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Question Number 82431 by Power last updated on 21/Feb/20

Commented by Power last updated on 21/Feb/20

solution sir please prove it

$$\mathrm{solution}\:\mathrm{sir}\:\mathrm{please}\:\mathrm{prove}\:\:\mathrm{it} \\ $$

Commented by mr W last updated on 21/Feb/20

what′s your result sir? f(x)=?

$${what}'{s}\:{your}\:{result}\:{sir}?\:{f}\left({x}\right)=? \\ $$

Commented by mr W last updated on 21/Feb/20

i got f(x)=(1/2)(x−(1/(1−x))+((x−1)/x)), right?

$${i}\:{got}\:{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({x}−\frac{\mathrm{1}}{\mathrm{1}−{x}}+\frac{{x}−\mathrm{1}}{{x}}\right),\:{right}? \\ $$

Commented by Power last updated on 21/Feb/20

f(x)=((x^3 −x+1)/(2x(x−1)))

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}^{\mathrm{3}} −\mathrm{x}+\mathrm{1}}{\mathrm{2x}\left(\mathrm{x}−\mathrm{1}\right)} \\ $$

Commented by mr W last updated on 21/Feb/20

mister power: f(x)=(1/2)(x−(1/(1−x))+((x−1)/x)) and f(x)=((x^3 −x+1)/(2x(x−1))) are the same!

$${mister}\:{power}: \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({x}−\frac{\mathrm{1}}{\mathrm{1}−{x}}+\frac{{x}−\mathrm{1}}{{x}}\right)\:{and}\: \\ $$$${f}\left({x}\right)=\frac{{x}^{\mathrm{3}} −{x}+\mathrm{1}}{\mathrm{2}{x}\left({x}−\mathrm{1}\right)} \\ $$$${are}\:{the}\:{same}! \\ $$

Answered by mr W last updated on 21/Feb/20

f(x)+f((1/(1−x)))=x ...(i) replace x in (i) with (1/(1−x)): f((1/(1−x)))+f(((x−1)/x))=(1/(1−x)) ...(ii) replace x in (i) with ((x−1)/x): f(((x−1)/x))+f(x)=((x−1)/x) ...(iii) (ii)−(iii): f((1/(1−x)))−f(x)=(1/(1−x))−((x−1)/x) ...(iv) (i)−(iv): 2f(x)=x−(1/(1−x))+((x−1)/x) ⇒f(x)=(1/2)(x−(1/(1−x))+((x−1)/x))=((x^3 −x+1)/(2x(x−1)))

$${f}\left({x}\right)+{f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)={x}\:\:\:...\left({i}\right) \\ $$$${replace}\:{x}\:{in}\:\left({i}\right)\:{with}\:\frac{\mathrm{1}}{\mathrm{1}−{x}}: \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)+{f}\left(\frac{{x}−\mathrm{1}}{{x}}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\:\:...\left({ii}\right) \\ $$$${replace}\:{x}\:{in}\:\left({i}\right)\:{with}\:\frac{{x}−\mathrm{1}}{{x}}: \\ $$$${f}\left(\frac{{x}−\mathrm{1}}{{x}}\right)+{f}\left({x}\right)=\frac{{x}−\mathrm{1}}{{x}}\:\:\:...\left({iii}\right) \\ $$$$ \\ $$$$\left({ii}\right)−\left({iii}\right): \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)−{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}−{x}}−\frac{{x}−\mathrm{1}}{{x}}\:\:\:...\left({iv}\right) \\ $$$$\left({i}\right)−\left({iv}\right): \\ $$$$\mathrm{2}{f}\left({x}\right)={x}−\frac{\mathrm{1}}{\mathrm{1}−{x}}+\frac{{x}−\mathrm{1}}{{x}} \\ $$$$\Rightarrow{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({x}−\frac{\mathrm{1}}{\mathrm{1}−{x}}+\frac{{x}−\mathrm{1}}{{x}}\right)=\frac{{x}^{\mathrm{3}} −{x}+\mathrm{1}}{\mathrm{2}{x}\left({x}−\mathrm{1}\right)} \\ $$

Commented by Power last updated on 21/Feb/20

yes thank you sir

$$\mathrm{yes}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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