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Question Number 177530 by mnjuly1970 last updated on 06/Oct/22
f (x ) + f ((( 1)/( ((1 −x^( 3) ))^(1/3) )) )= x^( 3) is given. find the value of: f (−1)=?
$$ \\ $$$${f}\:\left({x}\:\right)\:+\:{f}\:\left(\frac{\:\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{1}\:−{x}^{\:\mathrm{3}} }}\:\right)=\:{x}^{\:\mathrm{3}} \\ $$$$\:\:\:\:\:\:{is}\:{given}.\:{find}\:{the}\:{value}\:{of}: \\ $$$$\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:{f}\:\left(−\mathrm{1}\right)=? \\ $$$$ \\ $$
Answered by floor(10²Eta[1]) last updated on 06/Oct/22
f(x)+f((1/( ((1−x^3 ))^(1/3) )))=x^3 (1) x→(1/( ((1−x^3 ))^(1/3) )) f((1/( ((1−x^3 ))^(1/3) )))+f((((x^3 −1))^(1/3) /( x)))=(1/(1−x^3 )) (2) x→(1/( ((1−x^3 ))^(1/3) )) f((((x^3 −1))^(1/3) /x))+f(x)=((x^3 −1)/x^3 ) (3) (3)−(2): f(x)−f((1/( ((1−x^3 ))^(1/3) )))=((x^3 −1)/x^3 )−(1/(1−x^3 )) (4) (1)+(4): f(x)=(1/2)(((x^3 −1)/x^3 )−(1/(1−x^3 ))+x^3 ) ⇒f(−1)=(1/4)
$$\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{1}−\mathrm{x}^{\mathrm{3}} }}\right)=\mathrm{x}^{\mathrm{3}} \:\left(\mathrm{1}\right) \\ $$$$\mathrm{x}\rightarrow\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{1}−\mathrm{x}^{\mathrm{3}} }} \\ $$$$\mathrm{f}\left(\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{1}−\mathrm{x}^{\mathrm{3}} }}\right)+\mathrm{f}\left(\frac{\sqrt[{\mathrm{3}}]{\mathrm{x}^{\mathrm{3}} −\mathrm{1}}}{\:\mathrm{x}}\right)=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}^{\mathrm{3}} }\:\left(\mathrm{2}\right) \\ $$$$\mathrm{x}\rightarrow\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{1}−\mathrm{x}^{\mathrm{3}} }}\: \\ $$$$\mathrm{f}\left(\frac{\sqrt[{\mathrm{3}}]{\mathrm{x}^{\mathrm{3}} −\mathrm{1}}}{\mathrm{x}}\right)+\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}^{\mathrm{3}} −\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }\:\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{3}\right)−\left(\mathrm{2}\right): \\ $$$$\mathrm{f}\left(\mathrm{x}\right)−\mathrm{f}\left(\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{1}−\mathrm{x}^{\mathrm{3}} }}\right)=\frac{\mathrm{x}^{\mathrm{3}} −\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}^{\mathrm{3}} }\:\left(\mathrm{4}\right) \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{4}\right): \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{x}^{\mathrm{3}} −\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}^{\mathrm{3}} }+\mathrm{x}^{\mathrm{3}} \right) \\ $$$$\Rightarrow\mathrm{f}\left(−\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 07/Oct/22
thanks sir
$${thanks}\:{sir} \\ $$
Answered by mr W last updated on 06/Oct/22
f(−1)+f((1/( (2)^(1/3) )))=−1 ...(i) f((1/( (2)^(1/3) )))+f((2)^(1/3) )=(1/2) ...(ii) f((2)^(1/3) )+f(−1)=2 ...(iii) (i)+(iii)−(ii): 2f(−1)=2−1−(1/2)=(1/2) ⇒f(−1)=(1/4)
$${f}\left(−\mathrm{1}\right)+{f}\left(\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{2}}}\right)=−\mathrm{1}\:\:\:…\left({i}\right) \\ $$$${f}\left(\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{2}}}\right)+{f}\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:…\left({ii}\right) \\ $$$${f}\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}\right)+{f}\left(−\mathrm{1}\right)=\mathrm{2}\:\:\:…\left({iii}\right) \\ $$$$\left({i}\right)+\left({iii}\right)−\left({ii}\right): \\ $$$$\mathrm{2}{f}\left(−\mathrm{1}\right)=\mathrm{2}−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{f}\left(−\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by mnjuly1970 last updated on 07/Oct/22
thanks alot sir W
$${thanks}\:{alot}\:{sir}\:{W} \\ $$
Answered by a.lgnaoui last updated on 06/Oct/22
pour x=^3 (√2) f(^3 (√2) )+f(1/(^3 (√(1−2))))=2 (d apres l expressiin) donc f(^3 (√2) )+f(−1)=2 (1) pour x=−1 f(−1)+f((1/(^3 (√(1−(−1)^3 )))))=f(−1)+f((1/(^3 (√2))))=−1 (2) (1) et (2)⇒2−f(^3 (√2) )=−1−f((1/(^3 (√2)))) soit f(^3 (√2) )−f((1/(^3 (√2))))=3 (3) Calcul de f((1/(^3 (√2))))? f((1/(^3 (√2))))+f((1/(^3 (√(1−(1/2))))))=(1/2) f((1/(^3 (√2))))+f(^3 (√2))=(1/2) (4) (3) et (4) ⇒f((1/(^3 (√2))))=f(^3 (√2) )−3=(1/2)−f(−1) (3) et (4)⇒ { ((f(^3 (√2) )−f((1/(^3 (√2))))=3)),((f(^3 (√2) )+f((1/(^3 (√2))))=(1/2))) :} 2f(^3 (√2) )=(7/2) f((√2) )=(7/4) finalement en[remplacant dans(1) f((√2) )+f(−1)=2 f(−1)=2−f(^3 (√2) )=2−(7/4) f(−1)=(1/4)
$${pour}\:{x}=^{\mathrm{3}} \sqrt{\mathrm{2}}\:\:\: \\ $$$${f}\left(^{\mathrm{3}} \sqrt{\mathrm{2}}\:\right)+{f}\frac{\mathrm{1}}{\:^{\mathrm{3}} \sqrt{\mathrm{1}−\mathrm{2}}}=\mathrm{2}\:\:\left({d}\:{apres}\:{l}\:{expressiin}\right) \\ $$$$\:\:\:\:{donc}\:\:{f}\left(^{\mathrm{3}} \sqrt{\mathrm{2}}\:\right)+{f}\left(−\mathrm{1}\right)=\mathrm{2}\:\:\:\:\left(\mathrm{1}\right) \\ $$$${pour}\:\:\:{x}=−\mathrm{1} \\ $$$${f}\left(−\mathrm{1}\right)+{f}\left(\frac{\mathrm{1}}{\:^{\mathrm{3}} \sqrt{\mathrm{1}−\left(−\mathrm{1}\right)^{\mathrm{3}} }}\right)={f}\left(−\mathrm{1}\right)+{f}\left(\frac{\mathrm{1}}{\:^{\mathrm{3}} \sqrt{\mathrm{2}}}\right)=−\mathrm{1}\:\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)\:\:{et}\:\left(\mathrm{2}\right)\Rightarrow\mathrm{2}−{f}\left(^{\mathrm{3}} \sqrt{\mathrm{2}}\:\right)=−\mathrm{1}−{f}\left(\frac{\mathrm{1}}{\:^{\mathrm{3}} \sqrt{\mathrm{2}}}\right) \\ $$$${soit}\:\:{f}\left(^{\mathrm{3}} \sqrt{\mathrm{2}}\:\right)−{f}\left(\frac{\mathrm{1}}{\:^{\mathrm{3}} \sqrt{\mathrm{2}}}\right)=\mathrm{3}\:\:\:\:\:\:\:\:\left(\mathrm{3}\right) \\ $$$${Calcul}\:{de}\:\:\:\:\:\:\:{f}\left(\frac{\mathrm{1}}{\:^{\mathrm{3}} \sqrt{\mathrm{2}}}\right)? \\ $$$${f}\left(\frac{\mathrm{1}}{\:^{\mathrm{3}} \sqrt{\mathrm{2}}}\right)+{f}\left(\frac{\mathrm{1}}{\:^{\mathrm{3}} \sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${f}\left(\frac{\mathrm{1}}{\:^{\mathrm{3}} \sqrt{\mathrm{2}}}\right)+{f}\left(^{\mathrm{3}} \sqrt{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\left(\mathrm{4}\right)\: \\ $$$$\left(\mathrm{3}\right)\:\:\:{et}\:\left(\mathrm{4}\right)\:\:\Rightarrow{f}\left(\frac{\mathrm{1}}{\:^{\mathrm{3}} \sqrt{\mathrm{2}}}\right)={f}\left(^{\mathrm{3}} \sqrt{\mathrm{2}}\:\right)−\mathrm{3}=\frac{\mathrm{1}}{\mathrm{2}}−{f}\left(−\mathrm{1}\right) \\ $$$$\left(\mathrm{3}\right)\:\:{et}\:\:\left(\mathrm{4}\right)\Rightarrow\begin{cases}{{f}\left(\:^{\mathrm{3}} \sqrt{\mathrm{2}}\:\right)−{f}\left(\frac{\mathrm{1}}{\:^{\mathrm{3}} \sqrt{\mathrm{2}}}\right)=\mathrm{3}}\\{{f}\left(^{\mathrm{3}} \sqrt{\mathrm{2}}\:\right)+{f}\left(\frac{\mathrm{1}}{\:^{\mathrm{3}} \sqrt{\mathrm{2}}}\right)=\frac{\mathrm{1}}{\mathrm{2}}}\end{cases} \\ $$$$\mathrm{2}{f}\left(^{\mathrm{3}} \sqrt{\mathrm{2}}\:\right)=\frac{\mathrm{7}}{\mathrm{2}}\:\:\:\:\:{f}\left(\sqrt{\mathrm{2}}\:\right)=\frac{\mathrm{7}}{\mathrm{4}} \\ $$$${finalement}\:\:{en}\left[{remplacant}\:\:{dans}\left(\mathrm{1}\right)\right. \\ $$$${f}\left(\sqrt{\mathrm{2}}\:\right)+{f}\left(−\mathrm{1}\right)=\mathrm{2} \\ $$$$\:\:\:\:{f}\left(−\mathrm{1}\right)=\mathrm{2}−{f}\left(\:^{\mathrm{3}} \sqrt{\mathrm{2}}\:\right)=\mathrm{2}−\frac{\mathrm{7}}{\mathrm{4}} \\ $$$$\:\:\:{f}\left(−\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{4}} \\ $$

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