Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 83850 by Rio Michael last updated on 06/Mar/20

$$\mathrm{Find}\mathrm{the}\mathrm{maximum}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{function}f,\mathrm{defined}\mathrm{by}$$$$f\left(x\right)=\frac{x}{1+x^2},x\in \mathbb{R}$$

Commented by mathmax by abdo last updated on 06/Mar/20

$${lim}_{x\to \mathrm{\infty }}f\left(x\right)=0and{f}^{{}^{\prime }}\left(x\right)=\frac{1+x^2-x\left(2x\right)}{{(1+x^2)}^2}=\frac{1-x^2}{{(1+x^2)}^2}$$$$wehave{f}^{{}^{\prime }}\left(x\right)⩾0\iff -1⩽x⩽1$$$$x-\mathrm{\infty }-101+\mathrm{\infty }$$$$f^{{}^{\prime }}-++-$$$$f0decr-\frac{1}{2}inc0inc\frac{1}{2}dec0$$$$\Rightarrow {max}_{x\in R}f\left(x\right)=\frac{1}{2}$$

Answered by john santu last updated on 06/Mar/20

$$({\mathrm{x}}^2+1)\mathrm{y}-\mathrm{x}=0$$$${\mathrm{yx}}^2-\mathrm{x}+\mathrm{y}=0$$$$\Rightarrow \mathrm{\Delta }=1-4.{\mathrm{y}}^2⩾0$$$$(2\mathrm{y}-1)(2\mathrm{y}+1)⩽0$$$$-\frac{1}{2}⩽\mathrm{y}⩽\frac{1}{2}$$$$\{\begin{array}{l}\max =\frac{1}{2}\\ \min =-\frac{1}{2}\end{array}$$

Answered by mr W last updated on 07/Mar/20

$$f\left(x\right)=\frac{x}{1+x^2}=\frac{1}{x+\frac{1}{x}}$$$$⩽\frac{1}{2}\to max.$$$$⩾\frac{1}{-2}=-\frac{1}{2}\to min.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com