Ex algebra (13)

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Ex algebra (13)

  1. 1. Lista 10 de C´lculo I a 2010-2 18 UFES - Universidade Federal do Esp´ ırito Santo Fun¸oes inversas c˜ DMAT - Departamento de Matem´tica a Teorema da Fun¸ao Inversa c˜ LISTA 10 - 2010-2 Fun¸oes trigonom´tricas inversas c˜ e 1 1. Seja f (x) = − x3 , x > 0. x ′ (a) Mostre que f tem inversa em (0, ∞); (b) Calcule f −1 (0) e f −1 (0); (c) Determine a equa¸ao da reta tangente ao gr´fico de f −1 no ponto 0, f −1 (0) . c˜ a c˜ ıvel, deriv´vel, tal que f (1) = 2, f (2) = 7, f ′ (1) = 3 e f ′ (2) = 4, calcule 2. Sendo f uma fun¸ao invert´ a ′ f −1 (2). 1 − x3 , x ≤ 0 ′ 3. Seja f (x) = . Se f −1 existir, calcule f −1 (x) e esboce os gr´ficos de f e f −1 . a 1 − x2 , x > 0 Resolva as equa¸oes dos exerc´ c˜ ıcios 4. a 11. √ 3 1 4. sen x = 2 6. sen x = − 2 8. tan x = 0 10. tan x = −1 5. cos x = 0 7. cos x = −1 9. tan x = 1 11. sec x = −2 Nos exerc´ ıcios 12. a 19. encontre o valor de x. √ 3 1 12. x = arcsen 2 14. x = arcsen − 2 16. x = arctan 0 18. x = arctan −1 13. x = arccos 0 15. x = arccos −1 17. x = arctan 1 19. x = arcsec − 2 Deduza as f´rmulas dos exerc´ o ıcios 20. a 22. 1 1 √ 20. arcsec x − arccos =0 21. (arctan x)′ = 22. cos( arcsen x) = 1 − x2 x 1 + x2 Nos exerc´ ıcios 23. e 24. derive a fun¸ao. c˜ 1 23. f (x) = arcsen 3 (x + 1)2 + arccos √ x2+1 1 − cos x 24. g(x) = arctan 1 + cos x ıcios 25. e 26. encontre y ′ , se y = y(x) ´ definida implicitamente pela equa¸ao dada. Nos exerc´ e c˜ 25. x arctan y = x2 + y 2 26. arcsen (xy) = x + y Nos exerc´ ıcios 27. a 29. verifique a igualdade. d x3 x2 + 2 √ 27. arcsen x + 1 − x2 = x2 arcsen x dx 3 9 d x 1 28. arctan √ = √ dx 1 + 1 − x2 2 1 − x2 d 2 tan x 29. arctan =2 dx 1 − tan2 x 30. Seja f (x) = 2 x2 + 1 arctan x, x ∈ R. (a) Mostre que f ´ invert´ e ıvel; ′ (b) Verifique que f (−1) = −π e calcule f −1 (−π);
  2. 2. Lista 10 de C´lculo I a 2010-2 19 RESPOSTAS 1 + 3x4 π 1. (a) Como f ′ (x) = − < 0 em (0, ∞), 5. x = + kπ, k ∈ Z x2 2 f satisfaz as hip´teses do TFI o π 5π 6. x = − + 2kπ ou x = − + 2kπ, k ∈ Z (teorema da fun¸˜o inversa). ca 6 6 7. x = π + 2kπ , k ∈ Z Logo f ´ invert´ em (0, ∞); e ıvel −1 ` −1 ´′ 8. x = 2kπ , k ∈ Z (b) f (0) = 1 e f (0) = −1/4; π 9. x = + kπ, k ∈ Z (c) x + 4y = 4 4 1 π 2. 10. x = − + kπ, k ∈ Z 3 4 2π 2π 8 −1 11. x = + 2kπ ou x = − + 2kπ, k ∈ Z > p 3 , x>1 3 3 3 (1 − x)2 < −1 ′ π ` ´ 3. f (x) = 12. x = > √−1 , : x<1 3 2 1−x π y 13. x = f 2 f–1 π 1 14. x = − 6 15. x = π 16. x = 0 0 x π –1 1 2 17. x = 4 f–1 18. x = − π –1 f 4 π 2π 2π 4. x = + 2kπ ou x= + 2kπ, k ∈ Z 19. x = 3 3 3 1 20. Sabemos que y = arcsec x ⇔ sec y = x, 0 ≤ y < π 2 ou π 2 <y≤π⇔ = x, 0 ≤ y < π 2 ou π 2 <y≤π⇔ cos y 1 = cos y. Substituindo a primeira e a ultima rela¸˜o na equa¸˜o dada, obtemos y − arccos (cos y) = y − y = 0. ´ ca ca x 21. Deduzida em aula. p 22. Sabemos que y = arcsen x ⇔ sen y = x, − π ≤ y ≤ π . Por outro lado, cos y = ± 1 − sen 2 y, mas no intervalo 2 p2 √ considerado cos y ≥ 0, logo cos( arcsen x) = cos y = 1 − sen 2 y = 1 − x2 . 2(x + 1) −1 −1 ` 2 ´− 3 23. f ′ (x) = 3 arcsen 2 (x + 1)2 p ` ´ +r · x + 1 2 (2x) = 1 − (x + 1)4 1 2 1− 2 x +1 6(x + 1) arcsen 2 (x + 1)2 ` ´ x = p + √ 1 − (x + 1)4 (x2 + 1) x2 1 1 (1 + cos x)( sen x) − (1 − cos x)(− sen x) sen x 24. g ′ (x) = 1 − cos x × r × = 1+ 1 − cos x (1 + cos x)2 2| sen x| 2 1 + cos x 1 + cos x 1 + y 2 (2x − arctan y) 1 + y 2 x2 − y 2 ` ´ ` ´` ´ dy 25. = = 2 dx x − 2y (1 + y 2 ) x − 2xy (1 + y 2 ) p dy 1 − x2 y 2 − y 26. = p dx x − 1 − x2 y 2 30. (a) f ′ (x) = 2 + 4x arctan x = 0 pois (i) f ′ (0) = 2; (ii) x > 0 ⇒ arctan x > 0 ⇒ x arctan x > 0 ⇒ f ′ (x) > 2 ⇒ f ′ (x) > 0; (iii) x < 0 ⇒ arctan x < 0 ⇒ x arctan x > 0 ⇒ f ′ (x) > 2 ⇒ f ′ (x) > 0. Logo aplicando o Teorema da Fun¸˜o Inversa, f possui inversa f −1 . ca ` ´′ 1 (b) f (−1) = 4 arctan(−1) = −π; f −1 (−π) = . 2+π

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