Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 158306 by mathlove last updated on 02/Nov/21

Commented by cortano last updated on 02/Nov/21

f(1−x)=(4^(1−x) /(4^(1−x) +2)) = ((4/4^x )/((4/4^x )+2))=(4/(4+2.4^x )) f(x)+f(1−x)=(4^x /(4^x +2))+(4/(2.4^x +4)) = ((2.4^x +4)/(2.4^x +4)) = 1 ⇒f((1/(1997)))+f((2/(1997)))+...+f(((1996)/(1997)))=((1998)/2)=999

$${f}\left(\mathrm{1}−{x}\right)=\frac{\mathrm{4}^{\mathrm{1}−{x}} }{\mathrm{4}^{\mathrm{1}−{x}} +\mathrm{2}}\:=\:\frac{\frac{\mathrm{4}}{\mathrm{4}^{{x}} }}{\frac{\mathrm{4}}{\mathrm{4}^{{x}} }+\mathrm{2}}=\frac{\mathrm{4}}{\mathrm{4}+\mathrm{2}.\mathrm{4}^{{x}} } \\ $$$${f}\left({x}\right)+{f}\left(\mathrm{1}−{x}\right)=\frac{\mathrm{4}^{{x}} }{\mathrm{4}^{{x}} +\mathrm{2}}+\frac{\mathrm{4}}{\mathrm{2}.\mathrm{4}^{{x}} +\mathrm{4}} \\ $$$$\:=\:\frac{\mathrm{2}.\mathrm{4}^{{x}} +\mathrm{4}}{\mathrm{2}.\mathrm{4}^{{x}} +\mathrm{4}}\:=\:\mathrm{1} \\ $$$$\Rightarrow{f}\left(\frac{\mathrm{1}}{\mathrm{1997}}\right)+{f}\left(\frac{\mathrm{2}}{\mathrm{1997}}\right)+...+{f}\left(\frac{\mathrm{1996}}{\mathrm{1997}}\right)=\frac{\mathrm{1998}}{\mathrm{2}}=\mathrm{999} \\ $$

Commented by mathlove last updated on 02/Nov/21

why faind f(1−x) and why the end ((1998)/2)

$${why}\:{faind}\:{f}\left(\mathrm{1}−{x}\right) \\ $$$${and}\:{why}\:{the}\:{end}\:\frac{\mathrm{1998}}{\mathrm{2}} \\ $$

Commented by tounghoungko last updated on 03/Nov/21

since ((1996)/(1997))=1−(1/(1997)) ((1995)/(1997))=1−(2/(1997)) . and so so on

$${since}\:\frac{\mathrm{1996}}{\mathrm{1997}}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1997}} \\ $$$$\:\frac{\mathrm{1995}}{\mathrm{1997}}=\mathrm{1}−\frac{\mathrm{2}}{\mathrm{1997}}\:.\:{and}\:{so}\:{so}\:{on} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com