Question

Determina el cociente y el resto de la siguiente división, utiliza el método de Horner.
Con procedimiento ​

Answer 1

$\mathrm{Dominio\:de\:}\:\frac{10x^4+25x^3+8x^2+6x-17}{2x^2+3x-5}:\\\begin{bmatrix}\mathrm{Solucion:}\:&\:x<-\frac{5}{2}\quad \mathrm{or}\quad \:-\frac{5}{2}<x<1\quad \mathrm{or}\quad \:x>1\:\\ \:\mathrm{Notacion\:intervalo}&\:\left(-\infty \:,\:-\frac{5}{2}\right)\cup \left(-\frac{5}{2},\:1\right)\cup \left(1,\:\infty \:\right)\end{bmatrix}$

$\mathrm{Puntos\:de\:interseccion\:con\:el\:eje\:de}\:\frac{10x^4+25x^3+8x^2+6x-17}{2x^2+3x-5}:\quad \mathrm{X\:intersecta}:\:\left(0.66766\dots ,\:0\right),$

$\left(-2.39438\dots ,\:0\right),\:\mathrm{Y\:intersecta}:\:\left(0,\:\frac{17}{5}\right)$

$\mathrm{X\:intersecta}:\:\left(0.66766\dots ,\:0\right),\:\left(-2.39438\dots ,\:0\right),\:\mathrm{Y\:intersecta}:\:\left(0,\:\frac{17}{5}\right)$

$\mathrm{Puntos\:extremos\:de}\:\frac{10x^4+25x^3+8x^2+6x-17}{2x^2+3x-5}:$

$\mathrm{Minimo}\left(-2.83000\dots ,\:41.49312\dots \right),\:\mathrm{Maximo}\left(-2.10558\dots ,\:12.64791\dots \right),$

$\mathrm{Minimo}\left(-0.26659\dots ,\:3.26181\dots \right),$

$\mathrm{Maximo}\left(0.22349\dots ,\:3.53587\dots \right),$

$\mathrm{Minimo}\left(1.47869\dots ,\:36.22964\dots \right)$

$\mathrm{Combinar\:el\left(los\right)\:punto\left(s\right)\:critico\left(s\right):}\:x\approx \:-2.83000\dots ,\:x=-2.10558\dots ,\:x\approx \:-0.26659\dots ,\:x\approx \:0.22349\dots ,\:x\approx \:1.47869\dots \:\mathrm{con\:el\:dominio}:$

$-\infty \:<x<-2.83000\dots ,\:-2.83000\dots <x<-\frac{5}{2},\:-\frac{5}{2}<x<-2.10558\dots ,\:-2.10558\dots <x<-0.26659\dots ,\:-0.26659\dots <x<0.22349\dots ,\:0.22349\dots <x<1,\:1<x<1.47869\dots ,\:1.47869\dots <x<\infty \:$

$\mathrm{Sustituir\:el\:punto\:extremo}\:x=-2.83000\dots \:\mathrm{en}\:\frac{10x^4+25x^3+8x^2+6x-17}{2x^2+3x-5}\quad \Rightarrow \quad \:y=41.49312\dots$

$\mathrm{Minimo}\left(-2.83000\dots ,\:41.49312\dots \right)$

$\mathrm{Sustituir\:el\:punto\:extremo}\:x=-2.10558\dots \:\mathrm{en}\:\frac{10x^4+25x^3+8x^2+6x-17}{2x^2+3x-5}\quad \Rightarrow \quad \:y=12.64791$

$\mathrm{Maximo}\left(-2.10558\dots ,\:12.64791\dots \right)$

$\mathrm{Sustituir\:el\:punto\:extremo}\:x=-0.26659\dots \:\mathrm{en}\:\frac{10x^4+25x^3+8x^2+6x-17}{2x^2+3x-5}\quad \Rightarrow \quad \:y=3.26181\dots$

$\mathrm{Minimo}\left(-0.26659\dots ,\:3.26181\dots \right)$

$\mathrm{Sustituir\:el\:punto\:extremo}\:x=0.22349\dots \:\mathrm{en}\:\frac{10x^4+25x^3+8x^2+6x-17}{2x^2+3x-5}\quad \Rightarrow \quad \:y=3.53587\dots$

$\mathrm{Maximo}\left(0.22349\dots ,\:3.53587\dots \right)$

$\mathrm{Sustituir\:el\:punto\:extremo}\:x=1.47869\dots \:\mathrm{en}\:\frac{10x^4+25x^3+8x^2+6x-17}{2x^2+3x-5}\quad \Rightarrow \quad \:y=36.22964$$\mathrm{minimo}\left(1.47869\dots ,\:36.22964\dots \right)$

solucion

$\mathrm{Minimo}\left(-2.83000\dots ,\:41.49312\dots \right),\:\mathrm{Maximo}\left(-2.10558\dots ,\:12.64791\dots \right),$

$\mathrm{Minimo}\left(-0.26659\dots ,\:3.26181\dots \right),\:\mathrm{Maximo}\left(0.22349\dots ,\:3.53587\dots \right),$

$\mathrm{Minimo}\left(1.47869\dots ,\:36.22964\dots \right)$