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Question Number 130894 by EDWIN88 last updated on 30/Jan/21

Find the function f(x) if 3f(x−1)−f(((1−x)/x)) = 2x

$${Find}\:{the}\:{function}\:{f}\left({x}\right)\:{if}\: \\ $$$$\:\mathrm{3}{f}\left({x}−\mathrm{1}\right)−{f}\left(\frac{\mathrm{1}−{x}}{{x}}\right)\:=\:\mathrm{2}{x}\: \\ $$

Commented by benjo_mathlover last updated on 30/Jan/21

այս հարցը շատ հետաքրքիր է

Commented by EDWIN88 last updated on 30/Jan/21

շնորհակալություն

Answered by mr W last updated on 30/Jan/21

3f(x−1)−f((1/x)−1)=2x ...(i) 3f((1/x)−1)−f(x−1)=(2/x) ...(ii) 3×(i)+(ii): 8f(x−1)=2(3x+(1/x)) f(x−1)=(1/4)(3x+(1/x)) ⇒f(x)=(1/4)(3(x+1)+(1/(x+1)))

$$\mathrm{3}{f}\left({x}−\mathrm{1}\right)−{f}\left(\frac{\mathrm{1}}{{x}}−\mathrm{1}\right)=\mathrm{2}{x}\:\:\:...\left({i}\right) \\ $$$$\mathrm{3}{f}\left(\frac{\mathrm{1}}{{x}}−\mathrm{1}\right)−{f}\left({x}−\mathrm{1}\right)=\frac{\mathrm{2}}{{x}}\:\:\:...\left({ii}\right) \\ $$$$\mathrm{3}×\left({i}\right)+\left({ii}\right): \\ $$$$\mathrm{8}{f}\left({x}−\mathrm{1}\right)=\mathrm{2}\left(\mathrm{3}{x}+\frac{\mathrm{1}}{{x}}\right) \\ $$$${f}\left({x}−\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{3}{x}+\frac{\mathrm{1}}{{x}}\right) \\ $$$$\Rightarrow{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{3}\left({x}+\mathrm{1}\right)+\frac{\mathrm{1}}{{x}+\mathrm{1}}\right) \\ $$

Commented by EDWIN88 last updated on 30/Jan/21

����nice

Answered by benjo_mathlover last updated on 30/Jan/21

replace ((1−x)/x) by t ⇒ x=(1/(t+1)) 3f((1/(t+1))−1)−f(t)=(2/(t+1)) 3f(((−t)/(t+1)))−f(t)=(2/(t+1))... (i) replace x−1 by t ⇒x=t+1 3f(t)−f(((1−t−1)/(t+1)))= 2t+2 3f(x)−f(((−t)/(t+1)))=2t+2 ... (ii) ⇔ 3×(ii)⇒ 9f(t)−3f(((−t)/(t+1)))= 6t+6 ⇔ 1×(i)⇒ −f(t)+3f(((−t)/(t+1)))= (2/(t+1)) __________________________ + ⇔ 8f(t) = 6t+6+(2/(t+1)) ⇔ f(x) = ((3x+3)/4) + (1/(4x+4)) = ((12(x+1)^2 +4)/(16(x+1))) = ((3x^2 +6x+4)/(4x+4))

$$\mathrm{replace}\:\frac{\mathrm{1}−\mathrm{x}}{\mathrm{x}}\:\mathrm{by}\:\mathrm{t}\:\Rightarrow\:\mathrm{x}=\frac{\mathrm{1}}{\mathrm{t}+\mathrm{1}} \\ $$$$\mathrm{3f}\left(\frac{\mathrm{1}}{\mathrm{t}+\mathrm{1}}−\mathrm{1}\right)−\mathrm{f}\left(\mathrm{t}\right)=\frac{\mathrm{2}}{\mathrm{t}+\mathrm{1}} \\ $$$$\:\mathrm{3f}\left(\frac{−\mathrm{t}}{\mathrm{t}+\mathrm{1}}\right)−\mathrm{f}\left(\mathrm{t}\right)=\frac{\mathrm{2}}{\mathrm{t}+\mathrm{1}}...\:\left(\mathrm{i}\right) \\ $$$$\mathrm{replace}\:\mathrm{x}−\mathrm{1}\:\mathrm{by}\:\mathrm{t}\:\Rightarrow\mathrm{x}=\mathrm{t}+\mathrm{1} \\ $$$$\mathrm{3f}\left(\mathrm{t}\right)−\mathrm{f}\left(\frac{\mathrm{1}−\mathrm{t}−\mathrm{1}}{\mathrm{t}+\mathrm{1}}\right)=\:\mathrm{2t}+\mathrm{2} \\ $$$$\mathrm{3f}\left(\mathrm{x}\right)−\mathrm{f}\left(\frac{−\mathrm{t}}{\mathrm{t}+\mathrm{1}}\right)=\mathrm{2t}+\mathrm{2}\:...\:\left(\mathrm{ii}\right) \\ $$$$\Leftrightarrow\:\mathrm{3}×\left(\mathrm{ii}\right)\Rightarrow\:\mathrm{9f}\left(\mathrm{t}\right)−\mathrm{3f}\left(\frac{−\mathrm{t}}{\mathrm{t}+\mathrm{1}}\right)=\:\mathrm{6t}+\mathrm{6} \\ $$$$\Leftrightarrow\:\mathrm{1}×\left(\mathrm{i}\right)\Rightarrow\:−\mathrm{f}\left(\mathrm{t}\right)+\mathrm{3f}\left(\frac{−\mathrm{t}}{\mathrm{t}+\mathrm{1}}\right)=\:\frac{\mathrm{2}}{\mathrm{t}+\mathrm{1}} \\ $$$$\:\:\:\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\:+ \\ $$$$\:\Leftrightarrow\:\mathrm{8f}\left(\mathrm{t}\right)\:=\:\mathrm{6t}+\mathrm{6}+\frac{\mathrm{2}}{\mathrm{t}+\mathrm{1}} \\ $$$$\:\Leftrightarrow\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\frac{\mathrm{3x}+\mathrm{3}}{\mathrm{4}}\:+\:\frac{\mathrm{1}}{\mathrm{4x}+\mathrm{4}}\:=\:\frac{\mathrm{12}\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4}}{\mathrm{16}\left(\mathrm{x}+\mathrm{1}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{3x}^{\mathrm{2}} +\mathrm{6x}+\mathrm{4}}{\mathrm{4x}+\mathrm{4}} \\ $$

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