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Question Number 150469 by mathdanisur last updated on 12/Aug/21

If   f(x) = (4^x /(4^x  + 2))   find   f((1/(17))) + f(((16)/(17))) =^(?)

$$\mathrm{If}\:\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\frac{\mathrm{4}^{\boldsymbol{\mathrm{x}}} }{\mathrm{4}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{2}}\:\:\:\mathrm{find}\:\:\:\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{17}}\right)\:+\:\mathrm{f}\left(\frac{\mathrm{16}}{\mathrm{17}}\right)\:\overset{?} {=} \\ $$

Answered by gsk2684 last updated on 12/Aug/21

f(x)+f(1−x) =(4^x /(4^x +2))+(4^(1−z) /(4^(1−x) +2))  =(4^x /(4^x +2))+(4/(4+2.4^x ))=(4^x /(4^x +2))+(2/(2+4^x ))=1  f(x)+f(1−x)=1⇒f((1/(17)))+f(((16)/(17)))=1

$${f}\left({x}\right)+{f}\left(\mathrm{1}−{x}\right)\:=\frac{\mathrm{4}^{{x}} }{\mathrm{4}^{{x}} +\mathrm{2}}+\frac{\mathrm{4}^{\mathrm{1}−{z}} }{\mathrm{4}^{\mathrm{1}−{x}} +\mathrm{2}} \\ $$$$=\frac{\mathrm{4}^{{x}} }{\mathrm{4}^{{x}} +\mathrm{2}}+\frac{\mathrm{4}}{\mathrm{4}+\mathrm{2}.\mathrm{4}^{{x}} }=\frac{\mathrm{4}^{{x}} }{\mathrm{4}^{{x}} +\mathrm{2}}+\frac{\mathrm{2}}{\mathrm{2}+\mathrm{4}^{{x}} }=\mathrm{1} \\ $$$${f}\left({x}\right)+{f}\left(\mathrm{1}−{x}\right)=\mathrm{1}\Rightarrow{f}\left(\frac{\mathrm{1}}{\mathrm{17}}\right)+{f}\left(\frac{\mathrm{16}}{\mathrm{17}}\right)=\mathrm{1} \\ $$

Commented by mathdanisur last updated on 12/Aug/21

Thank You Ser

$$\mathrm{Thank}\:\mathrm{You}\:\mathrm{Ser} \\ $$

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