1 / 10

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.

quemby-stanley
Download Presentation

CLASE 2

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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 .

More Related