Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 183669 by mnjuly1970 last updated on 28/Dec/22

$$$$$$f\left(x\right)=\frac{x}{1+x+x^2}$$$${min}_f=?$$

Answered by manolex last updated on 28/Dec/22

$$f\left(x\right)=\frac{1}{\frac{1}{x}+1+x}$$$$\frac{1}{x}+x\gg 2$$$$\frac{1}{x}+x+1\gg 3$$$$\frac{1}{\frac{1}{x}+x+1}\ll \frac{1}{3}ismax$$$$l_{x\to \mathrm{\infty }}imf\left(x\right)=\frac{1}{\frac{1}{\mathrm{\infty }}+1+\mathrm{\infty }}=\frac{1}{0+1+\mathrm{\infty }}=\frac{1}{\mathrm{\infty }}=0$$$$correction$$$$x⟨0$$$$x+\frac{1}{x}\ll -2$$$$x+\frac{1}{x}+1\ll -1$$$$\frac{1}{x+\frac{1}{x}+1}\gg -1$$$$-1\ll f\left(x\right)\ll \frac{1}{3}$$$$min=-1$$

Commented by manolex last updated on 28/Dec/22

$$thanks,SirW,forcorrection$$

Answered by mr W last updated on 28/Dec/22

$$ifx=0,f\left(x\right)=0$$$$forx\ne 0:$$$$f\left(x\right)=\frac{1}{x+\frac{1}{x}+1}$$$$forx>0:$$$$f\left(x\right)=\frac{1}{x+\frac{1}{x}+1}⩽\frac{1}{2+1}=\frac{1}{3}$$$$forx<0:$$$$f\left(x\right)=\frac{1}{x+\frac{1}{x}+1}=\frac{1}{-(-x+\frac{1}{-x})+1}⩾\frac{1}{-2+1}=-1$$$$\Rightarrow -1⩽f\left(x\right)⩽\frac{1}{3}$$$$i.e.max.f\left(x\right)=\frac{1}{3}andmin.f\left(x\right)=-1$$

Answered by Frix last updated on 28/Dec/22

$$f^{\prime }\left(x\right)=\frac{1-x^2}{{(1+x+x^2)}^2}$$$$f^{″}\left(x\right)=-\frac{2(1+3x-x^3)}{(1+x+x^2)}$$$$f^{\prime }\left(x\right)=0\Rightarrow x_1=-1\wedge x_2=1$$$$f^{″}(-1)=2>0\Rightarrow \mathrm{minimum}\mathrm{is}-1\mathrm{at}x=-1$$$$f^{″}\left(1\right)=-\frac{2}{9}\Rightarrow \mathrm{maximum}\mathrm{is}\frac{1}{3}\mathrm{at}x=1$$

Answered by manxsol last updated on 28/Dec/22

$$$$$$a\ll \frac{x}{1+x+x^2}\ll b$$$$a\ll \frac{x}{1+x+x^2}$$$$\frac{a+ax+{ax}^2-x}{1+x+x^2}\ll 0$$$${ax}^2+(a-1)x+a\ll 0$$$$△={(a-1)}^2-4a^2\gg 0$$$$(a-1-2a)(a-1+2a)\gg 0$$$$(-a-1)(3a-1)\gg 0$$$$(a+1)(3a-1)\ll 0$$$$-1\ll a\ll \frac{1}{3}$$$$minf\left(x\right)=-1$$$$$$

Commented by mr W last updated on 29/Dec/22

$$niceapproach!$$

Commented by manxsol last updated on 29/Dec/22

$$followinghisvision,thanks$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com