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Question Number 4261 by 123456 last updated on 06/Jan/16

f(x)f(1−x)=f(1)  f(x)=?

$${f}\left({x}\right){f}\left(\mathrm{1}−{x}\right)={f}\left(\mathrm{1}\right) \\ $$$${f}\left({x}\right)=? \\ $$

Commented by Yozzii last updated on 06/Jan/16

f(x)=0 for example.     f(x)f(1−x)=f(1)=constant.  ⇒f′(x)f(1−x)−f(x)f^′ (1−x)=0  ((f^′ (x))/(f^′ (1−x)))=((f(x))/(f(1−x)))=r(x)     x=1: f(1)f(0)=f(1)⇒f(0)=1 if f(1)≠0.

$${f}\left({x}\right)=\mathrm{0}\:{for}\:{example}.\: \\ $$$$ \\ $$$${f}\left({x}\right){f}\left(\mathrm{1}−{x}\right)={f}\left(\mathrm{1}\right)={constant}. \\ $$$$\Rightarrow{f}'\left({x}\right){f}\left(\mathrm{1}−{x}\right)−{f}\left({x}\right){f}^{'} \left(\mathrm{1}−{x}\right)=\mathrm{0} \\ $$$$\frac{{f}^{'} \left({x}\right)}{{f}^{'} \left(\mathrm{1}−{x}\right)}=\frac{{f}\left({x}\right)}{{f}\left(\mathrm{1}−{x}\right)}={r}\left({x}\right)\: \\ $$$$ \\ $$$${x}=\mathrm{1}:\:{f}\left(\mathrm{1}\right){f}\left(\mathrm{0}\right)={f}\left(\mathrm{1}\right)\Rightarrow{f}\left(\mathrm{0}\right)=\mathrm{1}\:{if}\:{f}\left(\mathrm{1}\right)\neq\mathrm{0}.\: \\ $$$$\: \\ $$

Answered by prakash jain last updated on 07/Jan/16

f(x)=e^(kx)   f(1−x)=e^(k−kx)   f(x)f(1−x)=e^k =f(1)

$${f}\left({x}\right)={e}^{{kx}} \\ $$$${f}\left(\mathrm{1}−{x}\right)={e}^{{k}−{kx}} \\ $$$${f}\left({x}\right){f}\left(\mathrm{1}−{x}\right)={e}^{{k}} ={f}\left(\mathrm{1}\right) \\ $$

Commented by prakash jain last updated on 07/Jan/16

Thanks. Deleted f(x)+f(1−x) part of answer.

$$\mathrm{Thanks}.\:\mathrm{Deleted}\:{f}\left({x}\right)+{f}\left(\mathrm{1}−{x}\right)\:\mathrm{part}\:\mathrm{of}\:\mathrm{answer}. \\ $$

Commented by Yozzii last updated on 07/Jan/16

f(x)f(1−x)=ax×a(1−x)=a^2 x(1−x)  f(1)=a≠a^2 x(1−x)⇒f(x)f(1−x)≠f(1)

$${f}\left({x}\right){f}\left(\mathrm{1}−{x}\right)={ax}×{a}\left(\mathrm{1}−{x}\right)={a}^{\mathrm{2}} {x}\left(\mathrm{1}−{x}\right) \\ $$$${f}\left(\mathrm{1}\right)={a}\neq{a}^{\mathrm{2}} {x}\left(\mathrm{1}−{x}\right)\Rightarrow{f}\left({x}\right){f}\left(\mathrm{1}−{x}\right)\neq{f}\left(\mathrm{1}\right) \\ $$

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