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Clase 205

Clase 205. 2. 1. 3. Ejercicios variados. x 2 + 6x – 27. g ( x ) =. 81 – x 2. Ejercicio 1. Sean las funciones: f(x) =  x y. a) Halla la función h(x) = (fog)(x). b) Calcula los ceros de h(x) y sus valores inadmisibles. f(x) =  x. x 2 + 6x – 27. g ( x ) =. x 2 + 6x – 27.

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Clase 205

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  1. Clase 205 2 1 3 Ejercicios variados

  2. x2 + 6x – 27 g(x) = 81 – x2 Ejercicio 1 Sean las funciones: f(x) =  x y a) Halla la función h(x) = (fog)(x) b) Calcula los ceros de h(x) y sus valores inadmisibles.

  3. f(x) =  x x2 + 6x – 27 g(x) = x2 + 6x – 27 x2 + 6x – 27 81 – x2 81 – x2 81 – x2 a) h(x) = (fog)(x) = f [g(x) ] = f =

  4. x2 + 6x – 27 x2 + 6x – 27 81 – x2 81 – x2 Ceros b) h(x) = h(x) = 0 x  : x   9 = 0 x2 + 6x – 27 = 0 (x + 9)(x – 3) = 0 N.S. x = –9 ó x = 3

  5. x2 + 6x – 27 x2 + 6x – 27 81 – x2 81 – x2 3 9 x – 3 < 0 9 – x x – 3 > 0 x – 9 Valores inadmisibles h(x) = C.N. x1 = 3 < 0 C.D. x2 = 9 (x + 9)(x – 3) < 0 (9 + x)(9 – x) – + + x –9 x < 3; x  –9 ó x > 9

  6. Dada la ecuación x2 + 4x – 24y + 100 = 0 Ejercicio 2 Escribe la ecuación y esboce el gráfico de la elipse que cumple:  su centro es el vértice de esta curva,  uno de los vértices no princi- pales coincide con el foco, la distancia focal es de 16 u.  a) Halla el área de la figura formada por los focos de ambas curvas.

  7. x2+ 4x – 24y + 100 = 0 x2+ 4x = 24y – 100 x2 + 4x + 4 = 24y – 100 + 4 = 24y – 96 (x + 2)2 (x + 2)2 = 24(y – 4) Parábola que abre hacia arriba de vértice V(–2;4) (–2;4) O 4p = 24 b = p = 6 6 B1 F(–2;10) (–2;10)

  8. O(–2;4) b = 6 2c = 16 B1(–2;10) c = 8 a2 = b2 + c2 a2 = 62 + 82 a2 = 36 + 64 a2 = 100 a = 10 (x – )2 (y – )2 4 + 2 k h + = 1 a2 b2 36 100

  9. O(–2;4) B1(–2;10) y B1 a = 10 B2(–2; –2) 10 b = 6 c = 8 A = 48 u2 O A2 A1 4 F2 F1 x 0 –12 –10 6 –2 8 –2 A2(8;4) A1(–12;4) B2 F2(6;4) F1(–10;4)

  10. g(x) = 1 – 3x Para el estudio individual 1. Sea f(x) = log5(x2 – 1) y Obtenga la ecuación de la inversa de f(x) con x > 1. a) Halla (fog)(x) y su dominio. 2. Resuelve la ecuación: log 10 = log2(4 – x) – (hop)(x) siendo h(x)= log2x y p(x)=x – 1

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