 𝑑𝑎 𝐶𝑎𝑑𝑒𝑖𝑎 𝑡𝑒𝑚𝑜𝑠: 25 𝑓′(𝑥) = 2𝑎𝑥 + 𝑏 𝑓′′(𝑥) = 2𝑎 𝑓′′(𝑥) > 0 → 2𝑎 > 0 ∴ 𝑎 > 0. 𝐿𝑜𝑔𝑜, 𝑞𝑢𝑎𝑛𝑑𝑜 𝑓′′(𝑥) > 0 𝑡𝑒𝑚𝑜𝑠 𝑐𝑜𝑛𝑐𝑎𝑣𝑖𝑑𝑎𝑑𝑒 𝑣𝑜𝑙𝑡𝑎𝑑𝑎 𝑝𝑎𝑟𝑎 𝑐𝑖𝑚𝑎.𝑁𝑒𝑠𝑡𝑒 𝑐𝑎𝑠𝑜, 𝑡𝑒𝑚𝑜𝑠 𝐶. 𝑉. 𝐶 𝑝𝑎𝑟𝑎 𝑎 > 0. 𝑓′′(𝑥) < 0 → 2𝑎 < 0 ∴ 𝑎 < 0. 𝐿𝑜𝑔𝑜, 𝑞𝑢𝑎𝑛𝑑𝑜 𝑓′′(𝑥) < 0 𝑡𝑒𝑚𝑜𝑠 𝑐𝑜𝑛𝑐𝑎𝑣𝑖𝑑𝑎𝑑𝑒 𝑣𝑜𝑙𝑡𝑎𝑑𝑎 𝑝𝑎𝑟𝑎 𝑏𝑎𝑖𝑥𝑜. 𝑁𝑒𝑠𝑡𝑒 𝑐𝑎𝑠𝑜, 𝑡𝑒𝑚𝑜𝑠 𝐶. 𝑉. 𝐵 𝑝𝑎𝑟𝑎 𝑎 < 0. 𝑏) lim 𝑥→0+ (𝑐𝑥 + 1)𝑐𝑜𝑡𝑔(𝑥) = 𝑒−𝜋 . 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟 𝑐. lim 𝑥→0+ (𝑐𝑥 + 1)𝑐𝑜𝑡𝑔(𝑥) = lim 𝑥→0+ 𝑒ln(𝑐𝑥+1) 𝑐𝑜𝑡𝑔(𝑥) = lim 𝑥→0+ 𝑒𝑐𝑜𝑡𝑔(𝑥).ln(𝑐𝑥+1) = lim 𝑥→0+ 𝑒 ln(𝑐𝑥+1) 1 𝑐𝑜𝑡𝑔(𝑥) = 𝑒 lim 𝑥→0+ ln(𝑐𝑥+1) 1 𝑐𝑜𝑡𝑔(𝑥) . lim 𝑥→0+ ln(𝑐𝑥 + 1) 1 𝑐𝑜𝑡𝑔(𝑥) = lim 𝑥→0+ ln(𝑐𝑥 + 1) 𝑡𝑔(𝑥) = lim 𝑥→0+ 𝑐 𝑐𝑥 + 1 𝑠𝑒𝑐²(𝑥) = 𝑐 𝑐 + 1 𝑒 lim 𝑥→0+ ln(𝑐𝑥+1) 1 𝑐𝑜𝑡𝑔(𝑥) = 𝑒−𝜋 → 𝑒 𝑐 𝑐+1 = 𝑒−𝜋 𝑐 𝑐 + 1 = −𝜋 → 𝑐 = −𝜋𝑐 − 𝜋 → 𝑐(𝜋 + 1) = −𝜋 → 𝑐 = − 𝜋 𝜋 + 1 4. 𝐸𝑠𝑏𝑜ç𝑎𝑟 𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑒 𝑓(𝑥) = 𝑥²𝑒−𝑥² (I) 𝐼𝑛𝑡𝑒𝑟𝑠𝑒çõ𝑒𝑠 𝑐𝑜𝑚 𝑜𝑠 𝑒𝑖𝑥𝑜𝑠 𝑐𝑜𝑜𝑟𝑑𝑒𝑛𝑎𝑑𝑜𝑠: 𝑓(𝑥) = 0 → 𝑥. 𝑒−𝑥 2 = 0 ; 𝑒−𝑥 2 ≠ 0 ∀x ∈ ℝ 𝑓(𝑥) = 0 → 𝑥2 = 0 → 𝑥 = 0. −𝐸𝑥𝑖𝑠𝑡𝑒 𝑎𝑝𝑒𝑛𝑎𝑠 𝑜 𝑝𝑜𝑛𝑡𝑜 (0, 0) 𝑐𝑜𝑚𝑜 𝑖𝑛𝑡𝑒𝑟𝑠𝑒çã𝑜 𝑐𝑜𝑚 𝑜𝑠 𝑒𝑖𝑥𝑜𝑠. (II) 𝐶𝑟𝑒𝑠𝑐𝑖𝑚𝑒𝑛𝑡𝑜 𝑒 𝐷𝑒𝑐𝑟𝑒𝑠𝑐𝑖𝑚𝑒𝑛𝑡𝑜: 𝑓′(𝑥) = 2𝑥. 𝑒−𝑥 2 + 𝑥2(−2𝑥). 𝑒−𝑥 2 𝑓′(𝑥) = 𝑒−𝑥 2 (−2𝑥3 + 2𝑥) 𝑓′(𝑥) = 𝑥. 𝑒−𝑥 2 (−2𝑥2 + 2) Obs: 𝑒−𝑥 2 > 0 , ∀𝑥 ∈ ℝ 𝐹𝑎𝑧𝑒𝑛𝑑𝑜 𝑜 𝑒𝑠𝑡𝑢𝑑𝑜 𝑑𝑜 𝑠𝑖𝑛𝑎𝑙 𝑑𝑒 𝑓′(𝑥), 𝑡𝑒𝑚𝑜𝑠: −−−−−−−−− 0 + + + + +++++ + 𝑥 −−−− (−1) + + + + + + 1 − − − − −− − (−2𝑥2 + 2) +++++++++ ++++++++++ + 𝑒−𝑥 2 ++++ (−1) − − − 0 + + + 1 − − − −−− 𝑓(𝑥) = 𝑥. 𝑒−𝑥 2 (−2𝑥2 + 2) ∗ 𝐿𝑜𝑔𝑜, 𝑓(𝑥) é 𝑐𝑟𝑒𝑠𝑐𝑒𝑛𝑡𝑒 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 (−∞,−1) ∪ (0, 1) 𝑒 𝑑𝑒𝑐𝑟𝑒𝑠𝑐𝑒𝑛𝑡𝑒 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 (−1, 0) ∪ (1,+∞). (𝐼𝐼𝐼) 𝑃𝑜𝑛𝑡𝑜𝑠 𝐶𝑟í𝑡𝑖𝑐𝑜𝑠: −𝑂𝑐𝑜𝑟𝑟𝑒𝑚 𝑜𝑛𝑑𝑒 𝑓′(𝑥) = 0 𝑜𝑢 𝑜𝑛𝑑𝑒 𝑓′(𝑥) 𝑛ã𝑜 𝑒𝑥𝑖𝑠𝑡𝑒. 𝑁𝑒𝑠𝑡𝑒 𝑐𝑎𝑠𝑜, 𝑜𝑠 𝑝𝑜𝑛𝑡𝑜𝑠 𝑐𝑟í𝑡𝑖𝑐𝑜𝑠 𝑜𝑐𝑜𝑟𝑟𝑒𝑚 𝑒𝑚 𝑥 = −1; 𝑥 = 0 𝑒 𝑥 = 1. 𝑓(−1) = 1 𝑒 ; 𝑓(0) = 0 ; 𝑓(1) = 1 𝑒 . 26 𝐿𝑜𝑔𝑜, 𝑜𝑠 𝑝𝑜𝑛𝑡𝑜𝑠 𝑐𝑟í𝑡𝑖𝑐𝑜𝑠 𝑠ã𝑜 (−1, 1 𝑒 ) , (0, 0) 𝑒 (1, 1 𝑒 ). (𝐼𝑉)𝐶𝑜𝑛𝑐𝑎𝑣𝑖𝑑𝑎𝑑𝑒𝑠 𝑓′′(𝑥) = (−2𝑥). 𝑒−𝑥 2 (−2𝑥3 + 2𝑥) + 𝑒−𝑥 2 (−6𝑥2 + 2) 𝑓′′(𝑥) = 𝑒−𝑥 2 (4𝑥4 − 4𝑥2 − 6𝑥2 + 2) 𝑓′′(𝑥) = 𝑒−𝑥 2 (4𝑥4 − 10𝑥2 + 2) 𝐸𝑠𝑡𝑢𝑑𝑜 𝑑𝑜 𝑠𝑖𝑛𝑎𝑙 𝑑𝑒 𝑓′′(𝑥): 𝑠𝑒𝑗𝑎 𝑦 = 𝑥2𝑒𝑛𝑡ã𝑜 4𝑦2 − 10𝑦 + 2 = 0 → ∆= 100 − 32 = 68 𝑦 = 10 ± √68 8 → 𝑥 = ±√ 10 ± √68 8 ; 𝐶𝑜𝑚𝑜 𝑎 𝑓𝑢𝑛çã𝑜 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑎𝑙 𝑔(𝑥) = 4𝑥4 − 10𝑥2 + 2 é 𝑢𝑚𝑎 𝑓𝑢𝑛çã𝑜 𝑝𝑎𝑟 𝑓(𝑥) = 𝑓(−𝑥) 𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑐𝑜𝑛𝑐𝑙𝑢𝑖𝑟 𝑞𝑢𝑒 𝑒𝑠𝑡𝑎 𝑓𝑢𝑛çã𝑜 é 𝑠𝑖𝑚é𝑡𝑟𝑖𝑐𝑎 𝑒𝑚 𝑟𝑒𝑙𝑎çã𝑜 𝑜 𝑒𝑖𝑥𝑜 𝑦. 𝑂𝑢 𝑠𝑒𝑗𝑎, 𝐶𝑜𝑚 𝑒𝑠𝑠𝑎 𝑎𝑛á𝑙𝑖𝑠𝑒 𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑐𝑜𝑛𝑐𝑙𝑢𝑖𝑟 𝑞𝑢𝑒: 1 − 𝑓(𝑥) 𝑝𝑜𝑠𝑠𝑢𝑖 𝑐𝑜𝑛𝑐𝑎𝑣𝑖𝑑𝑎𝑑𝑒 𝑣𝑜𝑙𝑡𝑎𝑑𝑎 𝑝𝑎𝑟𝑎 𝑐𝑖𝑚𝑎 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 (−∞,−√ 10 + √68 8 ) ∪ (−√ 10 − √68 8 ,√ 10 − √68 8 ) ∪ (√ 10 + √68 8 ,+∞) 2 − 𝑓(𝑥) 𝑝𝑜𝑠𝑠𝑢𝑖 𝑐𝑜𝑛𝑐𝑎𝑣𝑖𝑑𝑎𝑑𝑒 𝑣𝑜𝑙𝑡𝑎𝑑𝑎 𝑝𝑎𝑟𝑎 𝑏𝑎𝑖𝑥𝑜 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 (−√ 10 + √68 8 ,−√ 10 − √68 8 ) ∪ (√ 10 − √68 8 ,√ 10 + √68 8 ) (𝑉) 𝑃𝑜𝑛𝑡𝑜𝑠 𝑑𝑒 𝐼𝑛𝑓𝑙𝑒𝑥ã𝑜: −𝑂𝑐𝑜𝑟𝑟𝑒𝑚 𝑛𝑜𝑠 𝑝𝑜𝑛𝑡𝑜𝑠 𝑜𝑛𝑑𝑒 𝑓′′(𝑥) = 0, 𝑜𝑢 𝑠𝑒𝑗𝑎, 𝑒𝑚 𝑥 = ±√ 10 ± √68 8 . (𝑉𝐼)𝐴𝑠𝑠í𝑛𝑡𝑜𝑡𝑎𝑠: −𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑖𝑠: 𝑁ã𝑜 𝑒𝑥𝑖𝑠𝑡𝑒𝑚! 𝑃𝑜𝑖𝑠, 𝑠𝑒𝑛𝑑𝑜 𝑓(𝑥) = 𝑥2 𝑒−𝑥 2 𝑡𝑒𝑚𝑜𝑠 𝐷(𝑓) = ℝ. 𝐿𝑜𝑔𝑜, 𝑛ã𝑜 ℎá 𝑟𝑒𝑠𝑡𝑟𝑖𝑛çã𝑜 𝑝𝑎𝑟𝑎 𝑣𝑎𝑙𝑜𝑟𝑒𝑠 𝑑𝑒 𝑥 𝑞𝑢𝑒 𝑝𝑜𝑠𝑠𝑎𝑚 𝑔𝑒𝑟𝑎𝑟 𝑖𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎çã𝑜 𝑛𝑜 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑑𝑜𝑟, 𝑑𝑒 𝑚𝑜𝑑𝑜 𝑎 𝑜𝑏𝑡𝑒𝑟𝑚𝑜𝑠 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙. −𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑖𝑠: 𝐷𝑖𝑧𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑎 𝑟𝑒𝑡𝑎 𝑦 = 𝑎 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑠𝑒 𝑜𝑐𝑜𝑟𝑟𝑒𝑟 𝑢𝑚 𝑑𝑜𝑠 𝑠𝑒𝑔𝑢𝑖𝑛𝑡𝑒 𝑐𝑎𝑠𝑜𝑠: lim 𝑥→+∞ 𝑓(𝑥) = 𝑎 𝑜𝑢 lim 𝑥→−∞ 𝑓(𝑥) = 𝑎 lim 𝑥→+∞ 𝑓(𝑥) = lim 𝑥→+∞ 𝑥2 𝑒−𝑥 2 = lim 𝑥→+∞ 2𝑥 (−2𝑥)𝑒−𝑥 2 = lim 𝑥→+∞ −1 𝑒−𝑥 2 = 0. 27 lim 𝑥→−∞ 𝑓(𝑥) = lim 𝑥→−∞ 𝑥2 𝑒−𝑥 2 = lim 𝑥→−∞ 2𝑥 (−2𝑥)𝑒−𝑥 2 = lim 𝑥→−∞ −1 𝑒−𝑥 2 = 0. 𝐿𝑜𝑔𝑜, 𝑎 𝑟𝑒𝑡𝑎 𝑦 = 0 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑎𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑒 𝑓(𝑥). 5. 𝑎) 𝑥² − 𝑦2 = 1 . 𝐸𝑛𝑐𝑜𝑛𝑡𝑟𝑎𝑟 𝑜 𝑝𝑜𝑛𝑡𝑜 𝑑𝑎 ℎ𝑖𝑝é𝑟𝑏𝑜𝑙𝑒 𝑚𝑎𝑖𝑠 𝑝𝑟ó𝑥𝑖𝑚𝑜 𝑑𝑜 𝑝𝑜𝑛𝑡𝑜 (0, −1). 𝐷𝑎𝑑𝑜 𝑢𝑚 𝑝𝑜𝑛𝑡𝑜 (𝑥, 𝑦) 𝑝𝑒𝑟𝑡𝑒𝑛𝑐𝑒𝑛𝑡𝑒 à ℎ𝑖𝑝é𝑟𝑏𝑜𝑙𝑒, 𝑡𝑒𝑚𝑜𝑠: 𝐷 = √𝑥2 + (𝑦 + 1)2 = √𝑥2 + 𝑦2 + 2𝑦 + 1 = √2𝑦2 + 2𝑦 + 2 𝐷′(𝑦) = 4𝑦 + 2 2√2𝑦2 + 2𝑦 + 2 = 2𝑦 + 1 √2𝑦2 + 2𝑦 + 2 𝐷′(𝑦) = 0 → 2𝑦 + 1 = 0 → 𝑦 = − 1 2 𝐿𝑜𝑔𝑜, 𝑝𝑎𝑟𝑎 𝑦 = − 1 2 𝑡𝑒𝑚𝑜𝑠: 𝑥2 − 1 4 = 1 → 𝑥2 = 5 4 → 𝑥 = ± √5 2 . 𝑂𝑠 𝑝𝑜𝑛𝑡𝑜𝑠 𝑑𝑎 ℎ𝑖𝑝é𝑟𝑏𝑜𝑙𝑒 𝑚𝑎𝑖𝑠 𝑝𝑟ó𝑥𝑖𝑚𝑜𝑠 𝑑𝑜 𝑝𝑜𝑛𝑡𝑜 (0, −1) 𝑠ã𝑜 (− √5 2 ,− 1 2 ) 𝑒 ( √5 2 ,− 1 2 ). 𝑏) 𝐴 𝑟𝑒𝑡𝑎 𝑥 + 𝑦 = 0 𝑡𝑎𝑛𝑔𝑒𝑛𝑐𝑖𝑎 𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑒 𝑢𝑚𝑎 𝑓𝑢𝑛çã𝑜 𝑓 𝑒𝑚 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑑𝑜 𝑝𝑜𝑛𝑡𝑜 𝑒 𝑞𝑢𝑒 𝑒𝑠𝑠𝑎 𝑓𝑢𝑛çã𝑜 é 𝑡𝑎𝑙 𝑞𝑢𝑒 𝑓′′(𝑥) = 3𝑥2𝑒 𝑓′(0) = 0. 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑎 𝑓𝑢𝑛çã𝑜 𝑓. ∗ 𝑃𝑟𝑖𝑚𝑒𝑖𝑟𝑜𝑠 𝑑𝑒𝑣𝑒𝑚𝑜𝑠 𝑝𝑟𝑜𝑐𝑢𝑟𝑎𝑟 𝑢𝑚𝑎 𝑎𝑛𝑡𝑖𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎 𝑑𝑒 𝑓′′(𝑥) = 3𝑥2. 𝐷𝑎í 𝑡𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑓′(𝑥) = 𝑥3 + 𝐶 , 𝑜𝑛𝑑𝑒 𝐶 é 𝑢𝑚𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑒 𝑎 𝑠𝑒𝑟 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑑𝑎. 𝐶𝑜𝑚𝑜 𝑓′(0) = 0, 𝑐𝑜𝑛𝑐𝑙𝑢í𝑚𝑜𝑠 𝑞𝑢𝑒 𝐶 = 0. 𝐿𝑜𝑔𝑜, 𝑓′(𝑥) = 𝑥3. 𝐴 𝑟𝑒𝑡𝑎 𝑦 = −𝑥 𝑝𝑜𝑠𝑠𝑢𝑖 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑖𝑔𝑢𝑎𝑙 𝑎 − 1, 𝑣𝑎𝑙𝑜𝑟 𝑎𝑠𝑠𝑢𝑚𝑖𝑑𝑜 𝑝𝑜𝑟 𝑓′(𝑥) 𝑒𝑚 𝑥 = −1. 𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑜 𝑝𝑜𝑛𝑡𝑜 𝑑𝑒 𝑡𝑎𝑛𝑔𝑒𝑛𝑐𝑖𝑎 é (−1, 1). ∗ 𝐸𝑛𝑐𝑜𝑛𝑡𝑟𝑎𝑛𝑑𝑜 𝑢𝑚𝑎 𝑎𝑛𝑡𝑖𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎 𝑝𝑎𝑟𝑎 𝑓′(𝑥) = 𝑥3𝑡𝑒𝑚𝑜𝑠 𝑓(𝑥) = 1 4 𝑥4 + 𝐾, 𝑜𝑛𝑑𝑒 𝐾 é 𝑢𝑚𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑒 𝑎 𝑠𝑒𝑟 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑑𝑎. 𝐶𝑜𝑚𝑜 𝑓(−1) = 1, 𝑒𝑛𝑡ã𝑜 𝑡𝑒𝑚𝑜𝑠: 1 4 (−1)4 + 𝐾 = 1 → 𝐾 = 1 − 1 4 → 𝐾 = 3 4 . 30 2. 𝑃𝑜𝑟 𝑠𝑒𝑚𝑒𝑙ℎ𝑎𝑛ç𝑎 𝑑𝑒 𝑡𝑟𝑖â𝑛𝑔𝑢𝑙𝑜𝑠 𝑡𝑒𝑚𝑜𝑠: 𝐻 𝑅 = 𝐻 − ℎ 𝑟 → 𝑅𝐻 − 𝑅ℎ = 𝑟𝐻 → ℎ = 𝐻(𝑅 − 𝑟) 𝑅 . 𝑉 = 𝜋𝑟2ℎ → 𝑉 = 𝜋𝑟2. 𝐻 𝑅 (𝑅 − 𝑟) → 𝑉 = 𝜋𝐻 𝑅 (𝑅𝑟2 − 𝑟3). 𝑃𝑜𝑟 𝑚𝑒𝑖𝑜 𝑑𝑎 𝑝𝑟𝑖𝑚𝑒𝑖𝑟𝑎 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎 𝑒𝑛𝑐𝑜𝑛𝑡𝑟𝑎𝑟𝑒𝑚𝑜𝑠 𝑜 𝑝𝑜𝑛𝑡𝑜 𝑜𝑛𝑑𝑒 𝑉 é 𝑚á𝑥𝑖𝑚𝑜. 𝑉′(𝑟) = 𝜋𝐻 𝑅 (2𝑅𝑟 − 3𝑟2); 𝑉′(𝑟) = 0 → 2𝑅𝑟 − 3𝑟2 = 0 ; 𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑛𝑑𝑜, 𝑜𝑏𝑡𝑒𝑚𝑜𝑠 𝑟 = 2𝑅 3 ; 𝑂𝑏𝑠:𝑁𝑜𝑡𝑒 𝑞𝑢𝑒 𝑟 = 0 𝑛ã𝑜 𝑝𝑜𝑑𝑒 𝑠𝑒𝑟 𝑠𝑜𝑙𝑢çã𝑜 𝑑𝑜 𝑝𝑟𝑜𝑏𝑙𝑒𝑚𝑎! 𝑆𝑒𝑛𝑑𝑜 𝑟 = 2𝑅 3 , 𝑒𝑛𝑐𝑜𝑛𝑡𝑟𝑎𝑚𝑜𝑠 ℎ = 𝐻 (𝑅 − 2𝑅 3 ) 𝑅 → ℎ = 𝐻 ( 𝑅 3) 𝑅 → ℎ = 𝐻 3 ∗ 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑖𝑛𝑑𝑜 𝐻 = 1 𝑒 𝑅 = 1 𝑡𝑒𝑚𝑜𝑠: 𝑟 = 2 3 𝑒 ℎ = 1 3 . 𝐷𝑒𝑠𝑠𝑎 𝑓𝑜𝑟𝑚𝑎 𝑡𝑒𝑚𝑜𝑠 𝑉𝑚á𝑥 = 𝜋. 𝑟 2. ℎ = 𝜋. ( 2 3 ) 2 . ( 1 3 ) = 4 9 𝜋 𝑢. 𝑉 3. 𝑎) lim 𝑥→ 𝜋 2 (cos 𝑥)cos𝑥 = lim 𝑥→ 𝜋 2 𝑒ln(cos𝑥) cos𝑥 = lim 𝑥→ 𝜋 2 𝑒cos𝑥.ln(cos𝑥) = lim 𝑥→ 𝜋 2 𝑒 ln(cos𝑥) sec𝑥 = 𝑒 lim 𝑥→ 𝜋 2 ln(cos𝑥) sec𝑥 . 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑛𝑑𝑜 𝑜 𝑙𝑖𝑚𝑖𝑡𝑒 𝑑𝑜 𝑒𝑥𝑝𝑜𝑒𝑛𝑡𝑒 𝑡𝑒𝑚𝑜𝑠: lim 𝑥→ 𝜋 2 ln(cos 𝑥) sec 𝑥 = lim 𝑥→ 𝜋 2 −𝑠𝑒𝑛 𝑥 cos 𝑥 sec 𝑥 . 𝑡𝑔 𝑥 = lim 𝑥→ 𝜋 2 −𝑡𝑔 𝑥 sec 𝑥 . 𝑡𝑔 𝑥 = lim 𝑥→ 𝜋 2 −1 sec 𝑥 = lim 𝑥→ 𝜋 2 (− cos 𝑥) = 0. 𝐿𝑜𝑔𝑜, lim 𝑥→ 𝜋 2 (cos 𝑥)cos𝑥 = 𝑒 lim 𝑥→ 𝜋 2 ln(cos𝑥) sec𝑥 = 𝑒0 = 1. 𝑏) lim 𝑥→∞ (1 + 3 𝑥 + 5 𝑥2 ) 𝑥 = lim 𝑥→∞ 𝑒 ln(1+ 3 𝑥 + 5 𝑥2 ) 𝑥 = lim 𝑥→∞ 𝑒 𝑥.ln(1+ 3 𝑥 + 5 𝑥2 ) = lim 𝑥→∞ 𝑒 ln(1+ 3 𝑥 + 5 𝑥2 ) 1 𝑥 = 𝑒 lim 𝑥→∞ ln(1+ 3 𝑥 + 5 𝑥2 ) 1 𝑥 . 31 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑛𝑑𝑜 𝑜 𝑙𝑖𝑚𝑖𝑡𝑒 𝑑𝑜 𝑒𝑥𝑝𝑜𝑒𝑛𝑡𝑒 𝑡𝑒𝑚𝑜𝑠: lim 𝑥→∞ ln (1 + 3 𝑥 + 5 𝑥2 ) 1 𝑥 = lim 𝑥→∞ (− 3 𝑥2 − 10 𝑥3 ) (1 + 3 𝑥 + 5 𝑥2 ) − 1 𝑥2 = lim 𝑥→∞ −𝑥 2(− 3 𝑥2 − 10 𝑥3 ) 1 + 3 𝑥 + 5 𝑥2 = lim 𝑥→∞ 3 + 10 𝑥 1 + 3 𝑥 + 5 𝑥2 = 3 + 0 1 + 0 + 0 = 3 1 = 3. 𝐿𝑜𝑔𝑜, lim 𝑥→∞ (1 + 3 𝑥 + 5 𝑥2 ) 𝑥 = 𝑒 lim 𝑥→∞ ln(1+ 3 𝑥 + 5 𝑥2 ) 1 𝑥 = 𝑒3. 4. 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑟 𝑎 á𝑟𝑒𝑎 𝑑𝑎 𝑟𝑒𝑔𝑖ã𝑜 𝑠𝑜𝑏 𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑒 𝑓(𝑥) = 𝑥3 + 𝑥2, 𝑒𝑛𝑡𝑟𝑒 𝑥 = 0 𝑒 𝑥 = 1. 𝑈𝑡𝑖𝑙𝑖𝑧𝑎𝑛𝑑𝑜 𝑜 𝑚é𝑡𝑜𝑑𝑜 𝑑𝑎 𝑆𝑜𝑚𝑎 𝑑𝑒 𝑅𝑖𝑒𝑛𝑚𝑎𝑛: Á𝑟𝑒𝑎 = lim 𝑛→∞ ∑𝑓(𝑥𝑖)∆𝑥 𝑛 𝑖=1 𝑂𝑛𝑑𝑒 ∆𝑥 = 1 − 0 𝑛 = 1 𝑛 ; 𝑎𝑠𝑠𝑖𝑚 𝑥𝑖 = 𝑖 𝑛 . 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑖𝑛𝑑𝑜 𝑛𝑎 𝑒𝑥𝑝𝑟𝑒𝑠𝑠ã𝑜, 𝑡𝑒𝑚𝑜𝑠: Á𝑟𝑒𝑎 = lim 𝑛→∞ ∑𝑓( 𝑖 𝑛 ) 1 𝑛 𝑛 𝑖=1 = lim 𝑛→∞ 1 𝑛 ∑( 𝑖3 𝑛3 + 𝑖2 𝑛2 ) 𝑛 𝑖=1 = lim 𝑛→∞ 1 𝑛 [ 1 𝑛3 ∑𝑖3 𝑛 𝑖=1 + 1 𝑛2 ∑𝑖2 𝑛 𝑖=1 ] ∗ 𝑂𝑏𝑠1 : ∑𝑖 2 𝑛 𝑖=1 = 𝑛(2𝑛2 + 3𝑛 + 1) 6 ∗ 𝑂𝑏𝑠2 : ∑𝑖 3 𝑛 𝑖=1 = 𝑛²(𝑛² + 2𝑛 + 1) 4 Á𝑟𝑒𝑎 = lim 𝑛→∞ 1 𝑛 [ 1 𝑛3 𝑛²(𝑛2 + 2𝑛 + 1) 4 + 1 𝑛2 𝑛(2𝑛2 + 3𝑛 + 1) 6 ] = lim 𝑛→∞ 1 𝑛 [ (𝑛2 + 2𝑛 + 1) 4𝑛 + (2𝑛2 + 3𝑛 + 1) 6𝑛 ] 32 = lim 𝑛→∞ 1 𝑛 [ 6(𝑛2 + 2𝑛 + 1) + 4(2𝑛2 + 3𝑛 + 1) 24𝑛 ] = lim 𝑛→∞ 14𝑛2 + 24𝑛 + 10 24𝑛2 = lim 𝑛→∞ 14𝑛2 𝑛2 + 24𝑛 𝑛2 + 10 𝑛2 24𝑛2 𝑛2 = lim 𝑛→∞ 14 + 24 𝑛 + 10 𝑛2 24 = 14 + 0 + 0 24 Á𝑟𝑒𝑎 = 14 24 = 7 12 𝑢. 𝐴 5. 𝐹𝑎𝑧𝑒𝑟 𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑒 𝑓(𝑥) = 𝑥3 − 𝑥 + 1 𝑥2 ; 𝑎) 𝐷𝑜𝑚í𝑛𝑖𝑜 𝑑𝑎 𝑓𝑢𝑛çã𝑜; ∗ 𝑓 é 𝑢𝑚𝑎 𝑓𝑢𝑛çã𝑜 𝑟𝑎𝑐𝑖𝑜𝑛𝑎𝑙 𝑒, 𝑐𝑜𝑚𝑜 𝑠𝑒𝑢 𝑛𝑢𝑚𝑒𝑟𝑎𝑑𝑜𝑟 é 𝑢𝑚𝑎 𝑓𝑢𝑛çã𝑜 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑎𝑙 𝑐𝑜𝑚 𝑜 𝑑𝑜𝑚í𝑛𝑖𝑜 𝑠𝑒𝑛𝑑𝑜 𝑜 𝑐𝑜𝑛𝑗𝑢𝑛𝑡𝑜 𝑑𝑜𝑠 𝑛ú𝑚𝑒𝑟𝑜𝑠 𝑟𝑒𝑎𝑖𝑠, 𝑜 𝑑𝑜𝑚í𝑛𝑖𝑜 𝑑𝑎 𝑓𝑢𝑛çã𝑜 𝑓 𝑓𝑖𝑐𝑎 𝑟𝑒𝑠𝑡𝑟𝑖𝑡𝑜 𝑎𝑜 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑑𝑜𝑟, 𝑜 𝑞𝑢𝑎𝑙 𝑑𝑒𝑣𝑒 𝑠𝑒𝑟 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒 𝑑𝑒 𝑧𝑒𝑟𝑜. ∗ 𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑥2 ≠ 0 ⇒ 𝑥 ≠ 0. 𝐷(𝑓) = ℝ − {0}. 𝑏) 𝐼𝑛𝑡𝑒𝑟𝑠𝑒çõ𝑒𝑠 𝑐𝑜𝑚 𝑜𝑠 𝑒𝑖𝑥𝑜𝑠 𝑐𝑜𝑜𝑟𝑑𝑒𝑛𝑎𝑑𝑜𝑠; ∗ 𝑁ã𝑜 ℎá 𝑖𝑛𝑡𝑒𝑟𝑠𝑒çã𝑜 𝑐𝑜𝑚 𝑜 𝑒𝑖𝑥𝑜 𝑦, 𝑝𝑜𝑖𝑠 𝑥 = 0 𝑛ã𝑜 𝑝𝑒𝑟𝑡𝑒𝑛𝑐𝑒 𝑎𝑜 𝑑𝑜𝑚í𝑛𝑖𝑜 𝑑𝑒 𝑓; 𝑓(𝑥) = 0 → 𝑥3 − 𝑥 + 1 = 0 𝐶𝑎𝑙𝑐𝑢𝑙𝑒𝑚𝑜𝑠 𝑓(−2) 𝑒 𝑓(−1): 𝑓(−2) = −8 + 2 + 1 4 = − 5 4 𝑒 𝑓(−1) = −1 + 1 + 1 1 = 1. ∗ 𝐶𝑜𝑚𝑜 𝑓 é 𝑢𝑚𝑎 𝑓𝑢𝑛çã𝑜 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑎𝑙 𝑟𝑎𝑐𝑖𝑜𝑛𝑎𝑙 𝑒, 𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 [−2, −1], 𝑒 𝑓(−2) < 0 < 𝑓(−1), 𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑔𝑎𝑟𝑎𝑛𝑡𝑖𝑟 𝑝𝑒𝑙𝑜 𝑇𝑒𝑜𝑟𝑒𝑚𝑎 𝑑𝑜 𝑉𝑎𝑙𝑜𝑟 𝐼𝑛𝑡𝑒𝑟𝑚𝑒𝑑𝑖á𝑟𝑖𝑜 𝑞𝑢𝑒 𝑒𝑥𝑖𝑠𝑡𝑒 𝑎𝑙𝑔𝑢𝑚 𝑥 ∈ (−2,−1) 𝑡𝑎𝑙 𝑞𝑢𝑒 𝑓(𝑥) = 0. 𝑂𝑢 𝑠𝑒𝑗𝑎, 𝑓(𝑥) 𝑝𝑜𝑠𝑠𝑢𝑖 𝑢𝑚𝑎 𝑟𝑎í𝑧 𝑒𝑛𝑡𝑟𝑒 𝑥 = −2 𝑒 𝑥 = −1. 𝑐) 𝐸𝑥𝑡𝑟𝑒𝑚𝑜𝑠 𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑜𝑠; 𝑓′(𝑥) = (3𝑥2 − 1)𝑥2 − (𝑥3 − 𝑥 + 1)2𝑥 𝑥4 𝑓′(𝑥) = 3𝑥4 − 𝑥2 − 2𝑥4 + 2𝑥2 − 2𝑥 𝑥4 𝑓′(𝑥) = 𝑥4 + 𝑥2 − 2𝑥 𝑥4 ; 𝑐𝑜𝑚 𝑥 ≠ 0 𝑡𝑒𝑚𝑜𝑠: 𝑓′(𝑥) = 𝑥3 + 𝑥 − 2 𝑥3 𝐹𝑎𝑧𝑒𝑛𝑑𝑜 𝑓′(𝑥) = 0 → 𝑥3 + 𝑥 − 2 𝑥3 = 0 ⇔ 𝑥3 + 𝑥 − 2 = 0 𝐹𝑎𝑧𝑒𝑛𝑑𝑜 𝑜 𝑡𝑒𝑠𝑡𝑒 𝑑𝑎𝑠 𝑟𝑎í𝑧𝑒𝑠 𝑝𝑟𝑜𝑣á𝑣𝑒𝑖𝑠 − 2,−1, 1 𝑒 2 𝑡𝑒𝑚𝑜𝑠: 𝑓′(−2) = −8 − 2 − 2 −8 = 10 8 ; 𝑓′(−1) = −1 − 1 − 2 −1 = 4 ; 35 2.6 Reavaliação da 2ª média-07 de Dezembro de 2007 1. 𝑆𝑒𝑗𝑎 𝑉 𝑜 𝑣𝑜𝑙𝑢𝑚𝑒 𝑑𝑒 á𝑔𝑢𝑎 𝑞𝑢𝑒 𝑝𝑒𝑟𝑚𝑎𝑛𝑒𝑐𝑒 𝑛𝑜 𝑐𝑜𝑛𝑒, 𝑉𝑒 𝑜 𝑣𝑜𝑙𝑢𝑚𝑒 𝑑𝑒 á𝑔𝑢𝑎 𝑞𝑢𝑒 𝑒𝑛𝑡𝑟𝑎 𝑒 𝑉𝑠 𝑜 𝑣𝑜𝑙𝑢𝑚𝑒 𝑑𝑒 á𝑔𝑢𝑎 𝑞𝑢𝑒 𝑒𝑠𝑐𝑜𝑎 (𝑣𝑎𝑧𝑎) 𝑑𝑜 𝑐𝑜𝑛𝑒. 𝐸𝑛𝑡ã𝑜: 𝑉 = 𝑉𝑒 − 𝑉𝑠 𝑑𝑉 𝑑𝑡 = 𝑑𝑉𝑒 𝑑𝑡 − 𝑑𝑉𝑠 𝑑𝑡 𝐷𝑒 𝑚𝑜𝑑𝑜 𝑞𝑢𝑒, 𝑑𝑉𝑒 𝑑𝑡 é 𝑎 𝑡𝑎𝑥𝑎 𝑐𝑜𝑚 𝑎 𝑞𝑢𝑎𝑙 𝑒𝑛𝑡𝑟𝑎 á𝑔𝑢𝑎 𝑛𝑜 𝑐𝑜𝑛𝑒, 𝑞𝑢𝑒 é 8𝑐𝑚³/𝑚𝑖𝑛 . 𝐿𝑜𝑔𝑜, 𝑑𝑉 𝑑𝑡 = 8 − 𝑑𝑉𝑠 𝑑𝑡 𝑁𝑎 𝑞𝑢𝑒𝑠𝑡ã𝑜 𝑒𝑙𝑒 𝑝𝑒𝑑𝑒 𝑎 𝑣𝑒𝑙𝑜𝑐𝑖𝑑𝑎𝑑𝑒 𝑑𝑒 𝑒𝑠𝑐𝑜𝑎𝑚𝑒𝑛𝑡𝑜, 𝑜𝑢 𝑠𝑒𝑗𝑎, 𝑑𝑉𝑠 𝑑𝑡 . 𝐴𝑠𝑠𝑖𝑚, 𝑑𝑉𝑠 𝑑𝑡 = 8 − 𝒅𝑽 𝒅𝒕 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑛𝑑𝑜 𝑑𝑉 𝑑𝑡 𝑡𝑒𝑚𝑜𝑠: 𝑑𝑉 𝑑𝑡 = 𝑑ℎ 𝑑𝑡 . 𝑑𝑉 𝑑ℎ 𝑂𝑏𝑠: 𝑑ℎ 𝑑𝑡 = 1𝑚𝑚/𝑚𝑖𝑛 = 0,1𝑐𝑚/𝑚𝑖𝑛 𝑉 = 𝜋𝑟2ℎ 3 . 𝑃𝑜𝑟 𝑠𝑒𝑚𝑒𝑙ℎ𝑎𝑛ç𝑎 𝑑𝑒 𝑡𝑟𝑖â𝑛𝑔𝑢𝑙𝑜𝑠 𝑡𝑒𝑚𝑜𝑠: ℎ 𝑟 = 20 5 => 𝑟 = ℎ 4 . 𝐸𝑛𝑡ã𝑜: 𝑉 = 𝜋ℎ3 48 , 𝑙𝑜𝑔𝑜 𝑑𝑉 𝑑ℎ = 3𝜋ℎ2 48 = 𝜋ℎ2 16 . 𝐴𝑠𝑠𝑖𝑚, 𝑑𝑉 𝑑𝑡 = 0,1 × 𝜋ℎ² 16 => 0,1𝜋ℎ² 16 . 𝑑𝑉𝑒 𝑑𝑡 = 8 − 0,1𝜋ℎ2 16 . 𝑄𝑢𝑎𝑛𝑑𝑜 ℎ = 16𝑐𝑚 𝑡𝑒𝑚𝑜𝑠: 𝑑𝑉𝑒 𝑑𝑡 = 8 − 0,1𝜋(16)2 16 = (8 − 1,6𝜋)𝑐𝑚3/𝑚𝑖𝑛 2. 𝑎)𝐸𝑞𝑢𝑎çã𝑜 𝑝𝑎𝑟𝑎 𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 à 𝑐𝑢𝑟𝑣𝑎 𝑥𝑦 = 𝑦𝑥 𝑛𝑜 𝑝𝑜𝑛𝑡𝑜 𝑒𝑚 𝑞𝑢𝑒 𝑥 = 1. 𝑃𝑎𝑟𝑎 𝑥 = 1 𝑡𝑒𝑚𝑜𝑠: 1𝑦 = 𝑦1 → 1 = 𝑦. 𝑂 𝑝𝑜𝑛𝑡𝑜 𝑒𝑚 𝑞𝑢𝑒𝑠𝑡ã𝑜 é (1, 1). ∗ 𝐴𝑝𝑙𝑖𝑐𝑎𝑛𝑑𝑜 𝑜 𝑙𝑜𝑔𝑎𝑟í𝑡𝑚𝑜 𝑒𝑚 𝑎𝑚𝑏𝑜𝑠 𝑜𝑠 𝑚𝑒𝑚𝑏𝑟𝑜𝑠 𝑑𝑎 𝑖𝑔𝑢𝑎𝑙𝑑𝑎𝑑𝑒, 𝑡𝑒𝑚𝑜𝑠: ln 𝑥𝑦 = ln𝑦𝑥 𝑦. ln 𝑥 = 𝑥. ln 𝑦 → 𝑝𝑜𝑟 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎çã𝑜 𝑙𝑜𝑔𝑎𝑟í𝑡𝑚𝑖𝑐𝑎 𝑒 𝑖𝑚𝑝𝑙í𝑐𝑖𝑡𝑎, 𝑡𝑒𝑚𝑜𝑠: 𝑦′. ln 𝑥 + 𝑦. 1 𝑥 = ln 𝑦 + 𝑥. 𝑦′ 𝑦 36 𝑦′. ln 𝑥 − 𝑦′. 𝑥 𝑦 = ln 𝑦 − 𝑦 𝑥 𝑦′. [ln 𝑥 − 𝑥 𝑦 ] = ln 𝑦 − 𝑦 𝑥 𝑦′ = ln𝑦 − 𝑦 𝑥 ln 𝑥 − 𝑥 𝑦 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑛𝑑𝑜 𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑦′ 𝑛𝑜 𝑝𝑜𝑛𝑡𝑜 (1, 1) 𝑡𝑒𝑚𝑜𝑠: 𝑦′ = ln1 − 1 1 ln 1 − 1 1 = 0 − 1 0 − 1 = −1 −1 = 1. 𝐴𝑠𝑠𝑖𝑚, 𝑎 𝑒𝑥𝑝𝑟𝑒𝑠𝑠ã𝑜 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 𝑛𝑜 𝑝𝑜𝑛𝑡𝑜 (1, 1) 𝑒 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑖𝑔𝑢𝑎𝑙 𝑎 1: 𝑦 − 1 = 1(𝑥 − 1) 𝑦 − 1 = 𝑥 − 1 𝑦 = 𝑥 𝑏)𝐸𝑛𝑐𝑜𝑛𝑡𝑟𝑎𝑟 𝑜 𝑝𝑜𝑛𝑡𝑜 𝑜𝑛𝑑𝑒 𝑎 𝑐𝑢𝑟𝑣𝑎 𝑦 = cosh 𝑥 𝑝𝑜𝑠𝑠𝑢𝑖 𝑎 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎 𝑖𝑔𝑢𝑎𝑙 𝑎 1. 𝑦′ = 𝑠𝑒𝑛ℎ(𝑥) 𝑦′ = 1 → 𝑠𝑒𝑛ℎ(𝑥) = 1 (𝐼) 𝑠𝑒𝑛ℎ(𝑥) = 𝑒𝑥 − 𝑒−𝑥 2 ; 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑖𝑛𝑑𝑜 𝑛𝑎 𝑒𝑞𝑢𝑎çã𝑜 (𝐼): 𝑒𝑥 − 𝑒−𝑥 2 = 1 𝑒𝑥 − 𝑒−𝑥 = 2 𝑒𝑥 − 1 𝑒𝑥 − 2 = 0 𝑒2𝑥 − 2𝑒𝑥 − 1 = 0 𝐹𝑎𝑧𝑒𝑛𝑑𝑜 𝑎 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑖çã𝑜 𝑏 = 𝑒𝑥 𝑜𝑏𝑡𝑒𝑚𝑜𝑠 𝑎 𝑠𝑒𝑔𝑢𝑖𝑛𝑡𝑒 𝑒𝑥𝑝𝑟𝑒𝑠𝑠ã𝑜 𝑑𝑜 𝑠𝑒𝑔𝑢𝑛𝑑𝑜 𝑔𝑟𝑎𝑢: 𝑏2 − 2𝑏 − 1 = 0 ∗ 𝑂𝑏𝑠: 𝑠𝑒 𝑏 = 𝑒𝑥, 𝑒𝑛𝑡ã𝑜 𝑏 > 0 ∀𝑥 ∈ ℝ. 𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑛𝑑𝑜 𝑎 𝑒𝑞𝑢𝑎çã𝑜 𝑎𝑐𝑖𝑚𝑎, 𝑡𝑒𝑚𝑜𝑠: ∆= 4 + 4 = 8 𝑏 = 2 ± 2√2 2 → 𝑏 = 1 + √2 𝑜𝑢 𝑏 = 1 − √2 (𝑏 < 0) 𝑂 𝑡𝑒𝑟𝑚𝑜 𝑒𝑚 𝑑𝑒𝑠𝑡𝑎𝑞𝑢𝑒 𝑛ã𝑜 𝑠𝑎𝑡𝑖𝑠𝑓𝑎𝑧 𝑎 𝑐𝑜𝑛𝑑𝑖çã𝑜 𝑎𝑛𝑡𝑒𝑟𝑖𝑜𝑟𝑚𝑒𝑛𝑡𝑒 𝑐𝑖𝑡𝑎𝑑𝑎! 𝑉𝑜𝑙𝑡𝑎𝑛𝑑𝑜 à 𝑒𝑥𝑝𝑟𝑒𝑠𝑠ã𝑜: 𝑏 = 𝑒𝑥 1 + √2 = 𝑒𝑥 𝑥 = ln(1 + √2) ∗ 𝐷𝑒𝑠𝑠𝑎 𝑓𝑜𝑟𝑚𝑎, 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑛𝑑𝑜 𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑎 𝑐𝑢𝑟𝑣𝑎 𝑒𝑚 𝑥 = ln(1 + √2): 𝑦 = cosh(𝑙𝑛(1 + √2)) 𝑦 = 𝑒𝑙𝑛(1+√2) + 𝑒− 𝑙𝑛(1+√2) 2 𝑦 = 1 + √2 + 1 1 + √2 2 = 1 + 2√2 + 2 + 1 2 + 2√2 = 4 + 2√2 2 + 2√2 = 2 + √2 1 + √2 = √2. ∗ 𝐿𝑜𝑔𝑜, 𝑜 𝑝𝑜𝑛𝑡𝑜 é (ln(1 + √2) , √2 ). 37 3. 𝑎) 𝑓(𝑥) = 𝑎𝑥. 𝑒𝑏𝑥 2 𝑡𝑒𝑚 𝑢𝑚 𝑣𝑎𝑙𝑜𝑟 𝑚á𝑥𝑖𝑚𝑜 𝑒𝑚 (2, 1). 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟 𝑎 𝑒 𝑏. 𝑇𝑒𝑚𝑜𝑠, 𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑞𝑢𝑒 𝑓(2) = 1 𝑒 𝑓′(2) = 0. 𝐿𝑜𝑔𝑜, 𝑓(2) = 2𝑎. 𝑒4𝑏 = 1 𝑓′(𝑥) = 𝑎. 𝑒𝑏𝑥 2 + 2𝑎𝑏𝑥2. 𝑒𝑏𝑥 2 𝑓′(2) = 𝑎. 𝑒4𝑏 + 8𝑎𝑏. 𝑒4𝑏 = 0 ; 𝐶𝑜𝑚𝑜 𝑒4𝑏 ≠ 0 ∀𝑏 ∈ ℝ 𝑑𝑖𝑣𝑖𝑑𝑖𝑚𝑜𝑠 𝑎𝑚𝑏𝑜𝑠 𝑜𝑠 𝑡𝑒𝑟𝑚𝑜𝑠. { 2𝑎. 𝑒 4𝑏 = 1 𝑎 + 8𝑎𝑏 = 0 → 𝐷𝑎 𝑠𝑒𝑔𝑢𝑛𝑑𝑎 𝑒𝑥𝑝𝑟𝑒𝑠𝑠ã𝑜, 𝑡𝑒𝑚𝑜𝑠: 𝑎(1 + 8𝑏) = 0 𝑂𝑢 𝑎 = 0 𝑜𝑢 1 + 8𝑏 = 0 → 𝑏 = − 1 8 ; 𝑆𝑒 𝑎 = 0, 𝑒𝑛𝑡ã𝑜 𝑓(𝑥) = 0 . 𝑈𝑚 𝑎𝑏𝑠𝑢𝑟𝑑𝑜! 𝑃𝑜𝑖𝑠, 𝑓(2) = 1. 𝐿𝑜𝑔𝑜, 𝑏 = − 1 8 . 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑖𝑛𝑑𝑜 𝑛𝑎 𝑝𝑟𝑖𝑚𝑒𝑖𝑟𝑎 𝑒𝑥𝑝𝑟𝑒𝑠𝑠ã𝑜: 2𝑎. 𝑒−1 2⁄ = 1 2𝑎 √𝑒 = 1 → 2𝑎 = √𝑒 → 𝑎 = 1 2 √𝑒; 𝑓(𝑥) = 1 2 √𝑒. 𝑥. 𝑒−𝑥 2 8⁄ 𝑏) lim 𝑥→0 (1 + 𝑎𝑥)𝑏 𝑥⁄ . ∗ 𝐹𝑎ç𝑎𝑚𝑜𝑠 𝑥 = 1 𝑎𝑡 , 𝑒 𝑎𝑗𝑢𝑠𝑡𝑎𝑛𝑑𝑜 𝑎 𝑒𝑥𝑝𝑟𝑒𝑠𝑠ã𝑜 𝑑𝑜 𝑙𝑖𝑚𝑖𝑡𝑒, 𝑠𝑒 𝑥 → 0 𝑒𝑛𝑡ã𝑜 𝑡 → ∞. lim 𝑥→0 (1 + 𝑎𝑥)𝑏 𝑥⁄ = lim 𝑡→∞ (1 + 1 𝑡 ) 𝑎𝑏𝑡 = [lim 𝑡→∞ (1 + 1 𝑡 ) 𝑡 ] 𝑎𝑏 = 𝑒𝑎𝑏 ∗ 𝑂𝑏𝑠: 𝐿𝑖𝑚𝑖𝑡𝑒 𝐹𝑢𝑛𝑑𝑎𝑚𝑒𝑛𝑡𝑎𝑙 𝐸𝑥𝑝𝑜𝑛𝑒𝑛𝑐𝑖𝑎𝑙 lim 𝑛→∞ (1 + 1 𝑛 ) 𝑛 = 𝑒 4. 𝑃𝑒𝑟í𝑚𝑒𝑡𝑟𝑜 = 20 𝑢. 𝐶 𝑃 = 2(𝑏 + 𝑙) = 20 → 𝑏 + 𝑙 = 10 → 𝑙 = 10 − 𝑏 (𝐼) 𝑉 = 𝜋. 𝑏2. 𝑙 → 𝑉 = 𝜋. 𝑏2. (10 − 𝑏) 𝑉 = 𝜋(10𝑏2 − 𝑏3) 𝑉′(𝑏) = 𝜋(20𝑏 − 3𝑏2) 𝐹𝑎𝑧𝑒𝑛𝑑𝑜 𝑉′(𝑏) = 0 𝑜𝑏𝑡𝑒𝑚𝑜𝑠: 20𝑏 − 3𝑏2 = 0 → 𝑏 = 0 𝑜𝑢 𝑏 = 20 3 . 𝑁𝑜𝑡𝑒 𝑞𝑢𝑒 𝑏 = 0 𝑛ã𝑜 𝑠𝑒𝑟𝑣𝑒 𝑐𝑜𝑚𝑜 𝑠𝑜𝑙𝑢çã𝑜! 𝑙 = 10 − 𝑏 → 𝑙 = 10 − 20 3 = 10 3 . 𝐿𝑜𝑔𝑜, 𝑎𝑠 𝑑𝑖𝑚𝑒𝑛𝑠õ𝑒𝑠 𝑑𝑜 𝑐𝑖𝑙𝑖𝑛𝑑𝑟𝑜 𝑑𝑒 𝑚𝑎𝑖𝑜𝑟 𝑣𝑜𝑙𝑢𝑚𝑒 𝑠ã𝑜 𝑏 = 20 3 𝑒 𝑙 = 10 3 . 5. 𝐸𝑠𝑏𝑜ç𝑎𝑟 𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑒 𝑦 = 2𝑥2 9 − 𝑥2 . 𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑒𝑚𝑜𝑠 𝑓(𝑥) = 𝑦 = 2𝑥2 9 − 𝑥2 (𝐼)𝐷𝑜𝑚í𝑛𝑖𝑜: 𝐷(𝑓) = 𝑥 ∈ ℝ; 9 − 𝑥2 ≠ 0 𝐷(𝑓) = ℝ − {−3, 3}. 40 ∗ 𝐹𝑎𝑧𝑒𝑛𝑑𝑜 𝑜 𝑒𝑠𝑡𝑢𝑑𝑜 𝑑𝑜 𝑠𝑖𝑛𝑎𝑙 𝑑𝑒 𝑓′′(𝑥), 𝑡𝑒𝑚𝑜𝑠: +++++++++ ++++++++++ ++ 108(𝑥2 + 3) −−−−−(−3) + + + ++ +(3) − − −− − − (9 − 𝑥2)3 −−−−−(−3) + + + ++ +(3) − − −− − − 𝑓′′(𝑥) = 108(𝑥2+3) (9−𝑥2)3 𝐷𝑎 𝑎𝑛á𝑙𝑖𝑠𝑒 𝑑𝑜 𝑠𝑖𝑛𝑎𝑙 𝑑𝑒 𝑓′′(𝑥) 𝑐𝑜𝑛𝑐𝑙𝑢í𝑚𝑜𝑠 𝑞𝑢𝑒: 𝑓(𝑥) 𝑡𝑒𝑚 𝑐𝑜𝑛𝑐𝑎𝑣𝑖𝑑𝑎𝑑𝑒 𝑣𝑜𝑙𝑡𝑎𝑑𝑎 𝑝𝑎𝑟𝑎 𝑐𝑖𝑚𝑎 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 (−3, 3) 𝑒 𝑓(𝑥) 𝑡𝑒𝑚 𝑐𝑜𝑛𝑐𝑎𝑣𝑖𝑑𝑎𝑑𝑒 𝑣𝑜𝑙𝑡𝑎𝑑𝑎 𝑝𝑎𝑟𝑎 𝑏𝑎𝑖𝑥𝑜 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 (−∞,−3) ∪ (3,+∞) ∗ 𝑁ã𝑜 ℎá 𝑝𝑜𝑛𝑡𝑜𝑠 𝑑𝑒 𝑖𝑛𝑓𝑙𝑒𝑥ã𝑜 𝑒𝑚 𝑓(𝑥), 𝑝𝑜𝑖𝑠 ∄𝑥 ∈ ℝ; 𝑓′′(𝑥) = 0. 𝐿𝑜𝑔𝑜, 𝑜 𝑒𝑠𝑏𝑜ç𝑜 𝑑𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑒 𝑓(𝑥) = 𝑦 = 2𝑥² 9 − 𝑥² é: 41 Capítulo 3 2008 3.1 1ª Prova-14 de Março de 2008 1. 𝑓(𝑥) = 1 𝑥2 ; 𝑒𝑛𝑐𝑜𝑛𝑡𝑟𝑎𝑟 𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 𝑛𝑜 𝑝𝑜𝑛𝑡𝑜 (−2, 1 4⁄ ) ∗ 𝐴 𝑖𝑛𝑐𝑙𝑖𝑛𝑎çã𝑜 (𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒 𝑎𝑛𝑔𝑢𝑙𝑎𝑟) 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑠𝑒𝑟á 𝑑𝑎𝑑𝑎 𝑝𝑒𝑙𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑎 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎 𝑑𝑒 𝑓(𝑥) 𝑛𝑜 𝑝𝑜𝑛𝑡𝑜 𝑒𝑚 𝑞𝑢𝑒 𝑥 = −2. 𝑓′(𝑥) = lim ∆𝑥→0 𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥) ∆𝑥 = lim ∆𝑥→0 1 (𝑥 + ∆𝑥)2 − 1 𝑥2 ∆𝑥 = lim ∆𝑥→0 𝑥2 − (𝑥 + ∆𝑥)2 𝑥2(𝑥 + ∆𝑥)2 ∆𝑥 = lim ∆𝑥→0 𝑥2 − 𝑥2 − 2𝑥. ∆𝑥 − ∆𝑥2 ∆𝑥[𝑥2(𝑥 + ∆𝑥)2] = lim ∆𝑥→0 −2𝑥. ∆𝑥 − ∆𝑥2 ∆𝑥[𝑥2(𝑥 + ∆𝑥)2] = lim ∆𝑥→0 ∆𝑥(−2𝑥 − ∆𝑥) ∆𝑥[𝑥2(𝑥 + ∆𝑥)2] = lim ∆𝑥→0 −2𝑥 − ∆𝑥 𝑥2(𝑥 + ∆𝑥)2 = −2𝑥 𝑥4 . 𝑓′(𝑥) = −2𝑥 𝑥4 = −2 𝑥3 𝑓′(−2) = −2 (−2)3 = −2 −8 = 1 4 . 𝐷𝑎𝑑𝑜 𝑢𝑚 𝑝𝑜𝑛𝑡𝑜 𝑒 𝑜 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑎 𝑒𝑞𝑢𝑎çã𝑜 𝑑𝑎 𝑟𝑒𝑡𝑎 é 𝑑𝑎 𝑓𝑜𝑟𝑚𝑎: 𝑦 − 𝑦𝑜 = 𝑚(𝑥 − 𝑥𝑜) 𝑦 − 1 4 = 1 4 (𝑥 + 2) 𝑦 = 1 4 𝑥 + 3 4 2. 𝑎) lim 𝑥→0 𝑡𝑔(𝑥 − 𝜋) 𝑠𝑒𝑛𝑥 = lim 𝑥→0 𝑠𝑒𝑛(𝑥 − 𝜋) cos(𝑥 − 𝜋) 𝑠𝑒𝑛𝑥 = lim 𝑥→0 𝑠𝑒𝑛(𝑥 − 𝜋) 𝑠𝑒𝑛𝑥[cos(𝑥 − 𝜋)] = lim 𝑥→0 𝑠𝑒𝑛𝑥. cos 𝜋 − 𝑠𝑒𝑛𝜋. cos 𝑥 𝑠𝑒𝑛𝑥[cos(𝑥 − 𝜋)] = lim 𝑥→0 −𝑠𝑒𝑛𝑥 𝑠𝑒𝑛𝑥[cos(𝑥 − 𝜋)] = lim 𝑥→0 −1 [cos(𝑥 − 𝜋)] = −1 cos(−𝜋) = −1 −1 = 1. 𝑏) lim 𝑥→∞ (𝑥3 + 2𝑥)2 (2𝑥 − 1)3(𝑥3 − 1) = lim 𝑥→∞ 𝑥6 + 4𝑥4 + 4𝑥2 (8𝑥3 − 12𝑥2 + 6𝑥 − 1)(𝑥3 − 1) = lim 𝑥→∞ 𝑥6 + 4𝑥4 + 4𝑥2 8𝑥6 − 12𝑥5 + 6𝑥4 − 9𝑥3 + 12𝑥2 − 6𝑥 + 1 = lim 𝑥→∞ 𝑥6 + 4𝑥4 + 4𝑥2 𝑥6 8𝑥6 − 12𝑥5 + 6𝑥4 − 9𝑥3 + 12𝑥2 − 6𝑥 + 1 𝑥6 = lim 𝑥→∞ 1 + 4 𝑥2 + 4 𝑥4 8 − 12 𝑥 + 6 𝑥2 − 9 𝑥3 + 12 𝑥4 − 6 𝑥5 + 1 𝑥6 = 1 + 0 + 0 8 − 0 + 0 − 0 + 0 − 0 + 0 = 1 8 . 42 𝑐) lim 𝑥→ 𝜋 2 𝑡𝑔𝑥 − sec 𝑥 + cos 𝑥 cos 𝑥 = lim 𝑥→ 𝜋 2 𝑠𝑒𝑛𝑥 cos 𝑥 − 1 cos 𝑥 + cos 𝑥 cos 𝑥 = lim 𝑥→ 𝜋 2 𝑠𝑒𝑛𝑥 − 1 + cos2 𝑥 cos2 𝑥 = lim 𝑥→ 𝜋 2 𝑠𝑒𝑛𝑥 − 1 + 1 − 𝑠𝑒𝑛2𝑥 1 − 𝑠𝑒𝑛2𝑥 = lim 𝑥→ 𝜋 2 𝑠𝑒𝑛𝑥 − 𝑠𝑒𝑛2𝑥 1 − 𝑠𝑒𝑛2𝑥 = lim 𝑥→ 𝜋 2 𝑠𝑒𝑛𝑥(1 − 𝑠𝑒𝑛𝑥) (1 − 𝑠𝑒𝑛𝑥)(1 + 𝑠𝑒𝑛𝑥) = lim 𝑥→ 𝜋 2 𝑠𝑒𝑛𝑥 (1 + 𝑠𝑒𝑛𝑥) = 1 (1 + 1) = 1 2 . 3. 𝑆𝑒 𝑓 𝑒 𝑔 𝑠ã𝑜 𝑓𝑢𝑛çõ𝑒𝑠 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎𝑠 𝑛𝑢𝑚 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 𝑓𝑒𝑐ℎ𝑎𝑑𝑜 [𝑎, 𝑏] 𝑠ã𝑜 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑒𝑠 𝑎𝑠 𝑠𝑒𝑔𝑢𝑖𝑛𝑡𝑒𝑠 𝑝𝑟𝑜𝑝𝑜𝑠𝑖çõ𝑒𝑠: ∗ 𝑓 𝑒 𝑔 𝑠ã𝑜 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎𝑠 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 𝑎𝑏𝑒𝑟𝑡𝑜 (𝑎, 𝑏), 1) 𝑓(𝑎) 𝑒 𝑔(𝑎) 𝑒𝑥𝑖𝑠𝑡𝑒𝑚; 2) lim 𝑥→𝑎+ 𝑓(𝑥) 𝑒 lim 𝑥→𝑎+ 𝑔(𝑥) 𝑒𝑥𝑖𝑠𝑡𝑒𝑚; 3) lim𝑥→𝑎+ 𝑓(𝑥) = 𝑓(𝑎) 𝑒 lim𝑥→𝑎+ 𝑔(𝑥) = 𝑔(𝑎) 𝐷𝑒 𝑚𝑜𝑑𝑜 𝑠𝑖𝑚𝑖𝑙𝑎𝑟, 𝑡𝑒𝑚𝑜𝑠: 1) 𝑓(𝑏) 𝑒 𝑔(𝑏) 𝑒𝑥𝑖𝑠𝑡𝑒𝑚; 2) lim 𝑥→𝑏− 𝑓(𝑥) 𝑒 lim 𝑥→𝑏− 𝑔(𝑥) 𝑒𝑥𝑖𝑠𝑡𝑒𝑚; 3) lim 𝑥→𝑏− 𝑓(𝑥) = 𝑓(𝑏) 𝑒 lim 𝑥→𝑏− 𝑔(𝑥) = 𝑔(𝑏); ∗ 𝐹𝑎𝑧𝑒𝑛𝑑𝑜 𝑎𝑠 𝑖𝑛𝑡𝑒𝑟𝑟𝑒𝑙𝑎çõ𝑒𝑠 𝑐𝑜𝑚 𝑎 𝑞𝑢𝑒𝑠𝑡ã𝑜, 𝑡𝑒𝑚𝑜𝑠: 𝑓(−2) = 7 ; 𝑔(−2) = 2 ; 𝑓(5) = 4 𝑒 𝑔(5) = 5. 𝑎) lim 𝑥→5− 𝑔2(𝑥) − 25 𝑔(𝑥) − 5 = lim 𝑥→5− [𝑔(𝑥) − 5][𝑔(𝑥) + 5] 𝑔(𝑥) − 5 = lim 𝑥→5− [𝑔(𝑥) + 5] = lim 𝑥→5− 𝑔(𝑥) + lim 𝑥→5− 5 = 5 + 5 = 10. 𝑏) (𝑔 − 𝑓)(𝑥) = 𝑔(𝑥) − 𝑓(𝑥) (𝑔 − 𝑓)(−2) = 𝑔(−2) − 𝑓(−2) = 2 − 7 = −5. (𝑔 − 𝑓)(5) = 𝑔(5) − 𝑓(5) = 5 − 4 = 1. 𝑐) 𝑉𝑖𝑚𝑜𝑠 𝑛𝑜 𝐶á𝑙𝑐𝑢𝑙𝑜 1 𝑞𝑢𝑒 𝑎 𝑠𝑜𝑚𝑎 𝑑𝑒 𝑓𝑢𝑛çõ𝑒𝑠 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎𝑠 é 𝑢𝑚𝑎 𝑓𝑢𝑛çã𝑜 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒, 𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜, (𝑔 − 𝑓)(𝑥) é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 [−2, 5], 𝑒 𝑎𝑖𝑛𝑑𝑎, (𝑔 − 𝑓)(−2) < 0 < (𝑔 − 𝑓)(5). 𝐿𝑜𝑔𝑜, 𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑔𝑎𝑟𝑎𝑛𝑡𝑖𝑟 𝑝𝑒𝑙𝑜 𝑇𝑒𝑜𝑟𝑒𝑚𝑎 𝑑𝑜 𝑉𝑎𝑙𝑜𝑟 𝐼𝑛𝑡𝑒𝑟𝑚𝑒𝑑𝑖á𝑟𝑖𝑜 𝑞𝑢𝑒 𝑒𝑥𝑖𝑠𝑡𝑒 𝑎𝑙𝑔𝑢𝑚 𝑐 ∈ (−2, 5) 𝑡𝑎𝑙 𝑞𝑢𝑒 (𝑔 − 𝑓)(𝑐) = 0. 𝑂𝑢 𝑠𝑒𝑗𝑎, (𝑔 − 𝑓)(𝑐) = 0 → 𝑔(𝑐) − 𝑓(𝑐) = 0 → 𝑓(𝑐) = 𝑔(𝑐). 4. 𝑓(𝑥) = { − 3 2 𝑥 + 5, 𝑠𝑒 𝑥 ≤ 2, −𝑥2 + 𝑏𝑥 + (1 − 𝑏), 𝑠𝑒 𝑥 > 2 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟 𝑏 𝑝𝑎𝑟𝑎 𝑞𝑢𝑒 𝑓 𝑠𝑒𝑗𝑎 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 𝑥 = 2. ∗ 𝐵𝑎𝑠𝑡𝑎 𝑒𝑛𝑐𝑜𝑛𝑡𝑟𝑎𝑟𝑚𝑜𝑠 𝑏 𝑡𝑎𝑙 𝑞𝑢𝑒 𝑓(2) = lim 𝑥→2+ 𝑓(𝑥) , 𝑝𝑜𝑖𝑠 𝑜 𝑙𝑖𝑚𝑖𝑡𝑒 𝑙𝑎𝑡𝑒𝑟𝑎𝑙 à 45 𝑓(1) = lim 𝑥→1+ 𝑓(𝑥) , 𝑝𝑜𝑖𝑠 𝑜 𝑙𝑖𝑚𝑖𝑡𝑒 𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑒𝑠𝑞𝑢𝑒𝑟𝑑𝑜 𝑗á 𝑠𝑎𝑡𝑖𝑠𝑓𝑎𝑧 𝑎 𝑖𝑔𝑢𝑎𝑙𝑑𝑎𝑑𝑒 𝑓(1). 𝑓(1) = 3.12 = 3.1 = 3 lim 𝑥→1+ 𝑓(𝑥) = lim 𝑥→1+ [𝑎𝑥 + 𝑏] = 𝑎 + 𝑏 𝐿𝑜𝑔𝑜, 𝑡𝑒𝑚𝑜𝑠 𝑎 𝑟𝑒𝑙𝑎çã𝑜: 𝑎 + 𝑏 = 3 (𝐼) 𝑏) 𝐸𝑛𝑐𝑜𝑛𝑡𝑟𝑎𝑟 𝑎 𝑒 𝑏 𝑝𝑎𝑟𝑎 𝑞𝑢𝑒 𝑓 𝑠𝑒𝑗𝑎 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖á𝑣𝑒𝑙 𝑒𝑚 𝑥 = 1. ∗ 𝐽á 𝑡𝑒𝑚𝑜𝑠 𝑎 𝑟𝑒𝑙𝑎çã𝑜 𝑝𝑎𝑟𝑎 𝑎 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑖𝑑𝑎𝑑𝑒 𝑒𝑚 𝑥 = 1. 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑛𝑑𝑜 𝑎𝑠 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎𝑠 𝑙𝑎𝑡𝑒𝑟𝑎𝑖𝑠 𝑒𝑚 𝑥 = 1, 𝑡𝑒𝑚𝑜𝑠: 𝑓′ − (1) = lim 𝑥→1− 𝑓(𝑥) − 𝑓(1) 𝑥 − 1 = lim 𝑥→1− 3𝑥² − 3 𝑥 − 1 = lim 𝑥→1− 3(𝑥 − 1)(𝑥 + 1) 𝑥 − 1 = lim 𝑥→1− [3𝑥 + 3] = 6 𝑓′ + (1) = lim 𝑥→1+ 𝑓(𝑥) − 𝑓(1) 𝑥 − 1 = lim 𝑥→1+ 𝑎𝑥 + 𝑏 − 3 𝑥 − 1 = lim 𝑥→1+ 𝑎𝑥 + (3 − 𝑎) − 3 𝑥 − 1 = lim 𝑥→1+ 𝑎(𝑥 − 1) 𝑥 − 1 = lim 𝑥→1+ 𝑎 = 𝑎. 𝑃𝑎𝑟𝑎 𝑞𝑢𝑒 𝑓 𝑠𝑒𝑗𝑎 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖á𝑣𝑒𝑙 𝑒𝑚 𝑥 = 1 𝑎 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎𝑠 𝑙𝑎𝑡𝑒𝑟𝑎𝑖𝑠 𝑑𝑒𝑣𝑒𝑚 𝑠𝑒𝑟 𝑖𝑔𝑢𝑎𝑖𝑠. 𝐿𝑜𝑔𝑜, 𝑓′ − (1) = 𝑓′ + (1) ⇒ 𝑎 = 6. 𝐸, 𝑝𝑒𝑙𝑎 𝑒𝑞𝑢𝑎çã𝑜 (𝐼) 𝑡𝑒𝑚𝑜𝑠 𝑏 = −3. 3. 𝑎) 𝐴𝑐ℎ𝑎𝑟 𝑦′𝑜𝑛𝑑𝑒 𝑥² − 𝑥𝑦 + 3 2 𝑦2 = 15 2 𝑃𝑜𝑟 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎çã𝑜 𝑖𝑚𝑝𝑙í𝑐𝑖𝑡𝑎, 𝑡𝑒𝑚𝑜𝑠: 2𝑥 − 𝑦 − 𝑥𝑦′ + 3𝑦𝑦′ = 0 𝑦′(3𝑦 − 𝑥) = 𝑦 − 2𝑥 𝑦′ = 𝑦 − 2𝑥 3𝑦 − 𝑥 𝑏) 𝑂 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑛𝑜𝑟𝑚𝑎𝑙 𝑛𝑢𝑚 𝑝𝑜𝑛𝑡𝑜 é 𝑑𝑎𝑑𝑜 𝑝𝑒𝑙𝑜 𝑖𝑛𝑣𝑒𝑟𝑠𝑜 𝑠𝑖𝑚é𝑡𝑟𝑖𝑐𝑜 𝑑𝑜 𝑣𝑎𝑙𝑜𝑟 𝑎𝑠𝑠𝑢𝑚𝑖𝑑𝑜 𝑝𝑒𝑙𝑎 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎 𝑛𝑎𝑞𝑢𝑒𝑙𝑒 𝑝𝑜𝑛𝑡𝑜. 𝑂𝑢 𝑠𝑒𝑗𝑎, 𝑚𝑛 = − 1 𝑦′ = 𝑥 − 3𝑦 𝑦 − 2𝑥 ; 𝑛𝑜 𝑝𝑜𝑛𝑡𝑜 (−2, 1) 𝑡𝑒𝑚𝑜𝑠: 𝑚𝑛 = −2 − 3 1 + 4 = − 5 5 = −1. 𝐿𝑜𝑔𝑜, 𝑎 𝑒𝑞𝑢𝑎çã𝑜 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑛𝑜𝑟𝑚𝑎𝑙 𝑛𝑜 𝑝𝑜𝑛𝑡𝑜 (−2, 1) é: 𝑦 − 1 = −1(𝑥 + 2) 𝑦 − 1 = −𝑥 − 2 𝑦 = −𝑥 − 1 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑖𝑛𝑑𝑜 𝑦 = −𝑥 − 1 𝑛𝑎 𝑒𝑥𝑝𝑟𝑒𝑠𝑠ã𝑜 𝑑𝑎 𝑐𝑢𝑟𝑣𝑎 𝑒𝑛𝑐𝑜𝑛𝑡𝑟𝑎𝑟𝑒𝑚𝑜𝑠 𝑜 𝑝𝑜𝑛𝑡𝑜 𝑜𝑛𝑑𝑒 𝑎 𝑟𝑒𝑡𝑎 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑎 𝑎 𝑐𝑢𝑟𝑣𝑎 𝑢𝑚𝑎 𝑠𝑒𝑔𝑢𝑛𝑑𝑎 𝑣𝑒𝑧. 𝑥2 − 𝑥(−𝑥 − 1) + 3 2 (−𝑥 − 1)2 = 15 2 𝑥2 + 𝑥2 + 𝑥 + 3 2 (𝑥2 + 2𝑥 + 1) = 15 2 4𝑥2 + 2𝑥 + 3𝑥2 + 6𝑥 + 3 = 15 7𝑥2 + 8𝑥 − 12 = 0 ∆= 64 + 336 = 400 𝑥 = −8 ± 20 14 → 𝑥 = 12 14 𝑒 𝑥 = −2 .𝑀𝑎𝑠 𝑥 = −2 𝑗á 𝑓𝑜𝑖 𝑑𝑎𝑑𝑜 𝑛𝑜 𝑝𝑜𝑛𝑡𝑜 (−2,1) 𝑈𝑠𝑎𝑟𝑒𝑚𝑜𝑠, 𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑥 = 12 14 = 6 7 . 46 𝑦 = −𝑥 − 1 𝑦 = − 6 7 − 1 = − 13 7 . 𝐿𝑜𝑔𝑜, 𝑎 𝑟𝑒𝑡𝑎 𝑛𝑜𝑟𝑚𝑎𝑙 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑎 𝑎 𝑐𝑢𝑟𝑣𝑎, 𝑢𝑚𝑎 𝑠𝑒𝑔𝑢𝑛𝑑𝑎 𝑣𝑒𝑧, 𝑛𝑜 𝑝𝑜𝑛𝑡𝑜 ( 6 7 ,− 13 7 ). 4. 𝑎) 𝑓(𝑥) = 10𝑥 2+𝑥+1 . 𝐸𝑞𝑢𝑎çã𝑜 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 𝑛𝑜 𝑝𝑜𝑛𝑡𝑜 (0, 10). 𝑓′(𝑥) = 10𝑥 2+𝑥+1. (2𝑥 + 1). ln(10) 𝑓′(0) = 10. ln 10 𝐸𝑞𝑢𝑎çã𝑜 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 𝑛𝑜 𝑝𝑜𝑛𝑡𝑜 (0, 10): 𝑦 − 10 = 10. ln 10 (𝑥 − 0) 𝑦 = (10. ln 10)𝑥 + 10 𝑏) 𝐻(𝑥) = 𝑒[𝑓(𝑥)] 2 ; 𝑓(1) = 𝑓′(1) = 1. 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟 𝑎 𝑒𝑞𝑢𝑎çã𝑜 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 𝑎𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑒 𝐻(𝑥) 𝑛𝑜 𝑝𝑜𝑛𝑡𝑜 𝑒𝑚 𝑞𝑢𝑒 𝑥 = 1. 𝐻(1) = 𝑒[𝑓(1)] 2 = 𝑒1 2 = 𝑒. 𝑝𝑜𝑛𝑡𝑜 𝑑𝑒 𝑡𝑎𝑛𝑔𝑒𝑛𝑐𝑖𝑎 (1, 𝑒). 𝐻′(𝑥) = 2𝑓(𝑥). 𝑓′(𝑥). 𝑒[𝑓(𝑥)] 2 𝐻′(1) = 2𝑓(1). 𝑓′(1). 𝑒[𝑓(1)] 2 𝐻′(1) = 2.1.1. 𝑒 = 2𝑒 𝐸𝑞𝑢𝑎çã𝑜 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 𝑛𝑜 𝑝𝑜𝑛𝑡𝑜 (1, 𝑒) ∶ 𝑦 − 𝑒 = 2𝑒(𝑥 − 1) 𝑦 − 𝑒 = 2𝑒𝑥 − 2𝑒 𝑦 = 2𝑒𝑥 − 𝑒 5. 𝑎) 𝑓(𝑥) = 𝑐𝑜𝑠𝑠𝑒𝑐𝑥 + 𝑐𝑜𝑡𝑔𝑥 ; 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟 𝑎 𝑒𝑞𝑢𝑎çã𝑜 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 𝑒𝑚 𝑥 = 𝜋 4 . 𝑓 ( 𝜋 4 ) = 𝑐𝑜𝑠𝑠𝑒𝑐 𝜋 4 + 𝑐𝑜𝑡𝑔 𝜋 4 = √2 + 1. 𝑝𝑜𝑛𝑡𝑜 𝑑𝑒 𝑡𝑎𝑛𝑔𝑒𝑛𝑐𝑖𝑎: ( 𝜋 4 , √2 + 1) 𝑓′(𝑥) = −𝑐𝑜𝑠𝑠𝑒𝑐𝑥. 𝑐𝑜𝑡𝑔𝑥 − 𝑐𝑜𝑠𝑠𝑒𝑐2𝑥 𝑓′ ( 𝜋 4 ) = −√2 − 2 𝐸𝑞𝑢𝑎çã𝑜 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 𝑛𝑜 𝑝𝑜𝑛𝑡𝑜 ( 𝜋 4 , √2 + 1): 𝑦 − (√2 + 1) = −(√2 + 2) (𝑥 − 𝜋 4 ) 𝑦 = (−√2 − 2)𝑥 + (√2 + 2) 4 𝜋 + √2 + 1 𝑦 = (−√2 − 2)𝑥 + √2(𝜋 + 4) + 2(𝜋 + 2) 4 𝑏) 𝑔(𝑥) = 𝑠𝑒𝑛²(cos(𝑥2 + 1)) . 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟 𝑎 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎 𝑑𝑒 𝑔(𝑥). 𝑆𝑒𝑗𝑎 𝑢 = 𝑥2 + 1 ; 𝑣 = cos 𝑢 ; 𝑦 = 𝑔(𝑣) = 𝑠𝑒𝑛2𝑣 𝑃𝑒𝑙𝑎 𝑅𝑒𝑔𝑟𝑎 𝑑𝑎 𝐶𝑎𝑑𝑒𝑖𝑎, 𝑡𝑒𝑚𝑜𝑠: 𝑑𝑦 𝑑𝑥 = 𝑑𝑢 𝑑𝑥 . 𝑑𝑣 𝑑𝑢 . 𝑑𝑦 𝑑𝑣 𝑦′ = 𝑓′(𝑥) = (2𝑥). (−𝑠𝑒𝑛(𝑢)). 2𝑠𝑒𝑛𝑣. cos 𝑣 𝑓′(𝑥) = −2𝑥. 𝑠𝑒𝑛(𝑥2 + 1). 𝑠𝑒𝑛(2𝑣) 𝑓′(𝑥) = −2𝑥. 𝑠𝑒𝑛(𝑥2 + 1). 𝑠𝑒𝑛(2cos(𝑥2 + 1)) 47 3.3 3ª Avaliação-17 de Maio de 2008 1. 𝑎) 𝑓(𝑥) = 𝑥ln 𝑥 ; 𝑃𝑜𝑟 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎çã𝑜 𝑙𝑜𝑔𝑎𝑟í𝑡𝑚𝑖𝑐𝑎, 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟 𝑓′(𝑥). ln 𝑓(𝑥) = ln 𝑥ln 𝑥 ln 𝑓(𝑥) = (ln 𝑥)2 𝑓′(𝑥) 𝑓(𝑥) = 2. ln 𝑥 . 1 𝑥 𝑓′(𝑥) = 2𝑥ln 𝑥. ln 𝑥 . 𝑥−1 𝑓′(𝑥) = 2𝑥ln 𝑥−1. ln 𝑥 𝑏)𝐸𝑛𝑐𝑜𝑛𝑡𝑟𝑎𝑟 𝑜 𝑝𝑜𝑛𝑡𝑜 𝑜𝑛𝑑𝑒 𝑎 𝑐𝑢𝑟𝑣𝑎 𝑦 = cosh 𝑥 𝑝𝑜𝑠𝑠𝑢𝑖 𝑎 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎 𝑖𝑔𝑢𝑎𝑙 𝑎 1. ∗ 𝐸𝑥𝑝𝑙𝑖𝑐𝑎çã𝑜: 𝑜 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑑𝑎 𝑟𝑒𝑡𝑎 é 𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 𝑑𝑜 â𝑛𝑔𝑢𝑙𝑜 𝑓𝑜𝑟𝑚𝑎𝑑𝑜 𝑒𝑛𝑡𝑟𝑒 𝑎 𝑟𝑒𝑡𝑎 𝑒 𝑜 𝑠𝑒𝑛𝑡𝑖𝑑𝑜 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑜 𝑑𝑜 𝑒𝑖𝑥𝑜 𝑑𝑜𝑠 𝑥, 𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑡𝑔 𝜋 4 = 1. 𝐿𝑒𝑚𝑏𝑟𝑎𝑛𝑑𝑜 𝑎 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎 é 𝑛𝑢𝑚𝑒𝑟𝑖𝑐𝑎𝑚𝑒𝑛𝑡𝑒 𝑖𝑔𝑢𝑎𝑙 à 𝑖𝑛𝑐𝑙𝑖𝑛𝑎çã𝑜 𝑑𝑎 𝑟𝑒𝑡𝑎, 𝑜𝑢 𝑠𝑒𝑗𝑎, 𝑖𝑔𝑢𝑎𝑙 𝑎 1. 𝑦′ = 𝑠𝑒𝑛ℎ(𝑥) 𝑦′ = 1 → 𝑠𝑒𝑛ℎ(𝑥) = 1 (𝐼) 𝑠𝑒𝑛ℎ(𝑥) = 𝑒𝑥 − 𝑒−𝑥 2 ; 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑖𝑛𝑑𝑜 𝑛𝑎 𝑒𝑞𝑢𝑎çã𝑜 (𝐼): 𝑒𝑥 − 𝑒−𝑥 2 = 1 𝑒𝑥 − 𝑒−𝑥 = 2 𝑒𝑥 − 1 𝑒𝑥 − 2 = 0 𝑒2𝑥 − 2𝑒𝑥 − 1 = 0 𝐹𝑎𝑧𝑒𝑛𝑑𝑜 𝑎 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑖çã𝑜 𝑏 = 𝑒𝑥 𝑜𝑏𝑡𝑒𝑚𝑜𝑠 𝑎 𝑠𝑒𝑔𝑢𝑖𝑛𝑡𝑒 𝑒𝑥𝑝𝑟𝑒𝑠𝑠ã𝑜 𝑑𝑜 𝑠𝑒𝑔𝑢𝑛𝑑𝑜 𝑔𝑟𝑎𝑢: 𝑏2 − 2𝑏 − 1 = 0 ∗ 𝑂𝑏𝑠: 𝑠𝑒 𝑏 = 𝑒𝑥, 𝑒𝑛𝑡ã𝑜 𝑏 > 0 ∀𝑥 ∈ ℝ. 𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑛𝑑𝑜 𝑎 𝑒𝑞𝑢𝑎çã𝑜 𝑎𝑐𝑖𝑚𝑎, 𝑡𝑒𝑚𝑜𝑠: ∆= 4 + 4 = 8 𝑏 = 2 ± 2√2 2 → 𝑏 = 1 + √2 𝑜𝑢 𝑏 = 1 − √2 (𝑏 < 0) 𝑂 𝑡𝑒𝑟𝑚𝑜 𝑒𝑚 𝑑𝑒𝑠𝑡𝑎𝑞𝑢𝑒 𝑛ã𝑜 𝑠𝑎𝑡𝑖𝑠𝑓𝑎𝑧 𝑎 𝑐𝑜𝑛𝑑𝑖çã𝑜 𝑎𝑛𝑡𝑒𝑟𝑖𝑜𝑟𝑚𝑒𝑛𝑡𝑒 𝑐𝑖𝑡𝑎𝑑𝑎! 𝑉𝑜𝑙𝑡𝑎𝑛𝑑𝑜 à 𝑒𝑥𝑝𝑟𝑒𝑠𝑠ã𝑜: 𝑏 = 𝑒𝑥 1 + √2 = 𝑒𝑥 𝑥 = ln(1 + √2) ∗ 𝐷𝑒𝑠𝑠𝑎 𝑓𝑜𝑟𝑚𝑎, 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑛𝑑𝑜 𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑎 𝑐𝑢𝑟𝑣𝑎 𝑒𝑚 𝑥 = ln(1 + √2): 𝑦 = cosh(𝑙𝑛(1 + √2)) 𝑦 = 𝑒𝑙𝑛(1+√2) + 𝑒− 𝑙𝑛(1+√2) 2 𝑦 = 1 + √2 + 1 1 + √2 2 = 1 + 2√2 + 2 + 1 2 + 2√2 = 4 + 2√2 2 + 2√2 = 2 + √2 1 + √2 = √2. ∗ 𝐿𝑜𝑔𝑜, 𝑜 𝑝𝑜𝑛𝑡𝑜 é (ln(1 + √2) , √2 ). 50 𝛥 = 𝑐2 − 4(𝑐2 − 12) → 𝛥 = −3𝑐2 + 48. (1) 𝑆𝑒 𝛥 = 0 ⇒ 𝑐 = ±4 ⇒ 𝑎 = ±16 ; 𝐷𝑒𝑓𝑖𝑛𝑖𝑚𝑜𝑠 𝑎𝑛𝑡𝑒𝑟𝑖𝑜𝑟𝑚𝑒𝑛𝑡𝑒 𝑞𝑢𝑒 − 11 < 𝑎 < 11. 𝐿𝑜𝑔𝑜, 𝑛ã𝑜 ℎá 𝑜𝑢𝑡𝑟𝑜 𝑝𝑜𝑛𝑡𝑜 𝑐𝑟í𝑡𝑖𝑐𝑜 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 (−1,1) 𝑝𝑎𝑟𝑎 𝛥 = 0. (2) 𝑆𝑒 𝛥 > 0 , 𝑐 > 4 𝑜𝑢 𝑐 < −4 𝑒, 𝑐𝑜𝑛𝑠𝑒𝑞𝑢𝑒𝑛𝑡𝑒𝑚𝑒𝑛𝑡𝑒, 𝑎 > 16 𝑜𝑢 𝑎 < −16. 𝐸𝑠𝑠𝑎 𝑠𝑖𝑡𝑢𝑎çã𝑜 é 𝑎𝑛á𝑙𝑜𝑔𝑎 à 𝑎𝑛𝑡𝑒𝑟𝑖𝑜𝑟. (3) 𝑆𝑒 𝛥 < 0, 𝑒𝑛𝑡ã𝑜 𝑐 é 𝑎 ú𝑛𝑖𝑐𝑎 𝑟𝑎𝑖𝑧 𝑟𝑒𝑎𝑙 𝑑𝑒 𝑔(𝑥) 𝑑𝑒 𝑡𝑎𝑙 𝑚𝑜𝑑𝑜 𝑞𝑢𝑒 𝑓′(𝑥) = 0 𝑠𝑒, 𝑠𝑜𝑚𝑒𝑛𝑡𝑒 𝑠𝑒, 𝑥 = 𝑐. ∗ 𝐶𝑜𝑚 𝑒𝑠𝑠𝑎𝑠 𝑎𝑛á𝑙𝑖𝑠𝑒𝑠 𝑐𝑜𝑛𝑐𝑙𝑢í𝑚𝑜𝑠 𝑞𝑢𝑒 𝑓 𝑝𝑜𝑠𝑠𝑢𝑖, 𝑛𝑜 𝑚á𝑥𝑖𝑚𝑜, 𝑢𝑚 𝑝𝑜𝑛𝑡𝑜 𝑐𝑟í𝑡𝑖𝑐𝑜 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 (−1, 1) 𝑝𝑎𝑟𝑎 𝑞𝑢𝑎𝑙𝑞𝑢𝑒𝑟 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑎, 𝑡𝑎𝑙 𝑞𝑢𝑒 − 11 < 𝑎 < 11. 5. 𝑎) 𝑓(𝑥) = − 𝑥 (𝑥 − 2)2 ; (1)𝑂 𝑑𝑜𝑚í𝑛𝑖𝑜 𝑑𝑒 𝑓; ∗ 𝐷(𝑓) = ℝ − {2} (2)𝐼𝑛𝑡𝑒𝑟𝑠𝑒çõ𝑒𝑠 𝑐𝑜𝑚 𝑜𝑠 𝑒𝑖𝑥𝑜𝑠; 𝑓(0) = 0 𝑓(𝑥) = 0 ⇔ 𝑥 = 0. ∗ 𝐼𝑛𝑡𝑒𝑟𝑠𝑒çã𝑜 (0, 0). (3)𝐴𝑠𝑠í𝑛𝑡𝑜𝑡𝑎𝑠; −𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑖𝑠: 𝐷𝑖𝑧𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑎 𝑟𝑒𝑡𝑎 𝑥 = 𝑎 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑠𝑒 𝑜𝑐𝑜𝑟𝑟𝑒𝑟 𝑢𝑚 𝑑𝑜𝑠 𝑠𝑒𝑔𝑢𝑖𝑛𝑡𝑒𝑠 𝑐𝑎𝑠𝑜𝑠: lim 𝑥→𝑎+ 𝑓(𝑥) = ±∞ 𝑜𝑢 lim 𝑥→𝑎− 𝑓(𝑥) = ±∞ ∗ 𝑉𝑒𝑟𝑖𝑓𝑖𝑐𝑎𝑛𝑑𝑜 𝑠𝑒 𝑎 𝑟𝑒𝑡𝑎 𝑥 = 2 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙: lim 𝑥→2+ 𝑓(𝑥) = lim 𝑥→2+ − 𝑥 (𝑥 − 2)2 = lim 𝑥→2+ −𝑥⏞ −2 ↑ (𝑥 − 2)2⏟ ↓ 0+ = −∞ ∗ 𝑂𝑏𝑠: 𝑠𝑒 𝑥 → 2+, 𝑒𝑛𝑡ã𝑜 𝑥 > 2 𝑒, 𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑥 − 2 > 0 ⇒ 𝑥 − 2 → 0+. 𝐶𝑜𝑚𝑜 𝑜 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑑𝑜𝑟 𝑡𝑒𝑛𝑑𝑒 𝑎 𝑧𝑒𝑟𝑜 𝑝𝑜𝑟 𝑣𝑎𝑙𝑜𝑟𝑒𝑠 𝑚𝑎𝑖𝑜𝑟𝑒𝑠 𝑞𝑢𝑒 𝑧𝑒𝑟𝑜 𝑒 𝑜 𝑛𝑢𝑚𝑒𝑟𝑎𝑑𝑜𝑟 𝑡𝑒𝑛𝑑𝑒 𝑎 𝑢𝑚𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑒 𝑘 < 0, 𝑜 𝑙𝑖𝑚𝑖𝑡𝑒 𝑠𝑒 𝑎𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑎 𝑛𝑎 𝑓𝑜𝑟𝑚𝑎 𝑘 0+ 𝑐𝑜𝑚 𝑘 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑜. ∗ 𝐿𝑜𝑔𝑜, 𝑎 𝑟𝑒𝑡𝑎 𝑥 = 2 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑎𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑒 𝑓(𝑥). −𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑖𝑠: 𝐷𝑖𝑧𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑎 𝑟𝑒𝑡𝑎 𝑦 = 𝑎 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑠𝑒 𝑜𝑐𝑜𝑟𝑟𝑒𝑟 𝑢𝑚 𝑑𝑜𝑠 𝑠𝑒𝑔𝑢𝑖𝑛𝑡𝑒𝑠 𝑐𝑎𝑠𝑜𝑠: lim 𝑥→+∞ 𝑓(𝑥) = 𝑎 𝑜𝑢 lim 𝑥→−∞ 𝑓(𝑥) = 𝑎 lim 𝑥→+∞ 𝑓(𝑥) = lim 𝑥→+∞ − 𝑥 (𝑥 − 2)2 = lim 𝑥→+∞ − 𝑥 𝑥2 − 4𝑥 + 4 = lim 𝑥→+∞ − 𝑥 𝑥2 − 4𝑥 + 4 = lim 𝑥→+∞ − 𝑥 𝑥2 𝑥2 𝑥2 − 4𝑥 𝑥2 + 4 𝑥2 = lim 𝑥→+∞ − 1 𝑥 ⏞ 0 ↑ 1 − 4 𝑥⏟ ↓ 0 + 4 𝑥2⏟ ↓ 0 = 0 1 − 0 + 0 = 0 1 = 0. 51 ∗ 𝐿𝑜𝑔𝑜, 𝑎 𝑟𝑒𝑡𝑎 𝑦 = 0 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑎𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑒 𝑓(𝑥). (4)𝐶𝑟𝑒𝑠𝑐𝑖𝑚𝑒𝑛𝑡𝑜 𝑒 𝐷𝑒𝑐𝑟𝑒𝑠𝑐𝑖𝑚𝑒𝑛𝑡𝑜; 𝑓′(𝑥) = −(𝑥 − 2)2 + 𝑥. 2(𝑥 − 2) (𝑥 − 2)4 𝑓′(𝑥) = −𝑥 + 2 + 2𝑥 (𝑥 − 2)3 𝑓′(𝑥) = 𝑥 + 2 (𝑥 − 2)3 ∗ 𝐴𝑛𝑎𝑙𝑖𝑠𝑎𝑛𝑑𝑜 𝑜 𝑐𝑜𝑚𝑝𝑜𝑟𝑡𝑎𝑚𝑒𝑛𝑡𝑜 (𝑠𝑖𝑛𝑎𝑙) 𝑑𝑒 𝑓′(𝑥): −−−−−− (−2) + + + + + ++++++ +++ (𝑥 + 2) −−−−−−−−− −−−−−− (2) + + + + + + (𝑥 − 2)3 ++++++ (−2) − − − − − −− (2) + + + + + 𝑓′(𝑥) = (𝑥 + 2) (𝑥 − 2)³⁄ ∗ 𝐷𝑎 𝑎𝑛á𝑙𝑖𝑠𝑒 𝑎𝑐𝑖𝑚𝑎, 𝑐𝑜𝑛𝑐𝑙𝑢í𝑚𝑜𝑠 𝑞𝑢𝑒: 𝑓(𝑥) é 𝑐𝑟𝑒𝑠𝑐𝑒𝑛𝑡𝑒 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 (−∞,−2) ∪ (2,+∞) 𝑒 𝑓(𝑥) é 𝑑𝑒𝑐𝑟𝑒𝑠𝑐𝑒𝑛𝑡𝑒 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 (−2, 2) (5)𝐶𝑜𝑛𝑐𝑎𝑣𝑖𝑑𝑎𝑑𝑒𝑠 𝑒 𝑃𝑜𝑛𝑡𝑜𝑠 𝑑𝑒 𝐼𝑛𝑓𝑙𝑒𝑥ã𝑜; 𝑓′′(𝑥) = (𝑥 − 2)3 − 3(𝑥 + 2)(𝑥 − 2)2 (𝑥 − 2)6 𝑓′′(𝑥) = 𝑥 − 2 − 3𝑥 − 6 (𝑥 − 2)4 𝑓′′(𝑥) = −2𝑥 − 8 (𝑥 − 2)4 ∗ 𝐴𝑛𝑎𝑙𝑖𝑠𝑎𝑛𝑑𝑜 𝑜 𝑐𝑜𝑚𝑝𝑜𝑟𝑡𝑎𝑚𝑒𝑛𝑡𝑜 𝑑𝑒 𝑓′′(𝑥): ++++++ (−4) − − − − − −−−−−− −− (−2𝑥 − 8) +++++++++ +++++ (2) + ++ + + + (𝑥 − 2)4 ++++++ (−4) − − − − − −(2) − − − − −− 𝑓′′(𝑥) = (−2𝑥 − 8) (𝑥 − 2)4⁄ ∗ 𝐷𝑎 𝑎𝑛á𝑙𝑖𝑠𝑒 𝑎𝑐𝑖𝑚𝑎, 𝑐𝑜𝑛𝑐𝑙𝑢í𝑚𝑜𝑠 𝑞𝑢𝑒: 𝑓(𝑥) 𝑝𝑜𝑠𝑠𝑢𝑖 𝑐𝑜𝑛𝑐𝑎𝑣𝑖𝑑𝑎𝑑𝑒 𝑣𝑜𝑙𝑡𝑎𝑑𝑎 𝑝𝑎𝑟𝑎 𝑐𝑖𝑚𝑎 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 (−∞,−4) 𝑒 𝑓(𝑥) 𝑝𝑜𝑠𝑠𝑢𝑖 𝑐𝑜𝑛𝑐𝑎𝑣𝑖𝑑𝑎𝑑𝑒 𝑣𝑜𝑙𝑡𝑎𝑑𝑎 𝑝𝑎𝑟𝑎 𝑏𝑎𝑖𝑥𝑜 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 (−4,2) ∪ (2,+∞). 𝑂𝑠 𝑝𝑜𝑛𝑡𝑜𝑠 𝑑𝑒 𝑖𝑛𝑓𝑙𝑒𝑥ã𝑜 𝑜𝑐𝑜𝑟𝑟𝑒𝑚 𝑞𝑢𝑎𝑛𝑑𝑜 ℎá 𝑚𝑢𝑑𝑎𝑛ç𝑎 𝑛𝑎 𝑑𝑖𝑟𝑒çã𝑜 𝑑𝑎 𝑐𝑜𝑛𝑐𝑎𝑣𝑖𝑑𝑎𝑑𝑒, 𝑒𝑚 𝑢𝑚 𝑛ú𝑚𝑒𝑟𝑜 𝑛𝑜 𝑑𝑜𝑚í𝑛𝑖𝑜 𝑑𝑎 𝑓𝑢𝑛ç𝑎𝑜. ∗ 𝐴𝑛𝑎𝑙𝑖𝑠𝑎𝑛𝑑𝑜 𝑎 𝑠𝑒𝑔𝑢𝑛𝑑𝑎 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎 𝑐𝑜𝑛𝑐𝑙𝑢í𝑚𝑜𝑠 𝑞𝑢𝑒 (−4, 1 9 ) é 𝑝𝑜𝑛𝑡𝑜 𝑑𝑒 𝑖𝑛𝑓𝑙𝑒𝑥ã𝑜! b) Esboço Gráfico! 52 55 𝐴 = 1 2 (2 − 1). (𝑓(2) − 𝑓(1)) = 1 2 . (1). (0 − (−1)) = 1 2 𝑢. 𝐴 Á𝑟𝑒𝑎 𝑒𝑛𝑡𝑟𝑒 𝑥 = 2 𝑒 𝑥 = 5: 𝐴 = 1 2 (5 − 2)(𝑓(5) − 𝑓(2)) = 1 2 . (3). (3 − 0) = 9 2 𝑢. 𝐴 ∗ 𝐿𝑜𝑔𝑜, 𝑎 á𝑟𝑒𝑎 𝑒𝑛𝑡𝑟𝑒 𝑥 = 1 𝑒 𝑥 = 5 é: 𝐴1→5 = 1 2 + 9 2 = 10 2 = 5 𝑢. 𝐴 𝑏) ∫ (𝑥 − 2) 𝑑𝑥 5 1 = [ 1 2 𝑥2 − 2𝑥| 1 5 ] = 1 2 (5)2 − 2. (5) − 1 2 (1)2 + 2. (1) = 25 2 − 10 − 1 2 + 2 = 24 2 − 8 = 12 − 8 = 4. ∗ 𝑂𝑏𝑠: 𝑁𝑜𝑡𝑒 𝑞𝑢𝑒, 𝑛𝑒𝑚 𝑠𝑒𝑚𝑝𝑟𝑒 𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑎 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑒 𝑎𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑎 á𝑟𝑒𝑎! 𝑐) ∫ (𝑥 − 2) 𝑑𝑥 2 0 ; 𝑈𝑠𝑎𝑛𝑑𝑜 𝑠𝑜𝑚𝑎𝑠 𝑑𝑒 𝑅𝑖𝑒𝑚𝑎𝑛𝑛 ∆𝑥 = 2 − 0 𝑛 = 2 𝑛 ; 𝑥𝑖 = 2𝑖 𝑛 ; 𝑒 𝑓(𝑥) = 𝑥 − 2 ∫ (𝑥 − 2) 𝑑𝑥 2 0 = lim 𝑛→∞ ∑𝑓(𝑥𝑖)∆𝑥 𝑛 𝑖=1 = lim 𝑛→∞ ∆𝑥∑𝑓(𝑥𝑖) 𝑛 𝑖=1 = lim 𝑛→∞ 2 𝑛 ∑( 2𝑖 𝑛 − 2) 𝑛 𝑖=1 = lim 𝑛→∞ 2 𝑛 [∑ 2𝑖 𝑛 𝑛 𝑖=1 −∑2 𝑛 𝑖=1 ] = lim 𝑛→∞ 2 𝑛 [ 2 𝑛 ∑𝑖 𝑛 𝑖=1 − 2𝑛] = lim 𝑛→∞ 2 𝑛 [ 2 𝑛 . 𝑛(𝑛 + 1) 2 − 2𝑛] = lim 𝑛→∞ ( 2(𝑛 + 1) 𝑛 − 4) = lim 𝑛→∞ ( 2𝑛 𝑛 + 2 𝑛 − 4) = lim 𝑛→∞ ( 2 𝑛 − 2) = −2. 5. 𝑓(𝑥) = 𝑥3 1 − 𝑥2 ; 𝑓′(𝑥) = 𝑥2(3 − 𝑥2) (1 − 𝑥2)2 ; 56 𝑓′′(𝑥) = 2𝑥(𝑥2 + 3) (1 − 𝑥2)3 (𝐼)𝐷𝑜𝑚í𝑛𝑖𝑜 𝑑𝑒 𝑓: ∗ 𝐷(𝑓) = ℝ − {−1, 1}; (𝐼𝐼) 𝐼𝑛𝑡𝑒𝑟𝑠𝑒çõ𝑒𝑠 𝑐𝑜𝑚 𝑜𝑠 𝑒𝑖𝑥𝑜𝑠 𝑐𝑜𝑜𝑟𝑑𝑒𝑛𝑎𝑑𝑜𝑠: 𝑓(0) = 0 ; 𝑓(𝑥) = 0 ⇒ 𝑥3 = 0 ⇒ 𝑥 = 0. 𝑖𝑛𝑡𝑒𝑟𝑠𝑒çã𝑜 (0, 0). (𝐼𝐼𝐼) 𝐴𝑠𝑠í𝑛𝑡𝑜𝑡𝑎𝑠: −𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑖𝑠: 𝐷𝑖𝑧𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑎 𝑟𝑒𝑡𝑎 𝑥 = 𝑎 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑠𝑒 𝑜𝑐𝑜𝑟𝑟𝑒𝑟 𝑢𝑚 𝑑𝑜𝑠 𝑠𝑒𝑔𝑢𝑖𝑛𝑡𝑒𝑠 𝑐𝑎𝑠𝑜𝑠: lim 𝑥→𝑎+ 𝑓(𝑥) = ±∞ 𝑜𝑢 lim 𝑥→𝑎− 𝑓(𝑥) = ±∞ ∗ 𝑉𝑒𝑟𝑖𝑓𝑖𝑐𝑎𝑛𝑑𝑜 𝑠𝑒 𝑎𝑠 𝑟𝑒𝑡𝑎 𝑥 = −1 𝑒 𝑥 = 1 𝑠ã𝑜 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎𝑠: lim 𝑥→−1+ 𝑓(𝑥) = lim 𝑥→−1+ 𝑥3 1 − 𝑥2 = lim 𝑥→−1+ 𝑥3⏞ −1 ↑ (1 + 𝑥)⏟ ↓ 0+ (1 − 𝑥)⏟ ↓ 2 = −∞ ∗ 𝐿𝑜𝑔𝑜, 𝑎 𝑟𝑒𝑡𝑎 𝑥 = −1 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑎𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑒 𝑓(𝑥). lim 𝑥→1+ 𝑓(𝑥) = lim 𝑥→1+ 𝑥3 1 − 𝑥2 = lim 𝑥→−1+ 𝑥3⏞ 1 ↑ (1 + 𝑥)⏟ ↓ 2 (1 − 𝑥)⏟ ↓ 0− = −∞ ∗ 𝐿𝑜𝑔𝑜, 𝑎 𝑟𝑒𝑡𝑎 𝑥 = 1 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑎𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑒 𝑓(𝑥). −𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑖𝑠: 𝐷𝑖𝑧𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑎 𝑟𝑒𝑡𝑎 𝑦 = 𝑎 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑠𝑒 𝑜𝑐𝑜𝑟𝑟𝑒𝑟 𝑢𝑚 𝑑𝑜𝑠 𝑠𝑒𝑔𝑢𝑖𝑛𝑡𝑒𝑠 𝑐𝑎𝑠𝑜𝑠: lim 𝑥→+∞ 𝑓(𝑥) = 𝑎 𝑜𝑢 lim 𝑥→−∞ 𝑓(𝑥) = 𝑎 lim 𝑥→+∞ 𝑓(𝑥) = lim 𝑥→+∞ 𝑥3 1 − 𝑥2 = lim 𝑥→+∞ 3𝑥2 −2𝑥 = lim 𝑥→+∞ − 3𝑥 2 = −∞. lim 𝑥→−∞ 𝑓(𝑥) = lim 𝑥→−∞ 𝑥3 1 − 𝑥2 = lim 𝑥→−∞ 3𝑥2 −2𝑥 = lim 𝑥→+∞ − 3𝑥 2 = +∞. ∗ 𝐿𝑜𝑔𝑜, 𝑛ã𝑜 ℎá 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎𝑠 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑖𝑠 𝑎𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑒 𝑓(𝑥). −𝑂𝑏𝑙í𝑞𝑢𝑎: 𝐷𝑖𝑧𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑎 𝑟𝑒𝑡𝑎 𝑦 = 𝑎𝑥 + 𝑏 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 𝑜𝑏𝑙𝑖𝑞𝑢𝑎 𝑠𝑒 lim 𝑥→±∞ [𝑓(𝑥) − (𝑎𝑥 + 𝑏)] = 0 𝑓(𝑥) = 𝑥3 1 − 𝑥2 = −𝑥 + 𝑥 1 − 𝑥2 𝑓(𝑥) − (−𝑥) = 𝑥 1 − 𝑥2 lim 𝑥→±∞ [𝑓(𝑥) − (−𝑥)] = lim 𝑥→±∞ 𝑥 1 − 𝑥2 = lim 𝑥→±∞ − 1 2𝑥 = 0. ∗ 𝐿𝑜𝑔𝑜, 𝑎 𝑟𝑒𝑡𝑎 𝑦 = −𝑥 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 𝑜𝑏𝑙í𝑞𝑢𝑎 𝑎𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑒 𝑓(𝑥). (𝐼𝑉) 𝐶𝑟𝑒𝑠𝑐𝑖𝑚𝑒𝑛𝑡𝑜 𝑒 𝐷𝑒𝑐𝑟𝑒𝑠𝑐𝑖𝑚𝑒𝑛𝑡𝑜: 57 𝑓′(𝑥) = 𝑥2(3 − 𝑥2) (1 − 𝑥2)2 ∗ 𝐹𝑎𝑧𝑒𝑛𝑑𝑜 𝑜 𝑒𝑠𝑡𝑢𝑑𝑜 𝑑𝑜 𝑠𝑖𝑛𝑎𝑙 𝑑𝑒 𝑓′(𝑥), 𝑡𝑒𝑚𝑜𝑠: +++++++++ +++++ 0 + ++ + ++++++++ 𝑥2 −−−− (−√3) + + + + ++++++++ + (√3) − − − − (3 − 𝑥2) +++++++++ + (−1) + + + + (1) + + + ++ +++ (1 − 𝑥2)2 −−−− (−√3) + + + (−1) + +0 + +(1) + +(√3) − − − − 𝑓′(𝑥) = 𝑥2(3−𝑥2) (1−𝑥2)2 ∗ 𝐷𝑎 𝑎𝑛á𝑙𝑖𝑠𝑒 𝑎𝑐𝑖𝑚𝑎, 𝑐𝑜𝑛𝑐𝑙𝑢í𝑚𝑜𝑠 𝑞𝑢𝑒: 𝑓(𝑥) é 𝑐𝑟𝑒𝑠𝑐𝑒𝑛𝑡𝑒 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 (−√3,−1) ∪ (−1,1) ∪ (1, √3) 𝑒 𝑓(𝑥) é 𝑑𝑒𝑐𝑟𝑒𝑠𝑐𝑒𝑛𝑡𝑒 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 (−∞,−√3) ∪ (√3,+∞). ∗ 𝑃𝑜𝑛𝑡𝑜𝑠 𝐶𝑟í𝑡𝑖𝑐𝑜𝑠 (𝑓′(𝑥) = 0 𝑜𝑢 𝑞𝑢𝑎𝑛𝑑𝑜 𝑓′(𝑥) 𝑛ã𝑜 𝑒𝑥𝑖𝑠𝑡𝑒) −𝑓′(𝑥) 𝑛ã𝑜 𝑒𝑥𝑖𝑠𝑡𝑒 𝑒𝑚 𝑥 = −1 𝑒 𝑒𝑚 𝑥 = 1. 𝑃𝑜𝑟é𝑚, 𝑥 = −1 𝑒 𝑥 = 1 𝑛ã𝑜 𝑝𝑒𝑟𝑡𝑒𝑛𝑐𝑒 𝑎𝑜 𝑑𝑜𝑚í𝑛𝑖𝑜 𝑑𝑎 𝑓𝑢𝑛çã𝑜 𝑒, 𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑛ã𝑜 𝑠ã𝑜 𝑝𝑜𝑛𝑡𝑜𝑠 𝑐𝑟í𝑡𝑖𝑐𝑜𝑠! −𝑓′(𝑥) = 0 𝑜𝑐𝑜𝑟𝑟𝑒 𝑒𝑚 𝑥 = ±√3 𝑒 𝑒𝑚 𝑥 = 0. 𝑓(√3) = (√3) 3 1 − (√3) 2 = 3√3 −2 = − 3√3 2 . 𝑃𝑜𝑛𝑡𝑜 𝑑𝑒 𝑀á𝑥𝑖𝑚𝑜 𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑜 (√3,− 3√3 2 ) 𝑓(−√3) = (−√3) 3 1 − (−√3) 2 = −3√3 −2 = 3√3 2 . 𝑃𝑜𝑛𝑡𝑜 𝑑𝑒 𝑀í𝑛𝑖𝑚𝑜 𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑜 (−√3, 3√3 2 ) (𝑉) 𝐶𝑜𝑛𝑐𝑎𝑣𝑖𝑑𝑎𝑑𝑒 𝑒 𝑃𝑜𝑛𝑡𝑜𝑠 𝑑𝑒 𝐼𝑛𝑓𝑙𝑒𝑥ã𝑜: 𝑓′′(𝑥) = 2𝑥(𝑥2 + 3) (1 − 𝑥2)3 ∗ 𝐴𝑛𝑎𝑙𝑖𝑠𝑎𝑛𝑑𝑜 𝑜 𝑐𝑜𝑚𝑝𝑜𝑟𝑡𝑎𝑚𝑒𝑛𝑡𝑜 (𝑠𝑖𝑛𝑎𝑙) 𝑑𝑒 𝑓′′(𝑥), 𝑡𝑒𝑚𝑜𝑠: −−−−−−−−− −−−−0 + + + ++ ++++++++ 2𝑥 +++++++++ ++++++++++ ++++++++ (𝑥2 + 3) −−−−−− (−1) + + + + + +++++(1) − − − − − −− (1 − 𝑥2)3 ++++++ (−1) − − − − 0 + + + + + (1) − − − − − −− 𝑓′′(𝑥) = 2𝑥(𝑥2+3) (1−𝑥2)3 ∗ 𝐷𝑎 𝑎𝑛á𝑙𝑖𝑠𝑒 𝑎𝑐𝑖𝑚𝑎, 𝑐𝑜𝑛𝑐𝑙𝑢í𝑚𝑜𝑠 𝑞𝑢𝑒: 𝑓(𝑥) 𝑝𝑜𝑠𝑠𝑢𝑖 𝑐𝑜𝑛𝑐𝑎𝑣𝑖𝑑𝑎𝑑𝑒 𝑣𝑜𝑙𝑡𝑎𝑑𝑎 𝑝𝑎𝑟𝑎 𝑐𝑖𝑚𝑎 𝑒𝑚 (−∞,−1) ∪ (0, 1) 𝑒 𝑓(𝑥) 𝑝𝑜𝑠𝑠𝑢𝑖 𝑐𝑜𝑛𝑐𝑎𝑣𝑖𝑑𝑎𝑑𝑒 𝑣𝑜𝑙𝑡𝑎𝑑𝑎 𝑝𝑎𝑟𝑎 𝑏𝑎𝑖𝑥𝑜 𝑒𝑚 (−1, 0) ∪ (1,+∞) ∗ 𝑃𝑜𝑛𝑡𝑜 𝑑𝑒 𝐼𝑛𝑓𝑙𝑒𝑥ã𝑜 𝑜𝑐𝑜𝑟𝑟𝑒 𝑞𝑢𝑎𝑛𝑑𝑜 ℎá 𝑚𝑢𝑑𝑎𝑛ç𝑎 𝑛𝑎 𝑑𝑖𝑟𝑒çã𝑜 𝑑𝑎 𝑐𝑜𝑛𝑐𝑎𝑣𝑖𝑑𝑎𝑑𝑒. 𝑁𝑒𝑠𝑠𝑒 𝑐𝑎𝑠𝑜, 𝑡𝑒𝑚𝑜𝑠 𝑒𝑚 𝑥 = 0 𝑢𝑚 𝑝𝑜𝑛𝑡𝑜 𝑑𝑒 𝑖𝑛𝑓𝑙𝑒𝑥ã𝑜. 𝑃. 𝐼(0,0) 60 𝑓′(𝑥) > 0 𝑠𝑒 𝑥 > 1 2 ; 𝑓′(𝑥) = 0 𝑠𝑒 𝑥 = 1 2 𝑒 𝑓′(𝑥) < 0 𝑠𝑒 𝑥 < 1 2 . −−−−−−−−− −−− 1 2 + + + +++ ++++++ 𝑓′(𝑥) ∗ 𝐿𝑜𝑔𝑜, 𝑡𝑒𝑚𝑜𝑠 𝑢𝑚 𝑝𝑜𝑛𝑡𝑜 𝑚í𝑛𝑖𝑚𝑜 𝑒𝑚 𝑥 = 1 2 . 𝑓 ( 1 2 ) = 𝑒 1 4 − 1 2 = 𝑒− 1 4 = 1 √𝑒 4 ; ∗ 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑛𝑑𝑜 𝑜𝑠 𝑣𝑎𝑙𝑜𝑟𝑒𝑠 𝑛𝑎𝑠 𝑒𝑥𝑡𝑟𝑒𝑚𝑖𝑑𝑎𝑑𝑒𝑠 𝑑𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜, 𝑡𝑒𝑚𝑜𝑠: 𝑓(0) = 𝑒0 = 1; 𝑓(1) = 𝑒0 = 1; ∗ 𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑜 𝑣𝑎𝑙𝑜𝑟 𝑚á𝑥𝑖𝑚𝑜 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 [0, 1] é 1 𝑒 𝑜 𝑣𝑎𝑙𝑜𝑟 𝑚í𝑛𝑖𝑚𝑜 é 1 √𝑒 4 . 3. 𝑓(𝑥) = 𝑎𝑥2 + 𝑏𝑥 − 𝑐 𝑓′(1) = 0 𝑒 𝑓(1) = 7 ; 𝑈𝑠𝑎𝑛𝑑𝑜 𝑒𝑠𝑠𝑎𝑠 𝑖𝑛𝑓𝑜𝑟𝑚𝑎çõ𝑒𝑠 𝑛𝑎 𝑎𝑛á𝑙𝑖𝑠𝑒 𝑑𝑎 𝑓𝑢𝑛çã𝑜, 𝑡𝑒𝑚𝑜𝑠: 𝑓(2) = −2 𝑓′(𝑥) = 2𝑎𝑥 + 𝑏 𝑓′(1) = 2𝑎 + 𝑏 = 0 (𝐼) 𝑓(1) = 𝑎 + 𝑏 − 𝑐 = 7 (𝐼𝐼) 𝑓(2) = 4𝑎 + 2𝑏 − 𝑐 = −2 (𝐼𝐼𝐼) { 2𝑎 + 𝑏 = 0 𝑎 + 𝑏 − 𝑐 = 7 4𝑎 + 2𝑏 − 𝑐 = −2 → { 2𝑎 + 𝑏 = 0 −𝑎 − 𝑐 = 7 −𝑐 = −2 → 𝑐 = 2; 𝑎 = −9 ; 𝑏 = 18 ∗ 𝐿𝑜𝑔𝑜, 𝑓(𝑥) = −9𝑥2 + 18𝑥 − 2 4. A 100 km 𝑆𝐴 = 100 − 20𝑡 B 𝑆𝐵 = 15𝑡 ∗ 𝐴 𝑑𝑖𝑠𝑡â𝑛𝑐𝑖𝑎 𝑒𝑛𝑡𝑟𝑒 𝐴 𝑒 𝐵 𝑠𝑒𝑟á 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑎𝑑𝑎 𝑝𝑒𝑙𝑎 ℎ𝑖𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑎 𝑑𝑜 𝑡𝑟𝑖â𝑛𝑔𝑢𝑙𝑜 𝑎𝑐𝑖𝑚𝑎, 𝑣𝑖𝑠𝑡𝑜 𝑞𝑢𝑒 𝐵 𝑠𝑒 𝑚𝑜𝑣𝑖𝑚𝑒𝑛𝑡𝑎 𝑛𝑎 𝑑𝑖𝑟𝑒çã𝑜 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 à 𝑡𝑟𝑎𝑗𝑒𝑡ó𝑟𝑖𝑎 𝑑𝑜 𝑛𝑎𝑣𝑖𝑜 𝐴. 𝐷(𝑡)2 = 𝑆𝐴(𝑡) 2 + 𝑆𝐵(𝑡) 2 2𝐷(𝑡). 𝐷′(𝑡) = 2𝑆𝐴(𝑡). 𝑆𝐴 ′ (𝑡) + 2𝑆𝐵(𝑡). 𝑆𝐵 ′ (𝑡) 𝐷(𝑡). 𝐷′(𝑡) = 𝑆𝐴(𝑡). 𝑆𝐴 ′ (𝑡) + 𝑆𝐵(𝑡). 𝑆𝐵 ′ (𝑡) 61 𝐷′(𝑡) = (100 − 20𝑡)(−20) + (15𝑡)(15) √(100 − 20𝑡)2 + (15𝑡)2 À𝑠 16ℎ 𝑡𝑒𝑟ã𝑜 𝑠𝑒 𝑝𝑎𝑠𝑠𝑎𝑑𝑜 3ℎ 𝑒𝑚 𝑟𝑒𝑙𝑎çã𝑜 𝑎𝑜 𝑚𝑜𝑚𝑒𝑛𝑡𝑜 𝑒𝑚 𝑞𝑢𝑒 𝐴 𝑒𝑠𝑡𝑎𝑣𝑎 𝑎 100𝑘𝑚 𝑑𝑒 𝐵. 𝐷′(3) = (100 − 20.3)(−20) + (15.3)(15) √(100 − 20.3)2 + (15.3)2 𝐷′(3) = −800 + 675 √402 + 225.9 𝐷′(3) = −125 √1600 + 2025 𝐷′(3) = −125 √3625 𝑘𝑚/ℎ ∗ 𝐿𝑜𝑔𝑜, 𝑜𝑠 𝑛𝑎𝑣𝑖𝑜𝑠 𝐴 𝑒 𝐵 𝑒𝑠𝑡ã𝑜 𝑠𝑒 𝑎𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑛𝑑𝑜!𝑁𝑜𝑡𝑒 𝑞𝑢𝑒 𝑎 𝑑𝑖𝑠𝑡â𝑛𝑐𝑖𝑎 𝑒𝑛𝑡𝑟𝑒 𝐴 𝑒 𝐵 𝑑𝑖𝑚𝑖𝑛𝑢𝑖𝑢, 𝑜 𝑞𝑢𝑒 𝑒𝑥𝑝𝑙𝑖𝑐𝑎 𝑜 𝑓𝑎𝑡𝑜 𝑑𝑒 𝑞𝑢𝑒 𝐷′(3) < 0. 5. 𝐷𝑎𝑑𝑜 𝑢𝑚 𝑝𝑜𝑛𝑡𝑜 𝑑𝑎 𝑒𝑙𝑖𝑝𝑠𝑒, 𝑎𝑠 𝑐𝑜𝑜𝑟𝑑𝑒𝑛𝑎𝑑𝑎𝑠 𝑑𝑒𝑙𝑒 𝑠𝑒𝑟á 𝑛𝑎 𝑓𝑜𝑟𝑚𝑎: (𝑥,± 3 2 √4 − 𝑥2) . 𝑃𝑒𝑙𝑎 𝑓𝑖𝑔𝑢𝑟𝑎 𝑎𝑜 𝑙𝑎𝑑𝑜, 𝑛𝑜𝑡𝑒 𝑞𝑢𝑒 𝑎 á𝑟𝑒𝑎 𝑑𝑜 𝑟𝑒𝑡â𝑛𝑔𝑢𝑙𝑜 𝑠𝑒𝑟á 𝑑𝑎𝑑𝑎 𝑐𝑜𝑚𝑜 𝐴 = 2𝑥. 2𝑦, 𝑜𝑢 𝑎𝑖𝑛𝑑𝑎, 𝐴 = |2𝑥. 2𝑦| 𝑒𝑚 𝑐𝑎𝑠𝑜 𝑑𝑒 𝑒𝑛𝑐𝑜𝑛𝑡𝑟𝑎𝑟𝑚𝑜𝑠 𝑣𝑎𝑙𝑜𝑟𝑒𝑠 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑜𝑠 𝑝𝑎𝑟𝑎 𝑥 𝑜𝑢 𝑝𝑎𝑟𝑎 𝑦. 𝐸𝑛𝑡ã𝑜: 𝐴 = |6𝑥√4 − 𝑥2| . 𝑆𝑒𝑗𝑎 𝑓(𝑥) = 6𝑥√4 − 𝑥2 ; 𝐴 = |𝑓(𝑥)| 𝑓′(𝑥) = 6√4 − 𝑥2 − 6𝑥2 √4 − 𝑥2 . 𝑓𝑎𝑧𝑒𝑛𝑑𝑜 𝑓′(𝑥) = 0 6√4 − 𝑥2 = 6𝑥2 √4 − 𝑥2 ⇒ 𝑥² = 4 − 𝑥2 ⇒ 2𝑥2 = 4 ⇒ 𝑥2 = 2 ∴ 𝑥 = ±√2 𝑃𝑎𝑟𝑎 𝑥 = ±√2 𝑡𝑒𝑟í𝑎𝑚𝑜𝑠 𝑢𝑚 𝑝𝑜𝑛𝑡𝑜 𝑑𝑒 𝑚á𝑥𝑖𝑚𝑜 𝑒 𝑚í𝑛𝑖𝑚𝑜 𝑝𝑎𝑟𝑎 𝑓(𝑥). 𝑁𝑜 𝑒𝑛𝑡𝑎𝑛𝑡𝑜, 𝑛𝑜𝑡𝑒 𝑞𝑢𝑒 𝑎 𝑓𝑢𝑛çã𝑜 𝑑𝑎 Á𝑅𝐸𝐴 é 𝑜 𝑚ó𝑑𝑢𝑙𝑜 𝑑𝑜𝑠 𝑣𝑎𝑙𝑜𝑟𝑒𝑠 𝑎𝑠𝑠𝑢𝑚𝑖𝑑𝑜𝑠 𝑝𝑜𝑟 𝑓(𝑥), 𝑠𝑒𝑛𝑑𝑜 𝑎𝑠𝑠𝑖𝑚, 𝑝𝑎𝑟𝑎 𝑥 = ±√2 𝑡𝑒𝑟𝑒𝑚𝑜𝑠 𝑜 𝑣𝑎𝑙𝑜𝑟 𝑚á𝑥𝑖𝑚𝑜 𝑑𝑒 𝐴. 𝐿𝑜𝑔𝑜: 𝐴 = |6. √2√4 − √2 2 | = |6√2√4 − 2| = |6√2√2| = |6.2| = |12| = 12 𝑢. 𝐴 𝐴 = |−6. √2√4 − (−√2) 2 | = |−6√2√4 − 2| = |−6√2√2| = |−6.2| = |−12| = 12𝑢. 𝐴 62 3.6 VPA 1-12 de Setembro de 2008 1. 𝑓(𝑥) = { 𝑥2, 𝑥 ≤ 1 𝑐𝑥 + 𝑘, 1 < 𝑥 < 4 −2𝑥, 𝑥 ≥ 4 𝑎) 𝐸𝑛𝑐𝑜𝑛𝑡𝑟𝑎𝑟 𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑐 𝑒 𝑘 𝑝𝑎𝑟𝑎 𝑞𝑢𝑒 𝑓 𝑠𝑒𝑗𝑎 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑛𝑜𝑠 𝑟𝑒𝑎𝑖𝑠. ∗ 𝑂𝑏𝑠: 𝑐𝑜𝑚𝑜 𝑡𝑜𝑑𝑎𝑠 𝑎𝑠 𝑠𝑒𝑛𝑡𝑒𝑛ç𝑎𝑠 𝑠ã𝑜 𝑓𝑢𝑛çõ𝑒𝑠 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑎𝑖𝑠 𝑒, 𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎𝑠 𝑜𝑛𝑑𝑒 𝑝𝑟𝑒𝑑𝑜𝑚𝑖𝑛𝑎𝑚. 𝐷𝑒𝑠𝑡𝑎 𝑜𝑏𝑠𝑒𝑟𝑣𝑎çã𝑜 𝑗á 𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑑𝑖𝑧𝑒𝑟 𝑞𝑢𝑒 𝑓 é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎𝑠 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 (−∞, 1) ∪ (1, 4) ∪ (4, +∞).𝐵𝑎𝑠𝑡𝑎 𝑒𝑛𝑐𝑜𝑛𝑡𝑟𝑎𝑟𝑚𝑜𝑠 𝑜𝑠 𝑣𝑎𝑙𝑜𝑟𝑒𝑠 𝑑𝑒 𝑐 𝑒 𝑘 𝑝𝑎𝑟𝑎 𝑞𝑢𝑒 𝑓 𝑠𝑒𝑗𝑎 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 𝑥 = 1 𝑒 𝑒𝑚 𝑥 = 4. ∗ 𝐸𝑚 𝑥 = 1, 𝑑𝑒𝑣𝑒𝑚𝑜𝑠 𝑡𝑒𝑟 𝑎 𝑠𝑒𝑔𝑢𝑖𝑛𝑡𝑒 𝑖𝑔𝑢𝑎𝑙𝑑𝑎𝑑𝑒 𝑠𝑎𝑡𝑖𝑠𝑓𝑒𝑖𝑡𝑎: 𝑓(1) = lim 𝑥→1+ 𝑓(𝑥) 1) 𝑓(1) = 1² = 1. 2) lim 𝑥→1+ 𝑓(𝑥) = lim 𝑥→1+ 𝑐𝑥 + 𝑘 = 𝑐 + 𝑘. 3) 𝑃𝑒𝑙𝑎 𝑖𝑔𝑢𝑎𝑙𝑑𝑎𝑑𝑒 𝑎𝑐𝑖𝑚𝑎, 𝑡𝑒𝑚𝑜𝑠: 𝑐 + 𝑘 = 1 (𝐼) ∗ 𝐸𝑚 𝑥 = 4, 𝑑𝑒𝑣𝑒𝑚𝑜𝑠 𝑡𝑒𝑟 𝑎 𝑠𝑒𝑔𝑢𝑖𝑛𝑡𝑒 𝑖𝑔𝑢𝑎𝑙𝑑𝑎𝑑𝑒 𝑠𝑎𝑡𝑖𝑠𝑓𝑒𝑖𝑡𝑎: 𝑓(4) = lim 𝑥→4− 𝑓(𝑥) 1)𝑓(4) = −2. (4) = −8. 2) lim 𝑥→4− 𝑓(𝑥) = lim 𝑥→4− 𝑐𝑥 + 𝑘 = 4𝑐 + 𝑘. 3) 𝑃𝑒𝑙𝑎 𝑖𝑔𝑢𝑎𝑙𝑑𝑎𝑑𝑒 𝑎𝑐𝑖𝑚𝑎, 𝑡𝑒𝑚𝑜𝑠: 4𝑐 + 𝑘 = −8 (𝐼𝐼) ∗ 𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑛𝑑𝑜 𝑜 𝑠𝑒𝑔𝑢𝑖𝑛𝑡𝑒 𝑠𝑖𝑠𝑡𝑒𝑚: { 𝑐 + 𝑘 = 1 4𝑐 + 𝑘 = −8 → { 𝑐 + 𝑘 = 1 3𝑐 = −9 → 𝑐 = −3; 𝑘 = 4. 2. 𝑎) 𝑓(𝑥) = 𝑥2 𝑥 − 1 ; 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟 𝑓′(𝑥) 𝑝𝑒𝑙𝑎 𝑑𝑒𝑓𝑖𝑛𝑖çã𝑜. 𝑃𝑒𝑙𝑎 𝑑𝑒𝑓𝑖𝑛𝑖çã𝑜 𝑑𝑒 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎 𝑑𝑒 𝑢𝑚𝑎 𝑓𝑢𝑛çã𝑜 𝑛𝑢𝑚 𝑝𝑜𝑛𝑡𝑜, 𝑡𝑒𝑚𝑜𝑠 𝑎 𝑠𝑒𝑔𝑢𝑖𝑛𝑡𝑒 𝑒𝑥𝑝𝑟𝑒𝑠𝑠ã𝑜: 𝑓′(𝑥) = lim ∆𝑥→0 𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥) ∆𝑥 65 𝑏) 𝑦 = 1 𝑥2 − 3𝑥 + 2 ; 𝑦 = 𝑓(𝑥) = 1 (𝑥 − 1)(𝑥 − 2) −𝐴𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑖𝑠: ∗ 𝐷𝑖𝑧𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑎 𝑟𝑒𝑡𝑎 𝑥 = 𝑎 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎𝑠 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑠𝑒 𝑜𝑐𝑜𝑟𝑟𝑒𝑟 𝑢𝑚 𝑑𝑜𝑠 𝑐𝑎𝑠𝑜𝑠 𝑎 𝑠𝑒𝑔𝑢𝑖𝑟: lim 𝑥→𝑎+ 𝑓(𝑥) = ±∞ 𝑜𝑢 lim 𝑥→𝑎− 𝑓(𝑥) = ±∞ ∗ 𝑉𝑒𝑟𝑖𝑓𝑖𝑐𝑎𝑛𝑑𝑜 𝑠𝑒 𝑎𝑠 𝑟𝑒𝑡𝑎𝑠 𝑥 = 1 𝑒 𝑥 = 2 𝑠ã𝑜 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎𝑠 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑖𝑠 𝑎𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑒 𝑓(𝑥), 𝑝𝑜𝑖𝑠 𝑠ã𝑜 𝑜𝑠 𝑝𝑜𝑛𝑡𝑜𝑠 𝑑𝑒 𝑑𝑒𝑠𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑖𝑑𝑎𝑑𝑒 𝑑𝑎 𝑓𝑢𝑛çã𝑜. lim 𝑥→1+ 𝑓(𝑥) = lim 𝑥→1+ 1 (𝑥 − 1)(𝑥 − 2) = lim 𝑥→1+ 1 (𝑥 − 1)⏟ ↓ 0+ (𝑥 − 2)⏟ ↓ −1 = −∞ 𝑂𝑏𝑠: 𝑠𝑒 𝑥 → 1+ , 𝑒𝑛𝑡ã𝑜 𝑥 > 1. 𝐿𝑜𝑔𝑜, 𝑥 − 1 > 0. lim 𝑥→1− 𝑓(𝑥) = lim 𝑥→1− 1 (𝑥 − 1)(𝑥 − 2) = lim 𝑥→1− 1 (𝑥 − 1)⏟ ↓ 0− (𝑥 − 2)⏟ ↓ −1 = +∞ 𝑂𝑏𝑠: 𝑠𝑒 𝑥 → 1− , 𝑒𝑛𝑡ã𝑜 𝑥 < 1. 𝐿𝑜𝑔𝑜, 𝑥 − 1 < 0. ∗ 𝐿𝑜𝑔𝑜, 𝑎 𝑟𝑒𝑡𝑎 𝑥 = 1 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙. lim 𝑥→2+ 𝑓(𝑥) = lim 𝑥→2+ 1 (𝑥 − 1)(𝑥 − 2) = lim 𝑥→1+ 1 (𝑥 − 1)⏟ ↓ 1 (𝑥 − 2)⏟ ↓ 0+ = +∞ 𝑂𝑏𝑠: 𝑠𝑒 𝑥 → 2+ , 𝑒𝑛𝑡ã𝑜 𝑥 > 2. 𝐿𝑜𝑔𝑜, 𝑥 − 2 > 0. lim 𝑥→2− 𝑓(𝑥) = lim 𝑥→2− 1 (𝑥 − 1)(𝑥 − 2) = lim 𝑥→2− 1 (𝑥 − 1)⏟ ↓ 1 (𝑥 − 2)⏟ ↓ 0− = −∞ 𝑂𝑏𝑠: 𝑠𝑒 𝑥 → 2− , 𝑒𝑛𝑡ã𝑜 𝑥 < 2. 𝐿𝑜𝑔𝑜, 𝑥 − 2 < 0. ∗ 𝐿𝑜𝑔𝑜, 𝑎 𝑟𝑒𝑡𝑎 𝑥 = 2 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙. 𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑎𝑠𝑠𝑖𝑛𝑡𝑜𝑡𝑎 𝑎𝑠 𝑟𝑒𝑡𝑎𝑠 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎𝑠 𝑝𝑜𝑟 𝑎𝑚𝑏𝑜𝑠 𝑜𝑠 𝑙𝑎𝑑𝑜𝑠, 𝑝𝑜𝑖𝑠 𝑜𝑠 𝑙𝑖𝑚𝑖𝑡𝑒𝑠 𝑙𝑎𝑡𝑒𝑟𝑎𝑖𝑠 𝑒𝑚 𝑥 = 1 𝑒 𝑒𝑚 𝑥 = 2 𝑛ã𝑜 𝑒𝑥𝑖𝑠𝑡𝑒𝑚. 4. 𝑎) lim 𝑥→2 |𝑥 − 2| 𝑥3 − 8 ; 𝑂𝑏𝑠: |𝑥 − 2| = { 𝑥 − 2, 𝑠𝑒 𝑥 ≥ 2 −(𝑥 − 2), 𝑠𝑒 𝑥 < 2 . lim 𝑥→2+ |𝑥 − 2| 𝑥3 − 8 = lim 𝑥→2+ 𝑥 − 2 𝑥3 − 8 = lim 𝑥→2+ (𝑥 − 2) (𝑥 − 2)(𝑥2 + 2𝑥 + 4) = lim 𝑥→2+ 1 (𝑥2 + 2𝑥 + 4) = 1 12 ∗ 𝑂𝑏𝑠: 𝑠𝑒 𝑥 → 2+, 𝑒𝑛𝑡ã𝑜 𝑥 > 2. 𝐿𝑜𝑔𝑜, |𝑥 − 2| = 𝑥 − 2 lim 𝑥→2+ |𝑥 − 2| 𝑥3 − 8 = lim 𝑥→2+ 𝑥 − 2 𝑥3 − 8 = lim 𝑥→2+ −(𝑥 − 2) (𝑥 − 2)(𝑥2 + 2𝑥 + 4) = lim 𝑥→2+ −1 (𝑥2 + 2𝑥 + 4) = − 1 12 ∗ 𝑂𝑏𝑠: 𝑠𝑒 𝑥 → 2−, 𝑒𝑛𝑡ã𝑜 𝑥 < 2. 𝐿𝑜𝑔𝑜, |𝑥 − 2| = −(𝑥 − 2). 𝐶𝑜𝑚𝑜 lim 𝑥→2+ |𝑥 − 2| 𝑥3 − 8 ≠ lim 𝑥→2− |𝑥 − 2| 𝑥3 − 8 , 𝑑𝑖𝑧𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 lim 𝑥→2 |𝑥 − 2| 𝑥3 − 8 ∄. 66 𝑏)𝑀𝑜𝑠𝑡𝑟𝑒 𝑞𝑢𝑒 lim 𝑥→0+ (√𝑥 3 ∙ 2𝑠𝑒𝑛 𝜋 2) = 0 −1 ≤ 𝑠𝑒𝑛 𝜋 2 ≤ 1 2−1 ≤ 2𝑠𝑒𝑛 𝜋 2 ≤ 21 1 2 ≤ 2𝑠𝑒𝑛 𝜋 2 ≤ 2 √𝑥 3 2 ≤ √𝑥 3 ∙ 2𝑠𝑒𝑛 𝜋 2 ≤ 2√𝑥 3 𝑆𝑒𝑗𝑎𝑚 𝑓(𝑥) = √𝑥 3 2 , 𝑔(𝑥) = √𝑥 3 ∙ 2𝑠𝑒𝑛 𝜋 2 𝑒 ℎ(𝑥) = 2√𝑥 3 . 𝐸𝑛𝑡ã𝑜, 𝑡𝑒𝑚𝑜𝑠: 𝑓(𝑥) ≤ 𝑔(𝑥) ≤ ℎ(𝑥) 𝐸 𝑎𝑖𝑛𝑑𝑎, lim 𝑥→0 𝑓(𝑥) = 0 𝑒 lim 𝑥→0 ℎ(𝑥) = 0, 𝑜𝑢 𝑠𝑒𝑗𝑎, lim 𝑥→0 𝑓(𝑥) = lim 𝑥→0 ℎ(𝑥) , 𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑔𝑎𝑟𝑎𝑛𝑡𝑖𝑟 𝑝𝑒𝑙𝑜 𝑇𝑒𝑜𝑟𝑒𝑚𝑎 𝑑𝑜 𝐶𝑜𝑛𝑓𝑟𝑜𝑛𝑡𝑜 𝑞𝑢𝑒 𝑠𝑒 𝑓(𝑥) ≤ 𝑔(𝑥) ≤ ℎ(𝑥) 𝑒 lim 𝑥→0 𝑓(𝑥) = lim 𝑥→0 ℎ(𝑥) , 𝑒𝑛𝑡ã𝑜 lim 𝑥→0 𝑔(𝑥) = lim 𝑥→0 𝑓(𝑥) = lim 𝑥→0 ℎ(𝑥) = 0. 𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜, lim 𝑥→0+ (√𝑥 3 ∙ 2𝑠𝑒𝑛 𝜋 2) = 0 5. 𝑦 = 1 𝑥 . 𝐸𝑞𝑢𝑎çã𝑜 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 𝑛𝑜 𝑝𝑜𝑛𝑡𝑜 𝑑𝑒 𝑎𝑏𝑠𝑐𝑖𝑠𝑠𝑎 𝑥 = 𝑎. 𝑎 > 0. ∗ 𝑃𝑜𝑛𝑡𝑜 𝑑𝑒 𝑡𝑎𝑛𝑔𝑒𝑛𝑐𝑖𝑎: 𝑃 = (𝑎, 1 𝑎 ). 𝑦′ = − 1 𝑥2 . 𝑦′(𝑎) = − 1 𝑎2 . 𝑦 − 𝑦𝑜 = 𝑚(𝑥 − 𝑥𝑜) 𝑦 − 1 𝑎 = − 1 𝑎2 (𝑥 − 𝑎) 𝑦 = − 1 𝑎2 𝑥 + 2 𝑎 ∗ 𝐼𝑛𝑡𝑒𝑟𝑠𝑒çõ𝑒𝑠 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 𝑐𝑜𝑚 𝑜𝑠 𝑒𝑖𝑥𝑜𝑠 𝑐𝑜𝑜𝑟𝑑𝑒𝑛𝑎𝑑𝑜𝑠: 𝐴 = (0, 2 𝑎 ) 𝑒 𝐵 = (2𝑎, 0) ∗ 𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑎 á𝑟𝑒𝑎 𝑑𝑜 𝑡𝑟𝑖â𝑛𝑔𝑢𝑙𝑜 𝑓𝑜𝑟𝑚𝑎𝑑𝑜 𝑝𝑒𝑙𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 𝑒 𝑜𝑠 𝑒𝑖𝑥𝑜𝑠 𝑐𝑜𝑜𝑟𝑑𝑒𝑛𝑎𝑑𝑜𝑠 é: Á𝑟𝑒𝑎 = 1 2 (𝑥𝐵 − 𝑥𝑂) ∙ (𝑦𝐴 − 𝑦𝑂) = 1 2 (2𝑎) ( 2 𝑎 ) = 2 𝑢. 𝐴 67 3.7 VPA 1-13 de Setembro de 2008 1. 𝑓(𝑥) = { 𝑥2 + 1 , 𝑥 ≤ 1 − 𝑥2 2 + 4, 1 < 𝑥 ≤ 4 2 , 𝑥 > 4 ∗ 𝑓 é 𝑢𝑚𝑎 𝑓𝑢𝑛çã𝑜 𝑠𝑒𝑛𝑡𝑒𝑛𝑐𝑖𝑎𝑙, 𝑛𝑎 𝑞𝑢𝑎𝑙 𝑠𝑢𝑎𝑠 𝑠𝑒𝑛𝑡𝑒𝑛ç𝑎𝑠 𝑠ã𝑜 𝑓𝑢𝑛çõ𝑒𝑠 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑎𝑖𝑠 𝑒, 𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎𝑠 𝑒𝑚 𝑠𝑒𝑢𝑠 𝑑𝑜𝑚í𝑛𝑖𝑜𝑠. 𝐿𝑜𝑔𝑜, 𝑓 é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 (−∞, 1) ∪ (1,4) ∪ (4,+∞). 𝑉𝑎𝑚𝑜𝑠 𝑣𝑒𝑟𝑖𝑓𝑖𝑐𝑎𝑟 𝑠𝑒 𝑓 é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑛𝑜𝑠 𝑝𝑜𝑛𝑡𝑜𝑠 𝑜𝑛𝑑𝑒 ℎá 𝑚𝑢𝑑𝑎𝑛ç𝑎 𝑑𝑒 𝑐𝑜𝑚𝑝𝑜𝑟𝑡𝑎𝑚𝑒𝑛𝑡𝑜 𝑑𝑎 𝑓𝑢𝑛çã𝑜, 𝑜𝑢 𝑠𝑒𝑗𝑎, 𝑒𝑚 𝑥 = 1 𝑒 𝑥 = 4. ∗ 𝐷𝑖𝑧𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑓 é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑛𝑢𝑚 𝑝𝑜𝑛𝑡𝑜 𝑥 = 𝑎 𝑠𝑒, 𝑒 𝑠𝑜𝑚𝑒𝑛𝑡𝑒 𝑠𝑒, 1) 𝑓(𝑎) 𝑒𝑥𝑖𝑠𝑡𝑒; 2) lim 𝑥→𝑎 𝑓(𝑥) 𝑒𝑥𝑖𝑠𝑡𝑒; 3) lim 𝑥→𝑎 𝑓(𝑥) = 𝑓(𝑎) −𝐸𝑚 𝑥 = 1: 1) 𝑓(1) = 1² + 1 = 1 + 1 = 2. 2) lim 𝑥→1+ 𝑓(𝑥) = lim 𝑥→1+ (− 𝑥2 2 + 4) = lim 𝑥→1+ − 𝑥2 2 + lim 𝑥→1+ 4 = − 1 2 + 4 = 7 2 . lim 𝑥→1− 𝑓(𝑥) = lim 𝑥→1− (𝑥2 + 1) = lim 𝑥→1− 𝑥2 + lim 𝑥→1− 1 = 1 + 1 = 2. ∗ 𝐶𝑜𝑚𝑜 lim 𝑥→1+ 𝑓(𝑥) ≠ lim 𝑥→1− 𝑓(𝑥) 𝑒𝑛𝑡ã𝑜 lim 𝑥→1 𝑓(𝑥) ∄. ∗ 𝐶𝑜𝑛𝑠𝑒𝑞𝑢𝑒𝑛𝑡𝑒𝑚𝑒𝑛𝑡𝑒, 𝑓 𝑛ã𝑜 é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 𝑥 = 1. −𝐸𝑚 𝑥 = 4: 1) 𝑓(4) = − 42 2 + 4 = −8 + 4 = −4. 2) lim 𝑥→4+ 𝑓(𝑥) = lim 𝑥→4+ 2 = 2. lim 𝑥→4− 𝑓(𝑥) = lim 𝑥→4− ( −𝑥2 2 + 4) = lim 𝑥→4− −𝑥2 2 + lim 𝑥→4− 4 = −8 + 4 = −4. ∗ 𝐶𝑜𝑚𝑜 lim 𝑥→4+ 𝑓(𝑥) ≠ lim 𝑥→4− 𝑓(𝑥) 𝑒𝑛𝑡ã𝑜 lim 𝑥→4 𝑓(𝑥) ∄. ∗ 𝐶𝑜𝑛𝑠𝑒𝑞𝑢𝑒𝑛𝑡𝑒𝑚𝑒𝑛𝑡𝑒, 𝑓 𝑛ã𝑜 é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 𝑥 = 4. ∗ 𝑆𝑒𝑛𝑑𝑜 𝑓 𝑑𝑒𝑠𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 𝑥 = 1 𝑒 𝑥 = 4 𝑡𝑒𝑚𝑜𝑠, 𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑞𝑢𝑒 𝑓 𝑛ã𝑜 é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑛𝑜𝑠 𝑟𝑒𝑎𝑖𝑠. 𝑏) 𝐸𝑠𝑏𝑜ç𝑜 𝑑𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑒 𝑓(𝑥): 70 4. 𝑎) lim 𝑥→𝑎 𝑥2 − 𝑎2 |𝑥 − 𝑎| = lim 𝑥→𝑎 (𝑥 − 𝑎)(𝑥 + 𝑎) |𝑥 − 𝑎| ; 𝑂𝑏𝑠: |𝑥 − 𝑎| = { 𝑥 − 𝑎, 𝑠𝑒 𝑥 ≥ 𝑎 −(𝑥 − 𝑎), 𝑠𝑒 𝑥 < 𝑎 . lim 𝑥→𝑎+ (𝑥 − 𝑎)(𝑥 + 𝑎) |𝑥 − 𝑎| = lim 𝑥→𝑎+ (𝑥 − 𝑎)(𝑥 + 𝑎) (𝑥 − 𝑎) = lim 𝑥→𝑎+ (𝑥 + 𝑎) = 2𝑎. lim 𝑥→𝑎− (𝑥 − 𝑎)(𝑥 + 𝑎) |𝑥 − 𝑎| = lim 𝑥→𝑎− (𝑥 − 𝑎)(𝑥 + 𝑎) −(𝑥 − 𝑎) = lim 𝑥→𝑎+ −(𝑥 + 𝑎) = −2𝑎. ∗ 𝐶𝑜𝑚𝑜 𝑜𝑠 𝑙𝑖𝑚𝑖𝑡𝑒𝑠 𝑙𝑎𝑡𝑒𝑟𝑎𝑖𝑠 𝑒𝑥𝑖𝑠𝑡𝑒𝑚,𝑚𝑎𝑠 𝑠ã𝑜 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒𝑠, 𝑒𝑛𝑡ã𝑜 lim 𝑥→𝑎 𝑥2 − 𝑎2 |𝑥 − 𝑎| ∄. 𝑏) lim 𝑥→2 √6 − 𝑥 − 2 √3 − 𝑥 − 1 = lim 𝑥→2 √6 − 𝑥 − 2 √3 − 𝑥 − 1 ∙ (√6 − 𝑥 + 2) (√6 − 𝑥 + 2) ∙ (√3 − 𝑥 + 1) (√3 − 𝑥 + 1) = lim 𝑥→2 −(𝑥 − 2)(√3 − 𝑥 + 1) −(𝑥 − 2)(√6 − 𝑥 + 2) = lim 𝑥→2 (√3 − 𝑥 + 1) (√6 − 𝑥 + 2) = √3 − 2 + 1 √6 − 2 + 2 = √1 + 1 √4 + 2 = 2 4 = 1 2 . 5. 𝑎) |𝑔(𝑥) + 4| < 2(𝑥 − 3)4 ; 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟 lim 𝑥→3 𝑔(𝑥). ∗ 𝑈𝑠𝑎𝑛𝑑𝑜 𝑎 𝑑𝑒𝑠𝑖𝑔𝑢𝑎𝑙𝑑𝑎𝑑𝑒 𝑚𝑜𝑑𝑢𝑙𝑎𝑟 𝑡𝑒𝑚𝑜𝑠: −2(𝑥 − 3)4 < 𝑔(𝑥) + 4 < 2(𝑥 − 3)4 −2(𝑥 − 3)4 − 4 < 𝑔(𝑥) < 2(𝑥 − 3)4 − 4 𝑆𝑒𝑗𝑎𝑚 𝑓(𝑥) = −2(𝑥 − 3)4 − 4 𝑒 ℎ(𝑥) = 2(𝑥 − 3)4 − 4, 𝑒𝑛𝑡ã𝑜 𝑡𝑒𝑚𝑜𝑠: 𝑓(𝑥) < 𝑔(𝑥) < ℎ(𝑥) 𝐸 𝑎𝑖𝑛𝑑𝑎, lim 𝑥→3 𝑓(𝑥) = 0 𝑒 lim 𝑥→3 ℎ(𝑥) = 0, 𝑜𝑢 𝑠𝑒𝑗𝑎, lim 𝑥→3 𝑓(𝑥) = lim 𝑥→3 ℎ(𝑥) , 𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑔𝑎𝑟𝑎𝑛𝑡𝑖𝑟 𝑝𝑒𝑙𝑜 𝑇𝑒𝑜𝑟𝑒𝑚𝑎 𝑑𝑜 𝐶𝑜𝑛𝑓𝑟𝑜𝑛𝑡𝑜 𝑞𝑢𝑒 𝑠𝑒 𝑓(𝑥) < 𝑔(𝑥) < ℎ(𝑥) 𝑒 lim 𝑥→0 𝑓(𝑥) = lim 𝑥→0 ℎ(𝑥) , 𝑒𝑛𝑡ã𝑜 lim 𝑥→3 𝑔(𝑥) = lim 𝑥→3 𝑓(𝑥) = lim 𝑥→3 ℎ(𝑥) = −4. 𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜, lim 𝑥→3 𝑔(𝑥) = −4 𝑏) ℎ(𝑥) = 1 𝑥2 + 5𝑥 + 6 ; ℎ(𝑥) = 1 (𝑥 + 2)(𝑥 + 3) . ∗ 𝑇𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝐷(ℎ) = ℝ − {−2,−3}. 𝐿𝑜𝑔𝑜, 𝑣𝑎𝑚𝑜𝑠 𝑣𝑒𝑟𝑖𝑓𝑖𝑐𝑎𝑟 𝑠𝑒, 𝑛𝑜𝑠 𝑝𝑜𝑛𝑡𝑜𝑠 𝑜𝑛𝑑𝑒 ℎ é 𝑑𝑒𝑠𝑐𝑜𝑛𝑡í𝑛𝑢𝑎, 𝑡𝑒𝑚𝑜𝑠 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙. −𝐷𝑖𝑧𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑎 𝑟𝑒𝑡𝑎 𝑥 = 𝑎 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑠𝑒 𝑜𝑐𝑜𝑟𝑟𝑒𝑟 𝑢𝑚 𝑑𝑜𝑠 𝑠𝑒𝑔𝑢𝑖𝑛𝑡𝑒𝑠 𝑐𝑎𝑠𝑜𝑠: lim 𝑥→𝑎+ ℎ(𝑥) = ±∞ 𝑜𝑢 lim 𝑥→𝑎− ℎ(𝑥) = ±∞ lim 𝑥→−2+ ℎ(𝑥) = lim 𝑥→−2+ 1 (𝑥 + 2)(𝑥 + 3) = lim 𝑥→−2+ 1 (𝑥 + 2)⏟ ↓ 0+ (𝑥 + 3)⏟ ↓ 1 = +∞ 𝑂𝑏𝑠: 𝑠𝑒 𝑥 → −2+, 𝑒𝑛𝑡ã𝑜 𝑥 > −2 ∴ 𝑥 + 2 > 0. lim 𝑥→−2− ℎ(𝑥) = lim 𝑥→−2− 1 (𝑥 + 2)(𝑥 + 3) = lim 𝑥→−2+ 1 (𝑥 + 2)⏟ ↓ 0− (𝑥 + 3)⏟ ↓ 1 = −∞ 71 𝑂𝑏𝑠: 𝑠𝑒 𝑥 → −2−, 𝑒𝑛𝑡ã𝑜 𝑥 < −2 ∴ 𝑥 + 2 < 0. ∗ 𝐿𝑜𝑔𝑜, 𝑎 𝑟𝑒𝑡𝑎 𝑥 = −2 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑎𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑒 ℎ(𝑥). lim 𝑥→−3+ ℎ(𝑥) = lim 𝑥→−3+ 1 (𝑥 + 2)(𝑥 + 3) = lim 𝑥→−3+ 1 (𝑥 + 2)⏟ ↓ −1 (𝑥 + 3)⏟ ↓ 0+ = −∞ 𝑂𝑏𝑠: 𝑠𝑒 𝑥 → −3+, 𝑒𝑛𝑡ã𝑜 𝑥 > −3 ∴ 𝑥 + 3 > 0. lim 𝑥→−3− ℎ(𝑥) = lim 𝑥→−3− 1 (𝑥 + 2)(𝑥 + 3) = lim 𝑥→−3+ 1 (𝑥 + 2)⏟ ↓ −1 (𝑥 + 3)⏟ ↓ 0− = +∞ 𝑂𝑏𝑠: 𝑠𝑒 𝑥 → −3−, 𝑒𝑛𝑡ã𝑜 𝑥 < −3 ∴ 𝑥 + 3 < 0. ∗ 𝐿𝑜𝑔𝑜, 𝑎 𝑟𝑒𝑡𝑎 𝑥 = −3 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑎𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑒 ℎ(𝑥). 72 3.8 2ª Prova-03 de Outubro de 2008 1. 𝑎) 𝑓(𝑥) = ln|cos sec(3𝑥) . cotg(3𝑥)| . 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟 𝑓′(𝑥). 𝑓′(𝑥) = 𝐷𝑥[cos sec(3𝑥) . cotg(3𝑥)] cos sec(3𝑥) . cotg(3𝑥) 𝑓′(𝑥) = −3 cos sec(3𝑥) . cotg2(3𝑥) − 3 cos sec3(3𝑥) cos sec(3𝑥) . cotg(3𝑥) 𝑓′(𝑥) = −3cos sec(3𝑥)[cotg2(3𝑥) + cos sec²(3𝑥)] cos sec(3𝑥) . cotg(3𝑥) ; 1 + cotg2(𝑥) = cos sec²(𝑥) 𝑓′(𝑥) = −3[2 cos sec2(3𝑥) − 1] cotg(3𝑥) . 𝑏) 𝑥 sen 𝑦 + cos 2𝑦 = sen 𝑥 ; 𝐸𝑞𝑢𝑎çã𝑜 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 𝑛𝑜 𝑝𝑜𝑛𝑡𝑜 (0, 𝜋 4⁄ ). ∗ 𝑃𝑜𝑟 𝑑𝑒𝑟𝑖𝑣𝑎çã𝑜 𝑖𝑚𝑝𝑙í𝑐𝑖𝑡𝑎, 𝑡𝑒𝑚𝑜𝑠: sen 𝑦 + 𝑥𝑦′. cos 𝑦 + 2𝑦′. (− sen 2𝑦) = cos 𝑥 𝑦′. (𝑥. cos 𝑦 − 2 cos 2𝑦) = cos 𝑥 − sen 𝑦 𝑦′ = cos 𝑥 − sen 𝑦 𝑥. cos 𝑦 − 2 sen 2𝑦 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑖𝑛𝑑𝑜 𝑜 𝑝𝑜𝑛𝑡𝑜 𝑒𝑚 𝑞𝑢𝑒𝑠𝑡ã𝑜 𝑛𝑎 𝑒𝑥𝑝𝑟𝑒𝑠𝑠ã𝑜 𝑑𝑎 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎, 𝑡𝑒𝑚𝑜𝑠: 𝑦′ = cos 0 − sen 𝜋 4 −2. sen 𝜋 2 = 1 − √2 2 −2 = √2 − 2 4 . 𝐸𝑞𝑢𝑎çã𝑜 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒: 𝑦 − 𝜋 4 = √2 − 2 4 (𝑥 − 0) 𝑦 = √2 − 2 4 𝑥 + 𝜋 4 𝑜𝑢 𝑦 = (√2 − 2)𝑥 + 𝜋 4 2. 𝑓(𝑥) = √𝑥 + 1(2 − 𝑥)5 (𝑥 + 3)4 ; 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟 𝑓′(0) 𝑝𝑜𝑟 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎çã𝑜 𝑙𝑜𝑔𝑎𝑟í𝑡𝑚𝑖𝑐𝑎. ln 𝑓(𝑥) = ln √𝑥 + 1(2 − 𝑥)5 (𝑥 + 3)4 ln 𝑓(𝑥) = ln √𝑥 + 1 + ln(2 − 𝑥)5 − ln(𝑥 + 3)4 ln 𝑓(𝑥) = 1 2 ln(𝑥 + 1) + 5 ln(2 − 𝑥) − 4 ln(𝑥 + 3) 𝑓′(𝑥) 𝑓(𝑥) = 1 2 ∙ 1 (𝑥 + 1) + 5 ∙ (−1) (2 − 𝑥) − 4 ∙ 1 (𝑥 + 3) 𝑓′(𝑥) = 𝑓(𝑥) ∙ [ 1 2(𝑥 + 1) − 5 (2 − 𝑥) − 4 (𝑥 + 3) ] 𝑓′(𝑥) = √𝑥 + 1(2 − 𝑥)5 (𝑥 + 3)4 [ 1 2(𝑥 + 1) − 5 (2 − 𝑥) − 4 (𝑥 + 3) ] 𝑓′(0) = √0 + 1(2 − 0)5 (0 + 3)4 [ 1 2(0 + 1) − 5 (2 − 0) − 4 (0 + 3) ] = √1(2)5 34 [ 1 2 − 5 2 − 4 3 ] = 32 81 [−2 − 4 3 ] = 32 81 ∙ (− 10 3 ) = − 320 243 . 75 3.9 2ª Avaliação-04 de Outubro de 2008 1. 𝑦 = cos2 𝑥 ∙ tg4 𝑥 (𝑒2𝑥 + 1)3 ; 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟 𝑦′𝑝𝑜𝑟 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎çã𝑜 𝑙𝑜𝑔𝑎𝑟í𝑡𝑚𝑖𝑐𝑎. ln 𝑦 = ln cos2 𝑥 ∙ tg4 𝑥 (𝑒2𝑥 + 1)3 ln 𝑦 = ln(cos2 𝑥) + ln(tg4 𝑥) − ln(𝑒2𝑥 + 1)3 ln 𝑦 = 2. ln(cos 𝑥) + 4. ln(tg 𝑥) − 3. ln(𝑒2𝑥 + 1) 𝑦′ 𝑦 = −2. sen 𝑥 cos 𝑥 + 4. sec2 𝑥 tg 𝑥 − 3 2. 𝑒2𝑥 (𝑒2𝑥 + 1) 𝑦′ = 𝑦 [−2. tg 𝑥 + 4(cotg 𝑥 + tg 𝑥) − 6. 𝑒2𝑥 (𝑒2𝑥 + 1) ] 𝑦′ = 𝑦 ∙ [−2. tg 𝑥 + 4(cotg 𝑥 + tg 𝑥) − 6. 𝑒2𝑥 (𝑒2𝑥 + 1) ] 𝑦′ = 𝑦 ∙ [2. tg 𝑥 + 4. cotg 𝑥 − 6. 𝑒2𝑥 (𝑒2𝑥 + 1) ] 𝑦′ = 2𝑦 ∙ [tg 𝑥 + 2. cotg 𝑥 − 3. 𝑒2𝑥 (𝑒2𝑥 + 1) ] 𝑦′ = 2 ∙ cos2 𝑥 ∙ tg4 𝑥 (𝑒2𝑥 + 1)3 ∙ [tg 𝑥 + 2. cotg 𝑥 − 3. 𝑒2𝑥 (𝑒2𝑥 + 1) ]. 2. 𝑥2𝑦2 + 𝑥𝑦 = 2 ; 𝐸𝑠𝑡𝑎 𝑐𝑢𝑟𝑣𝑎 𝑝𝑜𝑠𝑠𝑢𝑖 𝑎𝑙𝑔𝑢𝑚 𝑝𝑜𝑛𝑡𝑜 𝑜𝑛𝑑𝑒 𝑦′ = 0 ? ∗ 𝐷𝑒𝑟𝑖𝑣𝑎𝑛𝑑𝑜 𝑖𝑚𝑝𝑙𝑖𝑐𝑖𝑡𝑎𝑚𝑒𝑛𝑡𝑒, 𝑡𝑒𝑚𝑜𝑠: 2𝑥𝑦2 + 2𝑦𝑦′𝑥2 + 𝑦 + 𝑥𝑦′ = 0 𝑦′(2𝑥2𝑦 + 𝑥) = −(2𝑥𝑦2 + 𝑦) 𝑦′ = − 2𝑥𝑦2 + 𝑦 2𝑥2𝑦 + 𝑥 = − 𝑦(2𝑥𝑦 + 1) 𝑥(2𝑥𝑦 + 1) ∴ 𝑦′ = − 𝑦 𝑥 ; 𝑦′ = 0 ⇔ 𝑦 = 0. ∗ 𝑁𝑜𝑡𝑒 𝑞𝑢𝑒 𝑝𝑎𝑟𝑎 𝑦 = 0 𝑡𝑒𝑚𝑜𝑠 𝑎 𝑠𝑒𝑔𝑢𝑖𝑛𝑡𝑒 𝑒𝑥𝑝𝑟𝑒𝑠𝑠ã𝑜 𝑛𝑎 𝑐𝑢𝑟𝑣𝑎: 𝑥2. 02 + 𝑥. 0 = 2 0 = 2 (𝑈𝑚 𝑎𝑏𝑠𝑢𝑟𝑑𝑜!), 0 ≠ 2. ∗ 𝐼𝑠𝑡𝑜 𝑖𝑚𝑝𝑙𝑖𝑐𝑎 𝑑𝑖𝑧𝑒𝑟, 𝑞𝑢𝑒 𝑎 𝑐𝑢𝑟𝑣𝑎 𝑎𝑐𝑖𝑚𝑎 𝑛ã𝑜 𝑝𝑜𝑠𝑠𝑢𝑖 𝑎𝑙𝑔𝑢𝑚 𝑝𝑜𝑛𝑡𝑜 𝑜𝑛𝑑𝑒 𝑦′ = 0 𝑒, 𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑛ã𝑜 𝑒𝑥𝑖𝑠𝑡𝑒 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑞𝑢𝑒 𝑠𝑒𝑟𝑖𝑎 𝑝𝑎𝑟𝑎𝑙𝑒𝑙𝑎 𝑎 𝑟𝑒𝑡𝑎 𝑦 = 5. 3. 𝑎) 𝑦 = 3cos(2𝑥) ; 𝐸𝑞𝑢𝑎çã𝑜 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 𝑒𝑚 𝑦 = 1 𝑐𝑜𝑚 0 ≤ 𝑥 ≤ 𝜋 2 . 𝑦 = 1 ⇒ 3cos(2𝑥) = 1 → 3cos(2𝑥) = 30 ∴ cos(2𝑥) = 0. ∗ 2𝑥 = 𝜋 2 ± 2𝑘𝜋 𝑜𝑢 2𝑥 = 3𝜋 2 ± 2𝑘𝜋 ∗ 𝑥 = 𝜋 4 ± 𝑘𝜋 𝑜𝑢 𝑥 = 3𝜋 4 ± 𝑘𝜋. 𝐶𝑜𝑚𝑜 0 ≤ 𝑥 ≤ 𝜋 2 , 𝑜𝑢 𝑠𝑒𝑗𝑎, 𝑥 é 𝑢𝑚 𝑎𝑟𝑐𝑜 𝑑𝑜 1º 𝑞𝑢𝑎𝑑𝑟𝑎𝑛𝑡𝑒, 𝑒𝑛𝑡ã𝑜 𝑥 = 𝜋 4 . ∗ 𝑆𝑒𝑗𝑎 𝑢 = 2𝑥 𝑒 𝑣 = cos 𝑢, 𝑒𝑛𝑡ã𝑜 𝑦 = 3𝑣 . 76 𝑃𝑒𝑙𝑎 𝑅𝑒𝑔𝑟𝑎 𝑑𝑎 𝐶𝑎𝑑𝑒𝑖𝑎, 𝑡𝑒𝑚𝑜𝑠: 𝑑𝑦 𝑑𝑥 = 𝑑𝑢 𝑑𝑥 ∙ 𝑑𝑣 𝑑𝑢 ∙ 𝑑𝑦 𝑑𝑣 𝑦′ = 2. (− sen 𝑢). 3𝑣. ln(3) 𝑦′ = −2. 3cos(2𝑥). sen(2𝑥) . ln(3) 𝑦′ ( 𝜋 4 ) = −2. 3cos( 𝜋 2 ). sen ( 𝜋 2 ) . ln(3) 𝑦′ ( 𝜋 4 ) = −2. 30. 1. ln(3) 𝑦′ ( 𝜋 4 ) = −2. ln(3) ∗ 𝐸𝑞𝑢𝑎çã𝑜 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 𝑛𝑜 𝑝𝑜𝑛𝑡𝑜 ( 𝜋 4 , 1): 𝑦 − 𝑦𝑜 = 𝑚(𝑥 − 𝑥𝑜) 𝑦 − 1 = −2. ln(3) (𝑥 − 𝜋 4 ) 𝑦 = −2. ln(3) 𝑥 + 𝜋 2 ln(3) + 1 𝑏) 𝑦 = (ln 𝑥)sen(2𝑥) ; 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟 𝑑𝑦 𝑑𝑥 . ln 𝑦 = sen(2𝑥) . ln(ln 𝑥) 𝑦′ 𝑦 = 2. cos(2𝑥). ln(ln 𝑥) + sen(2𝑥) . 1 𝑥 . 1 ln 𝑥 𝑦′ = 𝑦 [2. cos(2𝑥). ln(ln 𝑥) + sen(2𝑥) 𝑥 ln 𝑥 ] 𝑦′ = (ln 𝑥)sen(2𝑥) [2. cos(2𝑥). ln(ln 𝑥) + sen(2𝑥) 𝑥 ln 𝑥 ]. 4. 𝑎) 𝑦 = arcsen (√1 − 𝑥2) ; 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟 𝑑𝑦 𝑑𝑥 , 𝑠𝑒𝑛𝑑𝑜 0 < 𝑥 < 1. ∗ 𝑆𝑒𝑗𝑎𝑚 𝑢 = 1 − 𝑥2 𝑒 𝑣 = √𝑢. 𝐸𝑛𝑡ã𝑜 𝑦 = arcsen 𝑣. 𝑃𝑒𝑙𝑎 𝑅𝑒𝑔𝑟𝑎 𝑑𝑎 𝐶𝑎𝑑𝑒𝑖𝑎, 𝑡𝑒𝑚𝑜𝑠: 𝑑𝑦 𝑑𝑥 = 𝑑𝑢 𝑑𝑥 ∙ 𝑑𝑣 𝑑𝑢 ∙ 𝑑𝑦 𝑑𝑣 𝑦′ = (−2𝑥) ∙ 1 2√𝑢 ∙ 1 √1 − 𝑣2 𝑦′ = − 𝑥 √1 − 𝑥2 ∙ √1 − 1 + 𝑥2 𝑦′ = − 𝑥 √1 − 𝑥2 ∙ √𝑥2 ; 𝑜𝑏𝑠: √𝑥2 = |𝑥| 𝑦′ = − 𝑥 |𝑥|√1 − 𝑥2 𝑏) 𝑓(𝑥) = log𝜋 𝑥 𝑥2 ; 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟 𝑓′(1). 𝑓′(𝑥) = 𝐷𝑥[log𝜋 𝑥]. 𝑥 2 − 𝐷𝑥[𝑥 2]. log𝜋 𝑥 (𝑥2)2 77 𝑓′(𝑥) = 1 𝑥 ln 𝜋 ∙ 𝑥2 − (2𝑥). log𝜋 𝑥 𝑥4 𝑓′(𝑥) = 1 − 2 log𝜋 𝑥 𝑥3. ln 𝜋 ∴ 𝑓′(1) = 1 − 2 log𝜋 1 13. ln 𝜋 = 1 − 2.0 ln 𝜋 = 1 ln 𝜋 . 5. 𝑎)𝑦 = 𝑥 tgh(√𝑥) ; 𝐸𝑞𝑢𝑎çã𝑜 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑛𝑜𝑟𝑚𝑎𝑙 𝑛𝑜 𝑝𝑜𝑛𝑡𝑜 (0,0). ∗ 𝐴 𝑖𝑛𝑐𝑙𝑖𝑛𝑎çã𝑜 (𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒 𝑎𝑛𝑔𝑢𝑙𝑎𝑟) 𝑚𝑛 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑛𝑜𝑟𝑚𝑎𝑙 é 𝑜 𝑖𝑛𝑣𝑒𝑟𝑠𝑜 𝑠𝑖𝑚é𝑡𝑟𝑖𝑐𝑜 𝑑𝑜 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 𝑛𝑎𝑞𝑢𝑒𝑙𝑒 𝑝𝑜𝑛𝑡𝑜. 𝐿𝑜𝑔𝑜, 𝑚𝑛 = 1 𝑦′ 𝑦′ = 𝐷𝑥[𝑥]. tgh(√𝑥) + 𝑥. 𝐷𝑥[tgh(√𝑥)] 𝑦′ = tgh(√𝑥) + 𝑥. ( 1 2√𝑥 ) . sech2(√𝑥) ∗ 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑛𝑑𝑜 𝑜 𝑣𝑎𝑙𝑜𝑟 𝑑𝑒 𝑦′ 𝑛𝑜 𝑝𝑜𝑛𝑡𝑜 (0, 0), 𝑡𝑒𝑚𝑜𝑠: 𝑦′(0) = tgh(√0) + 𝑥. ( 1 2√0 ) . sech2(√𝑥) ; ∗ 𝑁𝑜𝑡𝑒 𝑞𝑢𝑒 ℎá 𝑢𝑚𝑎 𝑖𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎çã𝑜 𝑛𝑎 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎 𝑝𝑎𝑟𝑎 𝑥 = 0. ∗ 𝐴𝑛𝑎𝑙𝑖𝑠𝑎𝑛𝑑𝑜 𝑐𝑜𝑚 𝑐𝑢𝑖𝑑𝑎𝑑𝑜, 𝑛𝑜𝑡𝑎𝑚𝑜𝑠 𝑞𝑢𝑒 𝑦′ = tgh(√𝑥) + 𝑥. ( 1 2√𝑥 ) . sech2(√𝑥) 𝑠ó 𝑒𝑥𝑖𝑠𝑡𝑒 𝑝𝑎𝑟𝑎 𝑥 > 0. 𝐿𝑜𝑔𝑜, 𝑦 𝑠ó 𝑝𝑜𝑠𝑠𝑢𝑖 𝑦′ + (0) 𝑒 𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑐𝑎𝑙𝑐𝑢𝑙á − 𝑙𝑜. lim 𝑥→0+ 𝑦′ = lim 𝑥→0+ [tgh(√𝑥) + 𝑥. ( 1 2√𝑥 ) . sech2(√𝑥)] = lim 𝑥→0+ tgh(√𝑥) + lim 𝑥→0+ 𝑥. sech2(√𝑥) 2√𝑥 ∗ 𝑂 𝑣𝑎𝑙𝑜𝑟 𝑑𝑎 𝑝𝑟𝑖𝑚𝑒𝑖𝑟𝑎 𝑝𝑎𝑟𝑐𝑒𝑙𝑎 𝑑𝑎 𝑒𝑥𝑝𝑟𝑒𝑠𝑠ã𝑜 𝑎𝑐𝑖𝑚𝑎 é 𝑧𝑒𝑟𝑜! ∗ 𝐴 𝑖𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎çã𝑜 𝑒𝑚 𝑦′ 𝑎𝑝𝑎𝑟𝑒𝑐𝑒 𝑑𝑒𝑣𝑖𝑑𝑜 à 𝑠𝑒𝑔𝑢𝑛𝑑𝑎 𝑝𝑎𝑟𝑐𝑒𝑙𝑎. lim 𝑥→0+ 𝑥. sech2(√𝑥) 2√𝑥 ; 𝑠𝑒 𝑥 → 0+ 𝑒𝑛𝑡ã𝑜 𝑥 > 0. 𝐿𝑜𝑔𝑜, 𝑥 √𝑥 = √𝑥. lim 𝑥→0+ 𝑥. sech2(√𝑥) 2√𝑥 = lim 𝑥→0+ √𝑥. sech2(√𝑥) 2 = 0.1 2 = 0. ∗ 𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑦′ + (0) = 0.𝑁𝑜 𝑒𝑛𝑡𝑎𝑛𝑡𝑜, 𝑦′ − (0) ∄. 𝐼𝑠𝑡𝑜 𝑖𝑚𝑝𝑙𝑖𝑐𝑎 𝑑𝑖𝑧𝑒𝑟 𝑞𝑢𝑒 𝑦′(0) ∄. 𝐿𝑜𝑔𝑜, 𝑛ã𝑜 ℎá 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑛𝑜 𝑝𝑜𝑛𝑡𝑜 (0, 0) 𝑒, 𝑐𝑜𝑛𝑠𝑒𝑞𝑢𝑒𝑛𝑡𝑒𝑚𝑒𝑛𝑡𝑒, 𝑛ã𝑜 ℎá 𝑟𝑒𝑡𝑎 𝑛𝑜𝑟𝑚𝑎𝑙 𝑛𝑒𝑠𝑡𝑒 𝑝𝑜𝑛𝑡𝑜! ∗ 𝑂𝑏𝑠: 𝐶𝑎𝑠𝑜 𝑝𝑒𝑛𝑠á𝑠𝑠𝑒𝑚𝑜𝑠 𝑒𝑚 𝑢𝑠𝑎𝑟 𝑐𝑜𝑚𝑜 𝑟𝑒𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎 𝑑𝑒 𝑐𝑜𝑒𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑒 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑜 𝑣𝑎𝑙𝑜𝑟 𝑎𝑠𝑠𝑢𝑚𝑖𝑑𝑜 𝑝𝑜𝑟 𝑦′ + (0), 𝑒𝑠𝑡𝑎𝑟í𝑎𝑚𝑜𝑠 𝑐𝑜𝑚𝑒𝑡𝑒𝑛𝑑𝑜 𝑢𝑚 𝑔𝑟𝑎𝑛𝑑𝑒 𝑒𝑞𝑢í𝑣𝑜𝑐𝑜, 𝑝𝑜𝑖𝑠, 𝑒𝑠𝑠𝑎 𝑓𝑢𝑛çã𝑜 𝑛ã𝑜 é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 𝑥 = 0, 𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑛ã𝑜 é 𝑑𝑒𝑟𝑖𝑣á𝑣𝑒𝑙 𝑛𝑒𝑠𝑡𝑒 𝑝𝑜𝑛𝑡𝑜. 𝑏) 𝑓(𝑥) = arccos (𝑥 − 𝑥3 3 ) ; 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟 𝑓′′ ( 1 2 ). 𝑓′(𝑥) = (1 − 𝑥2) ∙ −1 √1 − (𝑥 − 𝑥3 3 ) 2 80 2 cos 𝑥 = 0 ; 𝑥 = − 𝜋 2 𝑒 𝑥 = 𝜋 2 . 𝐿𝑒𝑚𝑏𝑟𝑒 𝑞𝑢𝑒 𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 é [−𝜋, 𝜋]. 1 − 2 sen 𝑥 = 0 → sen 𝑥 = 1 2 ; 𝑥 = 𝜋 6 𝑒 𝑥 = 5𝜋 6 . ∗ 𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑜𝑠 𝑛ú𝑚𝑒𝑟𝑜𝑠 𝑐𝑟í𝑡𝑖𝑐𝑜𝑠 𝑑𝑒 𝑓 𝑠ã𝑜: − 𝜋 2 , 𝜋 6 , 𝜋 2 𝑒 5𝜋 6 . 𝑏) 𝑃𝑒𝑙𝑜 𝑀é𝑡𝑜𝑑𝑜 𝑑𝑜 𝐼𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 𝐹𝑒𝑐ℎ𝑎𝑑𝑜, 𝑝𝑎𝑟𝑎 𝑒𝑛𝑐𝑜𝑛𝑡𝑟𝑎𝑟𝑚𝑜𝑠 𝑜𝑠 𝑣𝑎𝑙𝑜𝑟𝑒𝑠 𝑚á𝑥𝑖𝑚𝑜 𝑒 𝑚í𝑛𝑖𝑚𝑜 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑜𝑠 𝑑𝑒 𝑢𝑚𝑎 𝑓𝑢𝑛çã𝑜 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑓 𝑒𝑚 𝑢𝑚 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 𝑓𝑒𝑐ℎ𝑎𝑑𝑜 [𝑎, 𝑏]. 1) 𝐸𝑛𝑐𝑜𝑛𝑡𝑟𝑎𝑟 𝑜𝑠 𝑣𝑎𝑙𝑜𝑟𝑒𝑠 𝑑𝑒 𝑓 𝑛𝑜𝑠 𝑛ú𝑚𝑒𝑟𝑜𝑠 𝑐𝑟í𝑡𝑖𝑐𝑜𝑠 𝑑𝑒 𝑓 𝑒𝑚 (𝑎, 𝑏); 2) 𝐸𝑛𝑐𝑜𝑛𝑡𝑟𝑎𝑟 𝑜𝑠 𝑣𝑎𝑙𝑜𝑟𝑒𝑠 𝑑𝑒 𝑓 𝑛𝑜𝑠 𝑒𝑥𝑡𝑟𝑒𝑚𝑜𝑠 𝑑𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜; 3) 𝑂 𝑚𝑎𝑖𝑜𝑟 𝑣𝑎𝑙𝑜𝑟 𝑒𝑛𝑐𝑜𝑛𝑡𝑟𝑎𝑑𝑜 𝑒𝑚 1 𝑒 2 é 𝑜 𝑣𝑎𝑙𝑜𝑟 𝑚á𝑥𝑖𝑚𝑜 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑜, 𝑒 𝑜 𝑚𝑒𝑛𝑜𝑟 𝑑𝑜𝑠 𝑣𝑎𝑙𝑜𝑟𝑒𝑠 é 𝑜 𝑚í𝑛𝑖𝑚𝑜 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑜. ∗ 1) 𝑓 (− 𝜋 2 ) = 2 sen (− 𝜋 2 ) + cos (−𝜋) = −2 − 1 = −3. 𝑓 ( 𝜋 6 ) = 2 sen 𝜋 6 + cos 𝜋 3 = 1 + 1 2 = 3 2 . 𝑓 ( 𝜋 2 ) = 2 sen 𝜋 2 + cos 𝜋 = 2 − 1 = 1. 𝑓 ( 5𝜋 6 ) = 2 sen 5𝜋 6 + cos 5𝜋 3 = 1 + 1 2 = 3 2 . ∗ 2) 𝑓(𝜋) = 2sen 𝜋 + cos 2𝜋 = 0 + 1 = 1. 𝑓(−𝜋) = 2 sen(−𝜋) + cos(−2𝜋) = 0 + 1 = 1. ∗ 3) 𝐶𝑜𝑚 𝑎 𝑎𝑛á𝑙𝑖𝑠𝑒 𝑎𝑐𝑖𝑚𝑎 𝑡𝑒𝑚𝑜𝑠, 𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑜 𝑣𝑎𝑙𝑜𝑟 𝑚á𝑥𝑖𝑚𝑜 é 𝑓 ( 𝜋 6 ) = 𝑓 ( 5𝜋 6 ) = 3 2 𝑒 𝑜 𝑣𝑎𝑙𝑜𝑟 𝑚í𝑛𝑖𝑚𝑜 é 𝑓 (− 𝜋 2 ) = −3. 3. 𝑎) 𝑓(𝑥) = 𝑥 + 3 𝑥 − 2 ; 𝑝𝑟𝑜𝑣𝑎𝑟 𝑞𝑢𝑒 𝑛ã𝑜 𝑒𝑥𝑖𝑠𝑡𝑒 𝑐 ∈ [0,5] 𝑡𝑎𝑙 𝑞𝑢𝑒 𝑓′(𝑐) = 𝑓(5) − 𝑓(0) 5 − 0 . 𝑓′(𝑐) = 8 3 + 3 2 5 = 25 30 = 5 6 . 𝑓′(𝑥) = 𝑥 − 2 − 𝑥 − 3 (𝑥 − 2)2 = − 5 (𝑥 − 2)2 ∗ 𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑛𝑑𝑜 𝑎 𝑠𝑒𝑔𝑢𝑖𝑛𝑡𝑒 𝑖𝑔𝑢𝑎𝑙𝑑𝑎𝑑𝑒: − 5 (𝑐 − 2)2 = 5 6 → (𝑐 − 2)2 = −6 → (𝑐 − 2) = √−6 . ∗ 𝐿𝑜𝑔𝑜, 𝑛ã𝑜 ℎá 𝑐 ∈ ℝ 𝑞𝑢𝑒 𝑠𝑎𝑡𝑖𝑠𝑓𝑎ç𝑎 𝑎 𝑒𝑞𝑢𝑎çã𝑜 𝑎𝑐𝑖𝑚𝑎. ∗ 𝑃𝑜𝑟 𝑞𝑢𝑒 𝑖𝑠𝑠𝑜 𝑛ã𝑜 𝑐𝑜𝑛𝑡𝑟𝑎𝑑𝑖𝑧 𝑜 𝑇𝑒𝑜𝑟𝑒𝑚𝑎 𝑑𝑜 𝑉𝑎𝑙𝑜𝑟 𝑀é𝑑𝑖𝑜? −𝑆𝑒𝑗𝑎 𝑓 𝑢𝑚𝑎 𝑓𝑢𝑛çã𝑜 𝑞𝑢𝑒 𝑠𝑎𝑡𝑖𝑠𝑓𝑎ç𝑎 𝑎𝑠 𝑠𝑒𝑔𝑢𝑖𝑛𝑡𝑒 ℎ𝑖𝑝ó𝑡𝑒𝑠𝑒𝑠: 1) 𝑓 é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 𝑓𝑒𝑐ℎ𝑎𝑑𝑜 [𝑎, 𝑏]; 2) 𝑓 é 𝑑𝑒𝑟𝑖𝑣á𝑣𝑒𝑙 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 𝑎𝑏𝑒𝑟𝑡𝑜 (𝑎, 𝑏); 𝐸𝑛𝑡ã𝑜 𝑒𝑥𝑖𝑠𝑡𝑒 𝑎𝑙𝑔𝑢𝑚 𝑐 ∈ (𝑎, 𝑏) 𝑡𝑎𝑙 𝑞𝑢𝑒 𝑓′(𝑐) = 𝑓(𝑏) − 𝑓(𝑎) 𝑏 − 𝑎 . 1) 𝐴 𝑓𝑢𝑛çã𝑜 𝑓(𝑥) = 𝑥 + 3 𝑥 − 2 𝑛ã𝑜 é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 [0, 5], 𝑝𝑜𝑖𝑠, 𝑛𝑜𝑡𝑒 𝑞𝑢𝑒 𝑥 = 2 𝑛ã𝑜 81 𝑝𝑒𝑟𝑡𝑒𝑛𝑐𝑒 𝑎𝑜 𝑑𝑜𝑚í𝑛𝑖𝑜 𝑑𝑒 𝑓. 𝐿𝑜𝑔𝑜, 𝑓 é 𝑑𝑒𝑠𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 𝑥 = 2. 2) 𝐴 𝑓𝑢𝑛çã𝑜 𝑓(𝑥) = 𝑥 + 3 𝑥 − 2 𝑛ã𝑜 é 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖á𝑣𝑒𝑙 𝑒𝑚 (0,5), 𝑝𝑜𝑖𝑠, 𝑓′(𝑥) = − 5 (𝑥 − 2)2 𝑠ó 𝑒𝑥𝑖𝑠𝑡𝑒 𝑝𝑎𝑟𝑎 𝑥 ≠ 2, 𝑒 2 ∈ (0,5). ∗ 𝐿𝑜𝑔𝑜, 𝑜 𝑓𝑎𝑡𝑜 𝑑𝑒 𝑛ã𝑜 𝑒𝑥𝑖𝑠𝑡𝑖𝑟 𝑢𝑚 𝑐 ∈ [0,5] 𝑡𝑎𝑙 𝑞𝑢𝑒 𝑓′(𝑐) = 𝑓(5) − 𝑓(0) 5 − 0 𝑛ã𝑜 𝑐𝑜𝑛𝑡𝑟𝑎𝑑𝑖𝑧 𝑜 𝑇𝑒𝑜𝑟𝑒𝑚𝑎 𝑑𝑜 𝑉𝑎𝑙𝑜𝑟 𝑀é𝑑𝑖𝑜! 𝑏) 𝑓(𝑥) = 𝑥³ + 9𝑥² + 33𝑥 − 8 𝑝𝑜𝑠𝑠𝑢𝑖 𝑒𝑥𝑎𝑡𝑎𝑚𝑒𝑛𝑡𝑒 𝑢𝑚𝑎 𝑟𝑎í𝑧 𝑟𝑒𝑎𝑙. 𝑓(0) = −8 𝑒 𝑓(1) = 35. 𝑆𝑒𝑗𝑎 𝑐 ∈ ℝ 𝑡𝑎𝑙 𝑞𝑢𝑒 𝑓(𝑐) = 0. ∗ 𝑓 é 𝑢𝑚𝑎 𝑓𝑢𝑛çã𝑜 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑎𝑙 𝑒, 𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 [0, 1] , 𝑒 𝑓(0) < 0 < 𝑓(1), 𝑝𝑜𝑑𝑒𝑚𝑜𝑠 𝑔𝑎𝑟𝑎𝑛𝑡𝑖𝑟 𝑝𝑒𝑙𝑜 𝑇𝑒𝑜𝑟𝑒𝑚𝑎 𝑑𝑜 𝑉𝑎𝑙𝑜𝑟 𝐼𝑛𝑡𝑒𝑟𝑚𝑒𝑑𝑖á𝑟𝑖𝑜 𝑞𝑢𝑒 𝑒𝑥𝑖𝑠𝑡𝑒 𝑢𝑚 𝑛ú𝑚𝑒𝑟𝑜 𝑐 , 𝑐 ∈ (0, 1) 𝑡𝑎𝑙 𝑞𝑢𝑒 𝑓(𝑐) = 0. 𝐶𝑜𝑚 𝑖𝑠𝑠𝑜, 𝑝𝑟𝑜𝑣𝑎𝑚𝑜𝑠 𝑎 𝑒𝑥𝑖𝑠𝑡ê𝑛𝑐𝑖𝑎 𝑑𝑒 𝑢𝑚𝑎 𝑟𝑎í𝑧 𝑟𝑒𝑎𝑙 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 (0,1). ∗ 𝑆𝑢𝑝𝑜𝑛ℎ𝑎𝑚𝑜𝑠 𝑞𝑢𝑒 𝑓 𝑡𝑒𝑛ℎ𝑎 2 𝑟𝑎í𝑧𝑒𝑠 𝑟𝑒𝑎𝑖𝑠, 𝑜𝑢 𝑠𝑒𝑗𝑎, 𝑓(𝑐) = 𝑓(𝑏) = 0, 𝑐𝑜𝑚 𝑏 ≠ 𝑐. 𝐶𝑜𝑚𝑜 𝑓 é 𝑢𝑚𝑎 𝑓𝑢𝑛çã𝑜 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒 𝑑𝑒𝑟𝑖𝑣á𝑣𝑒𝑙 𝑛𝑜𝑠 𝑟𝑒𝑎𝑖𝑠, 𝑝𝑒𝑙𝑜 𝑇𝑒𝑜𝑟𝑒𝑚𝑎 𝑑𝑜 𝑅𝑜𝑙𝑙𝑒, 𝑒𝑥𝑖𝑠𝑡𝑒 𝑢𝑚 𝑛ú𝑚𝑒𝑟𝑜 𝑑 ∈ (𝑐, 𝑏) 𝑡𝑎𝑙 𝑞𝑢𝑒 𝑓′(𝑑) = 0. 𝑓′(𝑥) = 3𝑥2 + 18𝑥 + 33 𝑓′(𝑥) = 0 → 3𝑥2 + 18𝑥 + 33 = 0 𝑥2 + 6𝑥 + 11 = 0 ∆= 36 − 44 = −8. 𝐶𝑜𝑚 ∆< 0, 𝑡𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑓′(𝑥) 𝑛ã𝑜 𝑝𝑜𝑠𝑠𝑢𝑖 𝑟𝑎í𝑧 𝑟𝑒𝑎𝑙. ∗ 𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑓(𝑥) 𝑡𝑒𝑚 𝑛𝑜 𝑒𝑥𝑎𝑡𝑎𝑚𝑒𝑛𝑡𝑒 𝑢𝑚𝑎 𝑟𝑎í𝑧 𝑟𝑒𝑎𝑙. 4. 𝑦 = √1 + 𝑥³; 𝑑𝑦 𝑑𝑡 = 4𝑚/𝑠 𝑞𝑢𝑎𝑛𝑑𝑜 𝑦 = 3 𝑥³ = 𝑦² − 1 → 𝑥 = √𝑦2 − 1 3 . ∗ 𝑦 = 3 → √1 + 𝑥3 = 3 → 1 + 𝑥3 = 9 → 𝑥3 = 8 ∴ 𝑥 = 2. 𝑑𝑥 𝑑𝑡 = 𝑑𝑥 𝑑𝑦 ∙ 𝑑𝑦 𝑑𝑡 𝑑𝑥 𝑑𝑡 = 1 3 (2𝑦) 1 √(𝑦2 − 1)2 3 ∙ 4 𝑑𝑥 𝑑𝑡 = 8𝑦 3√(𝑦2 − 1)2 3 𝑑𝑥 𝑑𝑡 = 8.3 3√(32 − 1)2 3 𝑑𝑥 𝑑𝑡 = 8 √8² 3 = 8 4 = 2𝑚/𝑠 ∗ 𝐷𝑒 𝑓𝑜𝑟𝑚𝑎 𝑎𝑛á𝑙𝑜𝑔𝑎, 𝑡𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑦 é 𝑢𝑚𝑎 𝑓𝑢𝑛çã𝑜 𝑒𝑚 𝑥 𝑒 𝑎𝑚𝑏𝑜𝑠 𝑒𝑠𝑡ã𝑜 𝑒𝑚 𝑓𝑢𝑛çã𝑜 𝑑𝑜 𝑡𝑒𝑚𝑝𝑜, 𝑙𝑜𝑔𝑜: 𝑦(𝑡) = √1 + 𝑥(𝑡)3 → 𝑦′(𝑡) = 1 2 (3𝑥(𝑡)2. 𝑥′(𝑡)). 1 √1 + 𝑥(𝑡)3 82 4 = 1 2 (3. (2)2. 𝑥′(𝑡)). 1 √1 + 23 4 = 1 2 (12𝑥′(𝑡)). 1 3 24 = 12. 𝑥′(𝑡) → 𝑥′(𝑡) = 2𝑚/𝑠 5. ∗ 𝐷𝑒 𝑢𝑚𝑎 𝑒𝑠𝑓𝑒𝑟𝑎 𝑡𝑒𝑚𝑜𝑠 𝑎𝑠 𝑠𝑒𝑔𝑢𝑖𝑛𝑡𝑒𝑠 𝑒𝑥𝑝𝑟𝑒𝑠𝑠õ𝑒𝑠: 𝐴 = 4𝜋𝑅2 𝑒 𝑉 = 4 3 𝜋𝑅3 ∗ 𝑁𝑜 𝑒𝑛𝑢𝑛𝑐𝑖𝑎𝑑𝑜 𝑡𝑒𝑚𝑜𝑠 𝑑𝐴 𝑑𝑡 = 4𝑐𝑚2/𝑚𝑖𝑛 𝑒 𝑑𝑅 𝑑𝑡 = 0,1𝑐𝑚/𝑚𝑖𝑛 ∗ 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟 𝑑𝑉 𝑑𝑡 . 𝑑𝐴 𝑑𝑡 = 𝑑𝐴 𝑑𝑅 ∙ 𝑑𝑅 𝑑𝑡 4 = 8𝜋𝑅. (0,1) 𝑅 = 5 𝜋 𝑐𝑚 𝑑𝑉 𝑑𝑡 = 𝑑𝑉 𝑑𝑅 ∙ 𝑑𝑅 𝑑𝑡 𝑑𝑉 𝑑𝑡 = 4𝜋𝑅2. (0,1) 𝑑𝑉 𝑑𝑡 = 4𝜋. 25 𝜋2 . (0,1) 𝑑𝑉 𝑑𝑡 = 10 𝜋 𝑐𝑚³/𝑚𝑖𝑛 85 𝑏) lim 𝑥→+∞ (√𝑥2 + 25𝑥 − 𝑥) = lim 𝑥→+∞ (√𝑥2 + 25𝑥 − 𝑥)(√𝑥2 + 25𝑥 + 𝑥) (√𝑥2 + 25𝑥 + 𝑥) = lim 𝑥→+∞ 𝑥2 + 25𝑥 − 𝑥2 (√𝑥2 + 25𝑥 + 𝑥) = lim 𝑥→+∞ 25𝑥 √𝑥2 + 25𝑥 + 𝑥 = lim 𝑥→+∞ 25𝑥 |𝑥| √𝑥2 + 25𝑥 + 𝑥 |𝑥| = lim 𝑥→+∞ 25𝑥 𝑥 √𝑥2 + 25𝑥 √𝑥2 + 𝑥 𝑥 = lim 𝑥→+∞ 25 √1 + 25 𝑥 + 1 = 25 √1 + 0 + 1 = 25 1 + 1 = 25 2 . ∗ 𝑂𝑏𝑠: 𝑠𝑒 𝑥 → +∞ 𝑒𝑛𝑡ã𝑜 |𝑥| = 𝑥 ; |𝑥| = √𝑥2. 3. 𝑎) 𝑓(𝑥) = { 𝑎𝑥 + 𝑏, 𝑠𝑒 𝑥 < 2 2𝑥2 − 1, 𝑠𝑒 𝑥 ≥ 2 . ∗ 𝐷𝑖𝑧𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑢𝑚𝑎 𝑓𝑢𝑛çã𝑜 𝑓 é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑛𝑢𝑚 𝑝𝑜𝑛𝑡𝑜 𝑥 = 𝑎 𝑠𝑒, 𝑒 𝑠𝑜𝑚𝑒𝑛𝑡𝑒 𝑠𝑒, 𝑡𝑒𝑚𝑜𝑠: 1) 𝑓(𝑎) 𝑒𝑥𝑖𝑠𝑡𝑒; 2) lim 𝑥→𝑎 𝑓(𝑥) 𝑒𝑥𝑖𝑠𝑡𝑒; 3) lim 𝑥→𝑎 𝑓(𝑥) = 𝑓(𝑎). ∗ 𝑈𝑠𝑎𝑛𝑑𝑜 𝑒𝑠𝑠𝑎 𝑖𝑛𝑓𝑜𝑟𝑚𝑎çã𝑜 𝑒𝑚 𝑥 = 2, 𝑡𝑒𝑚𝑜𝑠: 1) 𝑓(2) = 2. (2)2 − 1 = 8 − 1 = 7. 2) lim 𝑥→2− 𝑓(𝑥) = lim 𝑥→2− (𝑎𝑥 + 𝑏) = 2𝑎 + 𝑏. 𝑂𝑏𝑠: 𝑁ã𝑜 ℎá 𝑛𝑒𝑐𝑒𝑠𝑠𝑖𝑑𝑎𝑑𝑒 𝑒𝑚 𝑠𝑒 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑟 𝑜 𝑙𝑖𝑚𝑖𝑡𝑒 𝑙𝑎𝑡𝑒𝑟𝑎𝑙 à 𝑑𝑖𝑟𝑒𝑖𝑡𝑎 𝑑𝑒 𝑥 = 2, 𝑝𝑜𝑖𝑠 lim 𝑥→2+ 𝑓(𝑥) = 𝑓(2). 3) lim 𝑥→2− 𝑓(𝑥) = 𝑓(2) → 2𝑎 + 𝑏 = 7 (𝐼) ∗ 𝐸𝑠𝑠𝑎 é 𝑎 𝑟𝑒𝑙𝑎çã𝑜 𝑒𝑛𝑡𝑟𝑒 𝑎 𝑒 𝑏 𝑝𝑎𝑟𝑎 𝑞𝑢𝑒 𝑓 𝑠𝑒𝑗𝑎 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 𝑥 = 2. 𝑏) 𝑃𝑎𝑟𝑎 𝑞𝑢𝑒 𝑓 𝑠𝑒𝑗𝑎 𝑑𝑒𝑟𝑖𝑣á𝑣𝑒𝑙 𝑒𝑚 𝑥 = 2: 𝑓′ + (2) = lim 𝑥→2+ 𝑓(𝑥) − 𝑓(2) 𝑥 − 2 = lim 𝑥→2+ 2𝑥2 − 1 − 7 𝑥 − 2 = lim 𝑥→2+ 2𝑥2 − 8 𝑥 − 2 = lim 𝑥→2+ 2(𝑥2 − 4) 𝑥 − 2 = lim 𝑥→2+ 2(𝑥 − 2)(𝑥 + 2) (𝑥 − 2) = lim 𝑥→2+ 2(𝑥 + 2) = 2(2 + 2) = 8. 𝑓′ − (2) = lim 𝑥→2− 𝑓(𝑥) − 𝑓(2) 𝑥 − 2 = lim 𝑥→2− 𝑎𝑥 + 𝑏 − 7 𝑥 − 2 = lim 𝑥→2− 𝑎𝑥 + 7 − 2𝑎 − 7 𝑥 − 2 = lim 𝑥→2− 𝑎(𝑥 − 2) (𝑥 − 2) = lim 𝑥→2− 𝑎 = 𝑎. ∗ 𝑃𝑎𝑟𝑎 𝑞𝑢𝑒 𝑓 𝑠𝑒𝑗𝑎 𝑑𝑒𝑟𝑖𝑣á𝑣𝑒𝑙 𝑒𝑚 𝑥 = 2 𝑑𝑒𝑣𝑒𝑚𝑜𝑠 𝑡𝑒𝑟 𝑓′ + (2) = 𝑓′ − (2). 𝐿𝑜𝑔𝑜, 𝑎 = 8. 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑖𝑛𝑑𝑜 𝑛𝑎 𝑒𝑞𝑢𝑎çã𝑜 (𝐼) 𝑡𝑒𝑚𝑜𝑠: 2. 𝑎 + 𝑏 = 7 → 16 + 𝑏 = 7 → 𝑏 = −9. ∗ 𝑆𝑜𝑙𝑢çã𝑜: 𝑎 = 8 𝑒 𝑏 = −9. 4. 𝑎) 𝑥³ + 𝑦³ = 6𝑥𝑦 ; 𝑒𝑞𝑢𝑎çã𝑜 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 𝑛𝑜 𝑝𝑜𝑛𝑡𝑜 (3,3). ∗ 𝑃𝑜𝑟 𝑑𝑒𝑟𝑖𝑣𝑎çã𝑜 𝑖𝑚𝑝𝑙í𝑐𝑖𝑡𝑎, 𝑡𝑒𝑚𝑜𝑠: 3𝑥2 + 3𝑦2𝑦′ = 6𝑦 + 6𝑥𝑦′ 86 𝑥2 + 𝑦2𝑦′ = 2𝑦 + 2𝑥𝑦′ 𝑦′. (𝑦2 − 2𝑥) = 2𝑦 − 𝑥2 𝑦′ = 2𝑦 − 𝑥² 𝑦² − 2𝑥 ; 𝑛𝑜 𝑝𝑜𝑛𝑡𝑜 (3,3) 𝑡𝑒𝑚𝑜𝑠 𝑦′ = 2.3 − 3² 3² − 2.3 = 6 − 9 9 − 6 = −3 3 = −1. ∗ 𝐸𝑞𝑢𝑎çã𝑜 𝑑𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒: 𝑦 − 𝑦𝑜 = 𝑚(𝑥 − 𝑥𝑜) 𝑦 − 3 = −1(𝑥 − 3) 𝑦 − 3 = −𝑥 + 3 𝑦 = −𝑥 + 6 𝑏)𝑂𝑛𝑑𝑒 𝑎 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 é ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙? (𝑜𝑛𝑑𝑒 𝑦′ = 0? ) 𝑦′ = 0 ⇔ 2𝑦 − 𝑥2 = 0 𝑒 𝑦2 − 2𝑥 ≠ 0 𝑥2 = 2𝑦 → 𝑦 = 𝑥2 2 . 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑖𝑛𝑑𝑜 𝑛𝑎 𝑒𝑥𝑝𝑟𝑒𝑠𝑠ã𝑜 𝑑𝑎 𝑐𝑢𝑟𝑣𝑎, 𝑡𝑒𝑚𝑜𝑠: 𝑥3 + ( 𝑥2 2 ) 3 = 6𝑥 ( 𝑥2 2 ) 𝑥3 + 𝑥6 8 = 3𝑥3 → 2𝑥3 = 𝑥6 8 → 𝑥6 − 16𝑥3 = 0 → 𝑥3(𝑥3 − 16) = 0 𝐿𝑜𝑔𝑜, { 𝑥 3 = 0 𝑥3 − 16 = 0 → 𝑥 = 0 𝑜𝑢 𝑥 = √16 3 . ∗ 𝑃𝑎𝑟𝑎 𝑥 = 0 𝑡𝑒𝑚𝑜𝑠: 03 + 𝑦3 = 6.0. 𝑦 → 𝑦3 = 0 → 𝑦 = 0. 𝑃𝑜𝑛𝑡𝑜 (0, 0). −𝑂𝑏𝑠: 𝑜 𝑝𝑜𝑛𝑡𝑜 (0, 0) 𝑛ã𝑜 𝑠𝑎𝑡𝑖𝑠𝑓𝑎𝑧 𝑎 𝑐𝑜𝑛𝑑𝑖çã𝑜 𝑦2 ≠ 2𝑥. 𝐿𝑜𝑔𝑜, 𝑛ã𝑜 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑒𝑚 𝑛𝑒𝑠𝑡𝑒 𝑝𝑜𝑛𝑡𝑜. ∗ 𝑃𝑎𝑟𝑎 𝑥 = √16 3 𝑡𝑒𝑚𝑜𝑠: (√16 3 ) 3 + 𝑦3 = 6. √16 3 . 𝑦 → 𝑦3 − 6√16 3 𝑦 + 16 = 0 −𝑁𝑎 𝑐𝑜𝑛𝑑𝑖çã𝑜 𝑎𝑐𝑖𝑚𝑎 𝑡𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑦2 ≠ 2𝑥. 𝑉𝑎𝑚𝑜 𝑠𝑢𝑝𝑜𝑟 𝑞𝑢𝑒 𝑦2 = 2𝑥. 𝐿𝑜𝑔𝑜, 𝑦 = √2√16 3 . 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑖𝑛𝑑𝑜 𝑛𝑎 𝑒𝑞𝑢𝑎çã𝑜 𝑎𝑐𝑖𝑚𝑎, 𝑡𝑒𝑚𝑜𝑠: (±√2√16 3 ) 3 − 6√16 3 (±√2√16 3 ) + 16 = 0 2√16 3 (±√2√16 3 ) − 6√16 3 (±√2√16 3 ) + 16 = 0 √2√16 3 (2√16 3 − 6√16 3 ) + 16 = 0 −4√16 3 √2√16 3 + 16 ≠ 0. 𝐿𝑜𝑔𝑜, 𝑡𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑦2 ≠ 2𝑥. 𝐴𝑠𝑠𝑖𝑚, 𝑡𝑒𝑚𝑜𝑠 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑒𝑚 𝑢𝑚 𝑝𝑜𝑛𝑡𝑜 𝑑𝑎 𝑐𝑢𝑟𝑣𝑎 𝑜𝑛𝑑𝑒 𝑥 = √16 3 . 𝐶𝑜𝑚𝑜𝑛𝑎 𝑒𝑥𝑝𝑟𝑒𝑠𝑠ã𝑜 𝑑𝑎 𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎 𝑡𝑖𝑛ℎ𝑎𝑚𝑜𝑠 𝑎 𝑐𝑜𝑛𝑑𝑖çã𝑜 𝑦 = 𝑥2 2 → 𝑦 = √162 3 2 . 87 ∗ 𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑡𝑒𝑚𝑜𝑠 𝑟𝑒𝑡𝑎 𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑒 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑛𝑜 𝑝𝑜𝑛𝑡𝑜 (√16 3 , √162 3 2 ). 5. 𝑎) 𝑦 = sen(tg√sen 𝑥) ; 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑟 𝑦′. ∗ 𝑆𝑒𝑗𝑎𝑚 𝑢 = sen 𝑥 , 𝑣 = √𝑢 𝑒 𝑧 = tg 𝑣 , 𝑒𝑛𝑡ã𝑜 𝑦 = sen 𝑧 𝑃𝑒𝑙𝑎 𝑅𝑒𝑔𝑟𝑎 𝑑𝑎 𝐶𝑎𝑑𝑒𝑖𝑎, 𝑡𝑒𝑚𝑜𝑠: 𝑑𝑦 𝑑𝑥 = 𝑑𝑢 𝑑𝑥 ∙ 𝑑𝑣 𝑑𝑢 ∙ 𝑑𝑧 𝑑𝑣 ∙ 𝑑𝑦 𝑑𝑧 𝑦′ = (cos 𝑥). 1 2√𝑢 . (sec2 𝑣). (cos 𝑧) 𝑦′ = cos 𝑥 . sec2(√sen 𝑥) . cos(tg √sen 𝑥) 2√sen 𝑥 𝑏) 𝑓(𝑥) = 23 𝑥2 ∗ 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟 𝑓′(1). 𝑆𝑒𝑗𝑎𝑚 𝑢 = 𝑥2, 𝑣 = 3𝑢 . 𝑒𝑛𝑡ã𝑜 𝑓(𝑣) = 2𝑣. 𝑃𝑒𝑙𝑎 𝑅𝑒𝑔𝑟𝑎 𝑑𝑎 𝐶𝑎𝑑𝑒𝑖𝑎, 𝑡𝑒𝑚𝑜𝑠: 𝑑𝑦 𝑑𝑥 = 𝑑𝑢 𝑑𝑥 ∙ 𝑑𝑣 𝑑𝑢 ∙ 𝑑𝑦 𝑑𝑣 𝑓′(𝑥) = (2𝑥) ∙ 3𝑢 ∙ ln(3) ∙ 2𝑣 ∙ ln(2) 𝑓′(𝑥) = (2𝑥) ∙ 3𝑥 2 ∙ ln(3) ∙ 23 𝑥2 ∙ ln(2) 𝑓′(1) = 2 ∙ 31 ∙ ln(3) ∙ 23 1 ∙ ln(2) 𝑓′(1) = 48 ∙ ln(3) ∙ ln(2) 90 é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 [0, 2], 𝑜 𝑇𝑒𝑜𝑟𝑒𝑚𝑎 𝑑𝑜 𝑉𝑎𝑙𝑜𝑟 𝐼𝑛𝑡𝑒𝑟𝑚𝑒𝑑𝑖á𝑟𝑖𝑜 𝑒𝑠𝑡𝑎𝑏𝑒𝑙𝑒𝑐𝑒 𝑞𝑢𝑒 𝑒𝑥𝑖𝑠𝑡𝑒 𝑢𝑚 𝑛ú𝑚𝑒𝑟𝑜 𝑐 𝑒𝑛𝑡𝑟𝑒 0 𝑒 2 𝑡𝑎𝑙 𝑞𝑢𝑒 𝑓(𝑐) = 0. 𝐸𝑚 𝑜𝑢𝑡𝑟𝑎𝑠 𝑝𝑎𝑙𝑎𝑣𝑟𝑎𝑠, 𝑎 𝑓𝑢𝑛çã𝑜 𝑓(𝑥) = 𝑥³ − 𝑥2 + 𝑥 − 1 𝑡𝑒𝑚 𝑝𝑒𝑙𝑜 𝑚𝑒𝑛𝑜𝑠 𝑢𝑚𝑎 𝑟𝑎í𝑧 𝑟𝑒𝑎𝑙 𝑐 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 (0, 2). ∗ 𝑆𝑢𝑝𝑜𝑛ℎ𝑎𝑚𝑜𝑠 𝑞𝑢𝑒 𝑓 𝑡𝑒𝑛ℎ𝑎 2 𝑟𝑎í𝑧𝑒𝑠 𝑟𝑒𝑎𝑖𝑠 𝑐 𝑒 𝑏. 𝐶𝑜𝑚𝑜 𝑓 é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖á𝑣𝑒𝑙 𝑒𝑚 (𝑐, 𝑏), 𝑢𝑚𝑎 𝑣𝑒𝑧 𝑞𝑢𝑒 é 𝑢𝑚 𝑝𝑜𝑙𝑖𝑛ô𝑚𝑖𝑜, 𝑝𝑒𝑙𝑜 𝑇𝑒𝑜𝑟𝑒𝑚𝑎 𝑑𝑒 𝑅𝑜𝑙𝑙𝑒 𝑒𝑥𝑖𝑠𝑡𝑒 𝑢𝑚 𝑛ú𝑚𝑒𝑟𝑜 𝑑 ∈ (𝑐, 𝑏) 𝑡𝑎𝑙 𝑞𝑢𝑒 𝑓′(𝑑) = 0. 𝑓′(𝑥) = 3𝑥2 − 2𝑥 + 1 ; ∆= 4 − 12 = −8. 𝐿𝑜𝑔𝑜, 𝑓′(𝑥) 𝑛ã𝑜 𝑝𝑜𝑠𝑠𝑢𝑖 𝑟𝑎í𝑧 𝑟𝑒𝑎𝑙. 𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑛ã𝑜 𝑒𝑥𝑖𝑠𝑡𝑒 𝑢𝑚 𝑛ú𝑚𝑒𝑟𝑜 𝑑 𝑡𝑎𝑙 𝑞𝑢𝑒 𝑓′(𝑑) = 0 𝑒, 𝑐𝑜𝑛𝑠𝑒𝑞𝑢𝑒𝑛𝑡𝑒𝑚𝑒𝑛𝑡𝑒, 𝑓(𝑥) 𝑡𝑒𝑚 𝑢𝑚𝑎 ú𝑛𝑖𝑐𝑎 𝑟𝑎í𝑧 𝑟𝑒𝑎𝑙. ∗ 𝑂𝑏𝑠: 𝐴 𝑟𝑎í𝑧 𝑒𝑚 𝑞𝑢𝑒𝑠𝑡ã𝑜 é 𝑢𝑚𝑎 𝑑𝑎𝑠 𝑟𝑎í𝑧𝑒𝑠 𝑛𝑜𝑡á𝑣𝑒𝑖𝑠 𝑑𝑒 𝑢𝑚 𝑝𝑜𝑙𝑖𝑛ô𝑚𝑖𝑜 𝑑𝑜 3º 𝑔𝑟𝑎𝑢. 𝑥 = 1 é 𝑟𝑎í𝑧 𝑑𝑒 𝑓(𝑥). 𝑂𝑢 𝑠𝑒𝑗𝑎, 𝑓(𝑥) = (𝑥 − 1)(𝑥2 + 1).𝑁𝑜𝑡𝑒 𝑞𝑢𝑒 𝑜 𝑠𝑒𝑔𝑢𝑛𝑑𝑜 𝑓𝑎𝑡𝑜𝑟 é 𝑖𝑟𝑟𝑒𝑑𝑢𝑡í𝑣𝑒𝑙, 𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑛ã𝑜 𝑝𝑜𝑠𝑠𝑢𝑖 𝑟𝑎í𝑧 𝑟𝑒𝑎𝑙. 𝑆𝑜𝑚𝑒𝑛𝑡𝑒 𝑝𝑎𝑟𝑎 𝑥 = 1, 𝑓(𝑥) = 0. 4. 𝑓(𝑥) = 𝑥3 (𝑥 − 1)2 ; 𝐸𝑠𝑏𝑜ç𝑎𝑟 𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜! 1) 𝐷𝑜𝑚í𝑛𝑖𝑜 𝑑𝑒 𝑓: 𝐷(𝑓) = ℝ − {1}. 2) 𝐼𝑛𝑡𝑒𝑟𝑠𝑒çõ𝑒𝑠 𝑐𝑜𝑚 𝑜𝑠 𝑒𝑖𝑥𝑜𝑠: 𝑃𝑜𝑛𝑡𝑜 (0, 0). 3) 𝐶𝑟𝑒𝑠𝑐𝑖𝑚𝑒𝑛𝑡𝑜 𝑒 𝐷𝑒𝑐𝑟𝑒𝑠𝑐𝑖𝑚𝑒𝑛𝑡𝑜: 𝑓′(𝑥) = 3𝑥2(𝑥 − 1)2 − 𝑥3. 2. (𝑥 − 1) (𝑥 − 1)4 = 3𝑥2(𝑥 − 1) − 2𝑥3 (𝑥 − 1)3 = 𝑥3 − 3𝑥2 (𝑥 − 1)3 = 𝑥2(𝑥 − 3) (𝑥 − 1)3 ; 𝑓′(𝑥) = 𝑥2(𝑥 − 3) (𝑥 − 1)3 ∗ 𝐹𝑎𝑧𝑒𝑛𝑑𝑜 𝑜 𝑒𝑠𝑡𝑢𝑑𝑜 𝑑𝑜 𝑐𝑜𝑚𝑝𝑜𝑟𝑡𝑎𝑚𝑒𝑛𝑡𝑜 (𝑠𝑖𝑛𝑎𝑙) 𝑑𝑒 𝑓′(𝑥), 𝑡𝑒𝑚𝑜𝑠: ++++ 0 + + + + ++ + + + + + +++ ++ 𝑥2 −−−−−−−−− −−−− 3 + + + + + ++ (𝑥 − 3) −−−−−−− 1 + + + + + ++++++ ++ (𝑥 − 1)3 ++++ 0 + + 1 − −− − − 3 + + + + + ++ 𝑓′(𝑥) = 𝑥²(𝑥 − 3)/(𝑥 − 1)³ ∗ 𝐷𝑎 𝑎𝑛á𝑙𝑖𝑠𝑒 𝑎𝑐𝑖𝑚𝑎, 𝑐𝑜𝑛𝑐𝑙𝑢í𝑚𝑜𝑠: 𝑓 é 𝑐𝑟𝑒𝑠𝑐𝑒𝑛𝑡𝑒 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 (−∞, 1) ∪ (3,+∞) 𝑓 é 𝑑𝑒𝑐𝑟𝑒𝑠𝑐𝑒𝑛𝑡𝑒 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 (1, 3) 4) 𝑃𝑜𝑛𝑡𝑜𝑠 𝐶𝑟í𝑡𝑖𝑐𝑜𝑠: 𝑂𝑛𝑑𝑒 𝑓′(𝑥) = 0 𝑜𝑢 𝑞𝑢𝑎𝑛𝑑𝑜 𝑓′(𝑥) 𝑛ã𝑜 𝑒𝑥𝑖𝑠𝑡𝑒. 𝑓′(𝑥) = 0 𝑝𝑎𝑟𝑎 𝑥 = 0 𝑒 𝑥 = 3. 𝑓(0) = 0; 𝑓(3) = 33 (3 − 1)2 = 27 4 . 𝑃𝑜𝑛𝑡𝑜𝑠 𝐶𝑟í𝑡𝑖𝑐𝑜𝑠 ∶ (0, 0) 𝑒 (3, 27 4 ). 𝑓′(𝑥) 𝑛ã𝑜 𝑒𝑥𝑖𝑠𝑡𝑒 𝑒𝑚 𝑥 = 1.𝑁𝑜 𝑒𝑛𝑡𝑎𝑛𝑡𝑜, 𝑥 = 1 𝑛ã𝑜 𝑝𝑒𝑟𝑡𝑒𝑛𝑐𝑒 𝑎𝑜 𝑑𝑜𝑚í𝑛𝑖𝑜 𝑑𝑒 𝑓 𝑒, 𝑝𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑛ã𝑜 é 𝑢𝑚 𝑛ú𝑚𝑒𝑟𝑜 𝑐𝑟í𝑡𝑖𝑐𝑜. 5) 𝐶𝑜𝑛𝑐𝑎𝑣𝑖𝑑𝑎𝑑𝑒: 𝑓′′(𝑥) = (3𝑥2 − 6𝑥)(𝑥 − 1)3 − (𝑥3 − 3𝑥2). 3. (𝑥 − 1)2 (𝑥 − 1)6 𝑓′′(𝑥) = (3𝑥2 − 6𝑥)(𝑥 − 1) − 3𝑥3 + 9𝑥2 (𝑥 − 1)4 91 𝑓′′(𝑥) = 6𝑥 (𝑥 − 1)4 𝐹𝑎𝑧𝑒𝑛𝑑𝑜 𝑜 𝑒𝑠𝑡𝑢𝑑𝑜 𝑑𝑜 𝑐𝑜𝑚𝑝𝑜𝑟𝑡𝑎𝑚𝑒𝑛𝑡𝑜 (𝑠𝑖𝑛𝑎𝑙) 𝑑𝑒 𝑓′′(𝑥) 𝑡𝑒𝑚𝑜𝑠: −−−−−−−−− −− 0 + ++ + +++ ++++ 6𝑥 +++++++++ +++++++ 1 + + + +++ (𝑥 − 1)4 −−−−−−−−− −− 0 + ++ + 1 + + + ++ + 𝑓′′(𝑥) = 6𝑥/(𝑥 − 1)4 ∗ 𝐷𝑎 𝑎𝑛á𝑙𝑖𝑠𝑒 𝑎𝑐𝑖𝑚𝑎, 𝑐𝑜𝑛𝑐𝑙𝑢í𝑚𝑜𝑠: 𝑓 𝑝𝑜𝑠𝑠𝑢𝑖 𝑐𝑜𝑛𝑐𝑎𝑣𝑖𝑑𝑎𝑑𝑒 𝑣𝑜𝑙𝑡𝑎𝑑𝑎 𝑝𝑎𝑟𝑎 𝑐𝑖𝑚𝑎 𝑒𝑚 (0, 1) ∪ (1,+∞) 𝑒 𝑓 𝑝𝑜𝑠𝑠𝑢𝑖 𝑐𝑜𝑛𝑐𝑎𝑣𝑖𝑑𝑎𝑑𝑒 𝑣𝑜𝑙𝑡𝑎𝑑𝑎 𝑝𝑎𝑟𝑎 𝑏𝑎𝑖𝑥𝑜 𝑒𝑚 (−∞, 0) 6) 𝑃𝑜𝑛𝑡𝑜𝑠 𝑑𝑒 𝐼𝑛𝑓𝑙𝑒𝑥ã𝑜: 𝑓′′(𝑥) = 0 𝑜𝑐𝑜𝑟𝑟𝑒 𝑎𝑝𝑒𝑛𝑎𝑠 𝑒𝑚 𝑥 = 0. 𝑓(0) = 0 . 𝑃𝑜𝑛𝑡𝑜 𝑑𝑒 𝑖𝑛𝑓𝑙𝑒𝑥ã𝑜 (0, 0). 7) 𝐴𝑠𝑠í𝑛𝑡𝑜𝑡𝑎𝑠: ∗ 𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑖𝑠: 𝐷𝑖𝑧𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑎 𝑟𝑒𝑡𝑎 𝑥 = 𝑎 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑠𝑒 𝑜𝑐𝑜𝑟𝑟𝑒𝑟 𝑢𝑚 𝑑𝑜𝑠 𝑠𝑒𝑔𝑢𝑖𝑛𝑡𝑒𝑠 𝑐𝑎𝑠𝑜𝑠: lim 𝑥→𝑎+ 𝑓(𝑥) = ±∞ 𝑜𝑢 lim 𝑥→𝑎− 𝑓(𝑥) = ±∞ ∗ 𝑆𝑎𝑏𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑓 é 𝑑𝑒𝑠𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 𝑥 = 1. 𝑉𝑒𝑟𝑖𝑓𝑖𝑐𝑎𝑛𝑑𝑜 𝑠𝑒 𝑎 𝑟𝑒𝑡𝑎 𝑥 = 1 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙: lim 𝑥→1+ 𝑓(𝑥) = lim 𝑥→1+ 𝑥3 (𝑥 − 1)2 = lim 𝑥→1+ 𝑥3⏞ 1 ↑ (𝑥 − 1)2⏟ ↓ 0+ = +∞ ∗ 𝑂𝑏𝑠: 𝑠𝑒 𝑥 → 1+ , 𝑥 > 1 𝑒𝑛𝑡ã𝑜 𝑥 − 1 > 0 ⇒ (𝑥 − 1)2 > 0 lim 𝑥→1− 𝑓(𝑥) = lim 𝑥→1− 𝑥3 (𝑥 − 1)2 = lim 𝑥→1+ 𝑥3⏞ 1 ↑ (𝑥 − 1)2⏟ ↓ 0+ = +∞ ∗ 𝑂𝑏𝑠: 𝑠𝑒 𝑥 → 1− , 𝑥 < 1 𝑒𝑛𝑡ã𝑜 𝑥 − 1 < 0 ⇒ (𝑥 − 1)2 > 0. −𝐿𝑜𝑔𝑜, 𝑎 𝑟𝑒𝑡𝑎 𝑥 = 1 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑎𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑒 𝑓(𝑥). ∗ 𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑖𝑠: 𝐷𝑖𝑧𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑎 𝑟𝑒𝑡𝑎 𝑦 = 𝑎 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑠𝑒 𝑜𝑐𝑜𝑟𝑟𝑒𝑟 𝑢𝑚 𝑑𝑜𝑠 𝑠𝑒𝑔𝑢𝑖𝑛𝑡𝑒𝑠 𝑐𝑎𝑠𝑜𝑠: lim 𝑥→+∞ 𝑓(𝑥) = 𝑎 𝑜𝑢 lim 𝑥→−∞ 𝑓(𝑥) = 𝑎 lim 𝑥→+∞ 𝑓(𝑥) = lim 𝑥→+∞ 𝑥3 (𝑥 − 1)2 = lim 𝑥→+∞ 𝑥3 𝑥2 − 2𝑥 + 1 = lim 𝑥→+∞ 3𝑥2 2𝑥 − 2 = lim 𝑥→+∞ 6𝑥 2 = lim 𝑥→+∞ 3𝑥 = +∞. 𝐿𝑜𝑔𝑜, 𝑓 𝑛ã𝑜 𝑝𝑜𝑠𝑠𝑢𝑖 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙. ∗ 𝑂𝑏𝑙í𝑞𝑢𝑎: 𝐷𝑖𝑧𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑎 𝑟𝑒𝑡𝑎 𝑦 = 𝑎𝑥 + 𝑏 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 𝑜𝑏𝑙í𝑞𝑢𝑎 𝑠𝑒, 𝑠𝑜𝑚𝑒𝑛𝑡𝑒 𝑠𝑒 lim 𝑥→±∞ [𝑓(𝑥) − (𝑎𝑥 + 𝑏)] = 0 92 𝑓(𝑥) = 𝑥3 𝑥2 − 2𝑥 + 1 = (𝑥 + 2) + 3𝑥 − 2 𝑥2 − 2𝑥 + 1 lim 𝑥→±∞ [𝑓(𝑥) − (𝑥 + 2)] = lim 𝑥→±∞ 3𝑥 − 2 𝑥2 − 2𝑥 + 1 = lim 𝑥→±∞ 3𝑥 − 2 𝑥2 𝑥2 − 2𝑥 + 1 𝑥2 = lim 𝑥→±∞ 3 𝑥 − 2 𝑥2 1 − 2 𝑥 + 1 𝑥2 = lim 𝑥→±∞ 3 𝑥 − lim𝑥→±∞ 2 𝑥2 lim 𝑥→±∞ 1 − lim 𝑥→±∞ 2 𝑥 + lim𝑥→±∞ 1 𝑥2 = 0 − 0 1 − 0 + 0 = 0 1 = 0. ∗ 𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑎 𝑟𝑒𝑡𝑎 𝑦 = 𝑥 + 2 é 𝑢𝑚𝑎 𝑎𝑠𝑠í𝑛𝑡𝑜𝑡𝑎 𝑖𝑛𝑐𝑙𝑖𝑛𝑎𝑑𝑎 (𝑜𝑏𝑙í𝑞𝑢𝑎) 𝑎𝑜 𝑔𝑟á𝑓𝑖𝑐𝑜 𝑑𝑒 𝑓(𝑥) 5. 𝑎) lim 𝑥→0 (2𝑥 + 1)1 sen(𝑥+𝑥 2)⁄ = lim 𝑥→0 𝑒ln(2𝑥+1) 1 sen(𝑥+𝑥2)⁄ = lim 𝑥→0 𝑒 ln(2𝑥+1) sen(𝑥+𝑥2) = 𝑒 lim 𝑥→0 ln(2𝑥+1) sen(𝑥+𝑥2); ∗ 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑛𝑑𝑜 𝑜 𝑙𝑖𝑚𝑖𝑡𝑒 𝑑𝑜 𝑒𝑥𝑝𝑜𝑒𝑛𝑡𝑒 𝑡𝑒𝑚𝑜𝑠: lim 𝑥→0 ln(2𝑥 + 1) sen(𝑥 + 𝑥2) = lim 𝑥→0 2 2𝑥 + 1 (2𝑥 + 1) cos(𝑥 + 𝑥²) = lim 𝑥→0 2 (2𝑥 + 1)² cos(𝑥 + 𝑥²) = 2 (0 + 1)2. cos(0 + 02) = 2 cos 0 = 2 1 = 2. ∗ lim 𝑥→0 (2𝑥 + 1)1 sen(𝑥+𝑥 2)⁄ = 𝑒 lim 𝑥→0 ln(2𝑥+1) sen(𝑥+𝑥2) = 𝑒2. 𝑏) 𝑓′′(𝑥) = −𝜋 + 𝑥2 3⁄ + 2 cos 𝑥 − 𝑒𝑥 ; 𝑓′(0) = 1 𝑒 𝑓(0) = 𝑒. ∗ 𝐴 𝑎𝑛𝑡𝑖𝑑𝑒𝑟𝑖𝑣𝑎𝑑𝑎 𝑚𝑎𝑖𝑠 𝑔𝑒𝑟𝑎𝑙 𝑝𝑎𝑟𝑎 𝑓′′(𝑥) é: 𝑓′(𝑥) = −𝜋𝑥 + 3 5 𝑥5 3⁄ + 2 sen 𝑥 − 𝑒𝑥 + 𝐶 𝑓′(0) = −1 + 𝐶 ; 𝑓′(0) = 1 ∴ −1 + 𝐶 = 1 → 𝐶 = 2. 95 𝑒𝑥𝑖𝑠𝑡ê𝑛𝑐𝑖𝑎 𝑑𝑎 𝑓𝑢𝑛çã𝑜 ln(1 − 𝑥2) , 𝑎𝑠𝑠𝑢𝑚𝑖𝑛𝑑𝑜 𝑞𝑢𝑒 (1 − 𝑥2) > 0. 𝐴𝑠𝑠𝑖𝑚, −−−−−−−− (−1) + + + +++++( 1 ) − − − − − −−− (1 − 𝑥2) ∗ 𝐶𝑜𝑚 𝑒𝑠𝑠𝑎 𝑎𝑛á𝑙𝑖𝑠𝑎, 𝑡𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑎 𝑓𝑢𝑛çã𝑜 ln(1 − 𝑥2) 𝑒𝑥𝑖𝑠𝑡𝑒 𝑝𝑎𝑟𝑎 − 1 < 𝑥 < 1. 𝐶𝑜𝑚𝑜 𝑒𝑚 𝑓(𝑥) 𝑒𝑠𝑡𝑎 𝑠𝑒𝑛𝑡𝑒𝑛ç𝑎 𝑝𝑟𝑒𝑑𝑜𝑚𝑖𝑛𝑎 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 − 1 < 𝑥 < 0, 𝑐𝑜𝑛𝑑𝑖𝑧 𝑐𝑜𝑚 𝑜 𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑑𝑜, 𝑙𝑜𝑔𝑜 𝑓 é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 (−1,0). ∗ 𝐴 𝑡𝑒𝑟𝑐𝑒𝑖𝑟𝑎 𝑠𝑒𝑛𝑡𝑒𝑛ç𝑎 é 𝑢𝑚𝑎 𝑓𝑢𝑛çã𝑜 𝑝𝑜𝑙𝑖𝑛𝑜𝑚𝑖𝑎𝑙 𝑟𝑎𝑐𝑖𝑜𝑛𝑎𝑙, 𝑜𝑛𝑑𝑒 𝑠𝑒𝑢 𝑑𝑜𝑚í𝑛𝑖𝑜, 𝑛𝑒𝑠𝑡𝑒 𝑐𝑎𝑠𝑜, 𝑑𝑒𝑝𝑒𝑛𝑑𝑒 𝑑𝑜 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑑𝑜𝑟 𝑞𝑢𝑒 𝑑𝑒𝑣𝑒 𝑠𝑒𝑟 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑡𝑒 𝑑𝑒 𝑧𝑒𝑟𝑜, 𝑗á 𝑞𝑢𝑒 𝑜 𝑚𝑒𝑚𝑏𝑟𝑜 𝑑𝑜 𝑛𝑢𝑚𝑒𝑟𝑎𝑑𝑜𝑟 é 𝑢𝑚𝑎 𝑓𝑢𝑛çã𝑜 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑒. 𝐴𝑛𝑎𝑙𝑖𝑠𝑎𝑛𝑑𝑜 𝑎 𝑡𝑒𝑟𝑐𝑒𝑖𝑟𝑎 𝑠𝑒𝑛𝑡𝑒𝑛ç𝑎, 𝑡𝑒𝑚𝑜𝑠: 1 1 + 𝑥2 ; 1 + 𝑥2 ≠ 0 −𝐹𝑎ç𝑎𝑚𝑜𝑠 1 + 𝑥2 = 0 , 𝑒𝑛𝑡ã𝑜 𝑥2 = −1 → 𝑥 = ±√−1 ; 𝐿𝑜𝑔𝑜, 𝑛ã𝑜 𝑡𝑒𝑚𝑜𝑠 𝑠𝑜𝑙𝑢çã𝑜 𝑛𝑜𝑠 𝑟𝑒𝑎𝑖𝑠 𝑝𝑎𝑟𝑎 𝑒𝑠𝑡𝑎 𝑒𝑥𝑝𝑟𝑒𝑠𝑠ã𝑜. 𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 1 + 𝑥2 ≠ 0, ∀𝑥 ∈ ℝ. 𝐴𝑠𝑠𝑖𝑚, 𝑎 𝑡𝑒𝑟𝑐𝑒𝑖𝑟𝑎 𝑠𝑒𝑛𝑡𝑒𝑛ç𝑎 é 𝑣á𝑙𝑖𝑑𝑎 𝑝𝑎𝑟𝑎 𝑞𝑢𝑎𝑙𝑞𝑢𝑒𝑟 𝑥 𝑝𝑒𝑟𝑡𝑒𝑛𝑐𝑒𝑛𝑡𝑒 𝑎𝑜𝑠 𝑟𝑒𝑖𝑎𝑠. 𝐸𝑚 𝑓(𝑥) 𝑒𝑠𝑡𝑎 𝑠𝑒𝑛𝑡𝑒𝑛ç𝑎 𝑣𝑎𝑙𝑒 𝑝𝑎𝑟𝑎 𝑥 ≤ −1. 𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑓 é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 (−∞,−1). ∗ 𝐶𝑜𝑚 𝑒𝑠𝑠𝑎𝑠 𝑎𝑛á𝑙𝑖𝑠𝑒𝑠, 𝑡𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑓 é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 (−∞,−1) ∪ (−1,0) ∪ (0,+∞). 𝐹𝑎𝑙𝑡𝑎 𝑣𝑒𝑟𝑖𝑓𝑖𝑐𝑎𝑟 𝑠𝑒 𝑓 é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑛𝑜𝑠 𝑝𝑜𝑛𝑡𝑜𝑠 𝑜𝑛𝑑𝑒 ℎá 𝑚𝑢𝑑𝑎𝑛ç𝑎 𝑑𝑒 𝑐𝑜𝑚𝑝𝑜𝑟𝑡𝑎𝑚𝑒𝑛𝑡𝑜 𝑑𝑎 𝑓𝑢𝑛çã𝑜, 𝑜𝑢 𝑠𝑒𝑗𝑎, 𝑒𝑚 𝑥 = −1 𝑒 𝑥 = 0. ∗ 𝐷𝑖𝑧𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑢𝑚𝑎 𝑓𝑢𝑛çã𝑜 𝑓 é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑛𝑢𝑚 𝑝𝑜𝑛𝑡𝑜 𝑥 = 𝑎 𝑠𝑒, 𝑠𝑜𝑚𝑒𝑛𝑡𝑒 𝑠𝑒, 1) 𝑓(𝑎) 𝑒𝑥𝑖𝑠𝑡𝑒; 2) lim 𝑥→𝑎 𝑓(𝑥) 𝑒𝑥𝑖𝑠𝑡𝑒; 3) lim 𝑥→𝑎 𝑓(𝑥) = 𝑓(𝑎) ∗ 𝑝𝑎𝑟𝑎 𝑥 = −1: 1) 𝑓(−1) = 1 1 + (−1)2 = 1 1 + 1 = 1 2 . 2) lim 𝑥→−1 𝑓(𝑥) ; lim 𝑥→−1+ 𝑓(𝑥) = lim 𝑥→−1+ ln(1 − 𝑥2) ; 𝑠𝑒 𝑥 → −1+ 𝑒𝑛𝑡ã𝑜 𝑥 > −1. 𝐿𝑜𝑔𝑜, (1 − 𝑥2) > 0. ∗ 𝐶𝑜𝑚𝑜 𝑜 𝑙𝑜𝑔𝑎𝑟𝑖𝑡𝑚𝑎𝑛𝑑𝑜 (1 − 𝑥2) → 0 , 𝑡𝑒𝑚𝑜𝑠 lim 𝑥→−1+ ln(1 − 𝑥2) = −∞; ∗ 𝐴𝑠𝑠𝑖𝑚, lim 𝑥→−1 𝑓(𝑥) ∄, 𝑝𝑜𝑖𝑠 𝑛ã𝑜 ℎá 𝑙𝑖𝑚𝑖𝑡𝑒 𝑙𝑎𝑡𝑒𝑟𝑎𝑙 à 𝑑𝑖𝑟𝑒𝑖𝑡𝑎 𝑑𝑒 − 1. −𝐿𝑜𝑔𝑜, 𝑓 𝑛ã𝑜 é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 𝑥 = −1. ∗ 𝑝𝑎𝑟𝑎 𝑥 = 0: 1)𝑓(0) = ln(0 + 1) = ln(1) = 0. 2) lim 𝑥→0 𝑓(𝑥) ; lim 𝑥→0− 𝑓(𝑥) = lim 𝑥→0− ln(1 − 𝑥2) = ln [ lim 𝑥→0− (1 − 𝑥2)] = ln(1) = 0. lim 𝑥→0+ 𝑓(𝑥) = lim 𝑥→0+ ln(𝑥 + 1) = ln [ lim 𝑥→0+ (𝑥 + 1)] = ln(1) = 0. ∗ 𝐿𝑜𝑔𝑜, lim 𝑥→0 𝑓(𝑥) = 0. 3) lim 𝑥→0 𝑓(𝑥) = 𝑓(0) = 0. −𝐿𝑜𝑔𝑜, 𝑓 é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 𝑥 = 0. 96 ∗ 𝐶𝑜𝑚 𝑒𝑠𝑠𝑎𝑠 𝑎𝑛á𝑙𝑖𝑠𝑒𝑠 𝑡𝑒𝑚𝑜𝑠 𝑞𝑢𝑒 𝑓 é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒𝑚 (−∞,−1) ∪ (−1,+∞). 𝑏) 𝑓(𝑥) = tg 𝑥 − cotg 𝑥 − 𝑥 ; 𝑝𝑟𝑜𝑣𝑎𝑟 𝑞𝑢𝑒 𝑓 𝑡𝑒𝑚 𝑢𝑚𝑎 ú𝑛𝑖𝑐𝑎 𝑟𝑎í𝑧 𝑟𝑒𝑎𝑙 𝑒𝑚 (0, 𝜋 2 ). ∗ 𝐶𝑎𝑙𝑐𝑢𝑙𝑒𝑚𝑜𝑠 𝑓 ( 𝜋 6 ) 𝑒 𝑓 ( 𝜋 3 ): 𝑓 ( 𝜋 6 ) = tg 𝜋 6 − cotg 𝜋 6 − 𝜋 6 = 1 √3 − √3 − 𝜋 6 = 6 − 18 − 𝜋√3 6√3 = − (12 + 𝜋√3) 6√3 ; 𝑓 ( 𝜋 3 ) = tg 𝜋 3 − cotg 𝜋 3 − 𝜋 3 = √3 − 1 √3 − 𝜋 3 = 9 − 3 − 𝜋√3 3√3 = 6 − 𝜋√3 3√3 ; ∗ 𝑓 é 𝑢𝑚𝑎 𝑠𝑜𝑚𝑎 𝑑𝑒 𝑓𝑢𝑛çõ𝑒𝑠, 𝑜𝑛𝑑𝑒 tg 𝑥 𝑒 cotg 𝑥 𝑠ã𝑜 𝑓𝑢𝑛çõ𝑒𝑠 𝑡𝑟𝑖𝑔𝑜𝑛𝑜𝑚é𝑡𝑟𝑖𝑐𝑎𝑠 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎𝑠 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 [ 𝜋 6 , 𝜋 3 ] , 𝑒 𝑎 𝑓𝑢𝑛çã𝑜 𝑖𝑑𝑒𝑛𝑡𝑖𝑑𝑎𝑑𝑒 𝑦 = 𝑥 𝑡𝑎𝑚𝑏é𝑚 é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑛𝑒𝑠𝑠𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜. 𝐿𝑜𝑔𝑜, 𝑓 é 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 [ 𝜋 6 , 𝜋 3 ] . 𝐴𝑠𝑠𝑖𝑚, 𝑓 ( 𝜋 6 ) < 0 < 𝑓 ( 𝜋 3 ), 𝑒𝑛𝑡ã𝑜, 𝑝𝑒𝑙𝑜 𝑇𝑒𝑜𝑟𝑒𝑚𝑎 𝑑𝑜 𝑉𝑎𝑙𝑜𝑟 𝐼𝑛𝑡𝑒𝑟𝑚𝑒𝑑𝑖á𝑟𝑖𝑜 𝑒𝑥𝑖𝑠𝑡𝑒 𝑎𝑙𝑔𝑢𝑚 𝑐 ∈ [ 𝜋 6 , 𝜋 3 ] , 𝑡𝑎𝑙 𝑞𝑢𝑒 𝑓(𝑐) = 0. 𝐸𝑚 𝑜𝑢𝑡𝑟𝑎𝑠 𝑝𝑎𝑙𝑎𝑣𝑟𝑎𝑠, 𝑓 𝑝𝑜𝑠𝑠𝑢𝑖 𝑢𝑚𝑎 𝑟𝑎í𝑧 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 (0, 𝜋 2 ). ∗ 𝑆𝑒𝑛𝑑𝑜 𝑓 𝑐𝑜𝑛𝑡í𝑛𝑢𝑎 𝑒 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖á𝑣𝑒𝑙 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 (0, 𝜋 2 ) , 𝑒 𝑠𝑢𝑝𝑜𝑛ℎ𝑎𝑚𝑜𝑠 𝑞𝑢𝑒 𝑓 𝑝𝑜𝑠𝑠𝑢𝑖 2 𝑟𝑎í𝑧𝑒𝑠 𝑐 𝑒 𝑑 𝑛𝑒𝑠𝑡𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜, 𝑝𝑒𝑙𝑜 𝑇𝑒𝑜𝑟𝑒𝑚𝑎 𝑑𝑒 𝑅𝑜𝑙𝑙𝑒, 𝑒𝑥𝑖𝑠𝑡𝑒 𝑎𝑙𝑔𝑢𝑚 𝑛ú𝑚𝑒𝑟𝑜 𝑎 ∈ (𝑐, 𝑑) 𝑡𝑎𝑙 𝑞𝑢𝑒 𝑓′(𝑎) = 0. 𝑓′(𝑥) = sec2 𝑥 + cossec2 𝑥 − 1 𝑓′(𝑥) = 0 → sec2 𝑥 + cossec2 𝑥 − 1 = 0 → sec2 𝑥 + cossec2 𝑥 = 1 1 cos2 𝑥 + 1 sen2 𝑥 = 1 → sen2 𝑥 + cos2 𝑥 sen2 𝑥 . cos2 𝑥 = 1 → 1 sen2 𝑥 . cos2 𝑥 = 1 ; sen2 𝑥 . cos2 𝑥 = 1 → sen2 𝑥 . (1 − sen2 𝑥) = 1 → −𝑠𝑒𝑛4𝑥 + sen2 𝑥 − 1 = 0 𝑠𝑒𝑛4𝑥 − sen2 𝑥 + 1 = 0 ∗ 𝑆𝑒𝑗𝑎 𝑦 = sen2 𝑥 𝑒𝑛𝑡ã𝑜: 𝑦2 − 𝑦 + 1 = 0 → ∆= 1 − 4 = −3. 𝐿𝑜𝑔𝑜, 𝑛ã𝑜 ℎá 𝑠𝑜𝑙𝑢çã𝑜 𝑟𝑒𝑎𝑙 𝑝𝑎𝑟𝑎 𝑎 𝑒𝑞𝑢𝑎çã𝑜! 𝐿𝑜𝑔𝑜, 𝑛ã𝑜 𝑒𝑥𝑖𝑠𝑡𝑒 (𝑎) ∈ (𝑐, 𝑑) 𝑡𝑎𝑙 𝑞𝑢𝑒 𝑓′(𝑎) = 0. 𝑃𝑜𝑟𝑡𝑎𝑛𝑡𝑜, 𝑓 𝑡𝑒𝑚 𝑢𝑚𝑎 ú𝑛𝑖𝑐𝑎 𝑟𝑎í𝑧 𝑟𝑒𝑎𝑙 𝑛𝑜 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑜 (0, 𝜋 2 ). 3. 𝑎) 𝑓(𝑥) = √sec√𝑥 ; 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑟 𝑓′ ( 𝜋2 16 ) . ∗ 𝑆𝑒𝑗𝑎 𝑢 = √𝑥 , 𝑣 = sec 𝑢 . 𝐸𝑛𝑡ã𝑜 𝑓(𝑣) = √𝑣 . 𝑃𝑒𝑙𝑎 𝑅𝑒𝑔𝑟𝑎 𝑑𝑎 𝐶𝑎𝑑𝑒𝑖𝑎, 𝑡𝑒𝑚𝑜𝑠: 𝑑𝑦 𝑑𝑥 = 𝑑𝑢 𝑑𝑥 ∙ 𝑑𝑣 𝑑𝑢 ∙ 𝑑𝑦 𝑑𝑣 𝑓′(𝑥) = 1 2√𝑥 ∙ sec 𝑢 ∙ tg 𝑢 ∙ 1 2√𝑣 𝑓′(𝑥) = 1 4 ∙ 1 √𝑥 sec√𝑥 ∙ tg √𝑥 ∙ 1 √sec√𝑥 97 𝑓′ ( 𝜋2 16 ) = 1 4 ∙ 1 √𝜋 2 16 sec√ 𝜋2 16 ∙ tg√ 𝜋2 16 ∙ 1 √sec√ 𝜋2 16 𝑓′ ( 𝜋2 16 ) = 1 4 ∙ 4 𝜋 ∙ sec 𝜋 4 ∙ tg 𝜋 4 ∙ 1 √sec 𝜋 4 𝑓′ ( 𝜋2 16 ) = 1 4 ∙ 4 𝜋 ∙ √2 ∙ 1 ∙ 1 √√2 𝑓′ ( 𝜋2 16 ) = √2 𝜋√√2 = 21 2⁄ 𝜋. 21 4⁄ = 21 4⁄ 𝜋 . 𝑏) 𝑓(𝑥) = 2arctg(senh𝑥) ; 𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑟 𝑓′(0). ∗ 𝑆𝑒𝑗𝑎𝑚 𝑢 = senh 𝑥 𝑒 𝑣 = arctg 𝑢 , 𝑒𝑛𝑡ã𝑜 𝑓(𝑣) = 2𝑣 . 𝑃𝑒𝑙𝑎 𝑅𝑒𝑔𝑟𝑎 𝑑𝑎 𝐶𝑎𝑑𝑒𝑖𝑎, 𝑡𝑒𝑚𝑜𝑠: 𝑑𝑦 𝑑𝑥 = 𝑑𝑢 𝑑𝑥 ∙ 𝑑𝑣 𝑑𝑢 ∙ 𝑑𝑦 𝑑𝑣 𝑓′(𝑥) = cosh 𝑥 ∙ 1 1 + 𝑢2 ∙ 2𝑣 ∙ ln(2) 𝑓′(𝑥) = cosh 𝑥 ∙ 1 1 + senh2 𝑥 ∙ 2arctg(senh𝑥) ∙ ln(2) 𝑓′(0) = cosh(0) ∙ 1 1 + senh2(0) ∙ 2arctg(senh(0)) ∙ ln(2) 𝑓′(0) = 1 ∙ 1 1 + 0 ∙ 2arctg(0) ∙ ln(2) 𝑓′(0) = 20 ∙ ln(2) 𝑓′(0) = ln(2) 4. 𝑓(𝑥) = ( 𝑥 cos 𝑥 ) 𝑥 ; 𝑎) 𝑓′(𝑥); ln 𝑓(𝑥) = 𝑥 ∙ ln ( 𝑥 cos 𝑥 ) ln 𝑓(𝑥) = 𝑥 ∙ [ln 𝑥 − ln(cos 𝑥)] 𝑓′(𝑥) 𝑓(𝑥) = ln 𝑥 − ln(cos 𝑥) + 𝑥 [ 1 𝑥 + sen 𝑥 cos 𝑥 ] 𝑓′(𝑥) 𝑓(𝑥) = ln ( 𝑥 cos 𝑥 ) + 1 + 𝑥. tg 𝑥 𝑓′(𝑥) = ( 𝑥 cos 𝑥 ) 𝑥 ∙ [ln ( 𝑥 cos 𝑥 ) + 1 + 𝑥. tg 𝑥] 𝑏) lim 𝑥→0 𝑓(𝑥) = lim 𝑥→0 ( 𝑥 cos 𝑥 ) 𝑥 = lim 𝑥→0 𝑒ln( 𝑥 cos𝑥 ) 𝑥 = lim 𝑥→0 𝑒𝑥.ln( 𝑥 cos𝑥 ) = 𝑒 lim 𝑥→0 𝑥.ln( 𝑥 cos𝑥 ) . ∗ 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑛𝑑𝑜 𝑜 𝑙𝑖𝑚𝑖𝑡𝑒 𝑑𝑜 𝑒𝑥𝑝𝑜𝑒𝑛𝑡𝑒, 𝑡𝑒𝑚𝑜𝑠: lim 𝑥→0 𝑥. ln ( 𝑥 cos 𝑥 ) = lim 𝑥→0 𝑥. [ln 𝑥 − ln(cos 𝑥)] = lim 𝑥→0 [ln 𝑥 − ln(cos 𝑥)] 1 𝑥 =