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Question Number 66104 by Rio Michael last updated on 09/Aug/19

$$f\left(x\right)=2x^3-x-4$$$$showthattheequationf\left(x\right)=0hasrootbetween1and2$$$$showthattheequationf\left(x\right)=0canbewrittenas$$$$x=\sqrt{(\frac{2}{x}+\frac{1}{2})}$$$$usetheiteration$$$$x_{n+1}=\sqrt{(\frac{2}{x_n}+\frac{1}{2}),}$$$$with{x}_0=1.385tofindto3decimalplacesthevalueof{x}_1.$$$$$$

Commented by mathmax by abdo last updated on 09/Aug/19

$$wehavef\left(x\right)=2x^3-x-4\Rightarrow f^{{}^{\prime }}\left(x\right)=6x^2-1=(\sqrt{6}x-1)(\sqrt{6}x+1)$$$$f^{{}^{\prime }}\left(x\right)=0\iff x=\stackrel{-}{+}\frac{1}{\sqrt{6}}variationoff\left(x\right)$$$$x-\mathrm{\infty }-\frac{1}{\sqrt{6}}\frac{1}{\sqrt{6}}+\mathrm{\infty }$$$$f^{{}^{\prime }}\left(x\right)+-+$$$$f\left(x\right)-\mathrm{\infty }incr.f(-\frac{1}{\sqrt{6}})decrf\left(\frac{1}{\sqrt{6}}\right)incr+\mathrm{\infty }$$$$fisincreasingon[\frac{1}{\sqrt{6}},+\mathrm{\infty }[$$$$f\left(1\right)=2-1-4=-5$$$$f\left(2\right)=16-2-4=10\Rightarrow f\left(1\right)f\left(2\right)<0\Rightarrow \mathrm{\exists }\alpha \in ]1,2[/f\left(\alpha \right)=0$$$$x=\sqrt{\frac{2}{x}+\frac{1}{2}}andx>0\Rightarrow x^2=\frac{4+x}{2x}\Rightarrow 2x^3=4+x\Rightarrow 2x^3-x-4=0\Rightarrow$$$$f\left(x\right)=0sof\left(x\right)=0andx>0\iff x=\sqrt{\frac{2}{x}+\frac{1}{2}}$$$$ifweconsidertheiteration{x}_{n+1}=\sqrt{\frac{2}{x_n}+\frac{1}{2}}$$$$weget{x}_1=\sqrt{\frac{2}{x_0}+\frac{1}{2}}=\sqrt{\frac{4+x_0}{2x_0}}=\sqrt{\frac{4+1,385}{2\times 1,385}}$$$$resttofinishthecalculus....$$

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