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CLASE 2

CLASE 2. CÁLCULO NUMÉRICO. ECUACIONES. = . Dom. • ¾ • √ 3. • ¾ • √ 3. • 1 • – 7. ECUACIÓN. S. • . x 2 = 3 x. • 0 • 3. • – 1,3. • 1 • – 7. = . Dom= S. IDENTIDA D. • . • 0 • 3. • – 1,3. x 2 – 1 =( x + 1 )( x – 1 ). Analizar si los siguientes pares.

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CLASE 2

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  1. CLASE 2

  2. CÁLCULO NUMÉRICO. ECUACIONES.

  3. = Dom •¾•√3 •¾•√3 •1 • –7 ECUACIÓN S • x2 = 3x •0 •3 • –1,3 •1 • –7 . = Dom=S IDENTIDAD • •0 •3 • –1,3 x2–1=(x+1)(x–1)

  4. Analizar si los siguientes pares x – 5x–2 f(x)= x–1 x·2 –12 1 4 D(2; ) ordenados pertenecen a la función f(x). A(3;4) C(1;0) B(6;–0,02)

  5. Tenemos que f(x)=y x – 5x–2 y x f(x)= x–1 3 – 51 x·2 –12 = 3 3 3 3 2 3·2 –12 3 – 5 –2 –2 NO DEF. = = = 3·4–12 0 12–12 3 A(3;4) f A(3;4) Si f(3)=4 entonces Af 3Domf .

  6. f(x)=y x – 5x–2 f(x)= x–1 6 – 54 x·2 –12 = 6 6 6 6 5 6·2 –12 6–5·2 –4 –1 6–10 = = = = 6·32–12 180 45 192–12 6 =–0,02 B(6;–0,02)f B(6;–0,02) y x .

  7. x – 5x–2 f(x)= x–1 –1 x·2 –12 1 1 1 1 Imposible en R 1 C(1;0)  f f(x)=y C(1;0) y x . 1Dom f

  8. x – 5x–2 f(x)= x–1 2 – 50 x·2 –12 = 1 4 2 2 2 2 D(2; ) 1 2·2 –12 2 –0 2 1 = = = –8 4 4–12 1 2 1 4 f f(2)= D(2; ) 4 f(x)=y x y .

  9. x –1 2 x+1 x2= ·(x+1) Dom=  –1 –bD 2a x= –4 No tiene solución en R . Resuelve la ecuación: (x–1)(x2+2x+2)=0 x2(x+1)=2 x–1=0 ó x2+2x+2=0 x3+x2=2 D=b2–4ac x=1 =22–4·1·2 x3+x2–2=0 =4–8 =–4 1 1 0 –2 . D=–4 0 1 1 2 2 0 1 2 2

  10. 1 2 1 4 ESTUDIO INDIVIDUAL Muestra que el punto M( ; ) pertenece a la función g. 3 –2 x – x g(x) = 31(log x + x) 2 .

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