Question

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2x³ - 20x² + 50 = 0

Answer 1

Definição do método

O método de Newton-Raphson utiliza um processo interativo para aproximar-se de uma raiz de uma função
$x_{n+1}=x_n\:-\:\frac{f\left(x_n\right)}{f\:'\left(x_n\right)}$


$\mathrm{Encontrar\:uma\:soluçao\:para\:}2x^3-20x^2+50=0\mathrm{\:utilizando\:o\:metodo\:de\:Newton-Raphson}$
$f\left(x\right)=2x^3-20x^2+50$
$\mathrm{Encontrar\:}f\:'\left(x\right):\quad 6x^2-40x$
$\frac{d}{dx}\left(2x^3-20x^2+50\right)$
$\mathrm{Aplicar\:a\:regra\:da\:soma/diferenca}:\quad \left(f\pm g\right)^'=f^'\pm g^'$


$\frac{d}{dx}\left(2x^3\right)=6x^2$
$\frac{d}{dx}\left(2x^3\right)$
$\mathrm{Retirar\:a\:constante}:\quad \left(a\cdot f\right)^'=a\cdot f^'$
$=2\frac{d}{dx}\left(x^3\right)$
$\mathrm{Aplicar\:a\:regra\:da\:potencia}:\quad \frac{d} {dx}\left(x^a\right)=a\cdot x^{a-1}$
$=2\cdot \:3x^{3-1}$
$Simplificar$
$=6x^2$


$\frac{d}{dx}\left(20x^2\right)=40x$
$\frac{d}{dx}\left(20x^2\right)$
$\mathrm{Retirar\:a\:constante}:\quad \left(a\cdot f\right)^'=a\cdot f^'$
$=20\frac{d}{dx}\left(x^2\right)$$\mathrm{Aplicar\:a\:regra\:da\:potencia}:\quad \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1}$
$=20\cdot \:2x^{2-1}$
$\mathrm{Simplificar}$
$=40x$


$\frac{d}{dx}\left(50\right)=0$
$\frac{d}{dx}\left(50\right)$
$\mathrm{Derivada\:de\:uma\:constante}:\quad \frac{d}{dx}\left(a\right)=0$
$=6x^2-40x+0$
$Simplificar$
$=6x^2-40x$


$\mathrm{Seja\:}x_0=-1$
$\mathrm{Calcular\:}x_{n+1}\mathrm{\:ate\:que\:}\Delta \:x_{n+1}\:\ \textless \ \:0.000001$

$x_1=-1.60870\dots \quad :\quad Δx_1=0.60870\dots$
$f\left(x_0\right)=2\left(-1\right)^3-20\left(-1\right)^2+50=28$
$f\left(x_0\right)=2\left(-1\right)^3-20\left(-1\right)^2+50=28$
$x_1=-1-\frac{28}{46}=-1.60870\dots$
$x_1=-1.60870\dots$