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f-x-f-1-x-f-x-




Question Number 197585 by Erico last updated on 23/Sep/23
f  (x)=f((1/x))  ⇒ f(x)=?
$$\mathrm{f}\: \left(\mathrm{x}\right)=\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}}\right) \\ $$$$\Rightarrow\:\mathrm{f}\left(\mathrm{x}\right)=? \\ $$
Answered by EdwarT last updated on 24/Sep/23
−(1/x^2 )
$$−\frac{\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$
Commented by mr W last updated on 24/Sep/23
have you checked your solution?  f(x)=−(1/x^2 )  f((1/x))=−x^2   but f′(x)=(2/x^3 ) ≠ f((1/x))
$${have}\:{you}\:{checked}\:{your}\:{solution}? \\ $$$${f}\left({x}\right)=−\frac{\mathrm{1}}{{x}^{\mathrm{2}} } \\ $$$${f}\left(\frac{\mathrm{1}}{{x}}\right)=−{x}^{\mathrm{2}} \\ $$$${but}\:{f}'\left({x}\right)=\frac{\mathrm{2}}{{x}^{\mathrm{3}} }\:\neq\:{f}\left(\frac{\mathrm{1}}{{x}}\right) \\ $$
Answered by Erico last updated on 24/Sep/23
May be like this  f(x)=2(√x)cos(((√3)/2)lnx−(π/3))
$$\mathrm{May}\:\mathrm{be}\:\mathrm{like}\:\mathrm{this} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{2}\sqrt{\mathrm{x}}\mathrm{cos}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{lnx}−\frac{\pi}{\mathrm{3}}\right) \\ $$
Answered by Frix last updated on 24/Sep/23
For x>0  f(x)=2(√x)sin ((2π+3(√3)ln x)/( 6))
$$\mathrm{For}\:{x}>\mathrm{0} \\ $$$${f}\left({x}\right)=\mathrm{2}\sqrt{{x}}\mathrm{sin}\:\frac{\mathrm{2}\pi+\mathrm{3}\sqrt{\mathrm{3}}\mathrm{ln}\:{x}}{\:\mathrm{6}} \\ $$
Answered by witcher3 last updated on 24/Sep/23
f′(x)=f((1/x))  f′′(x)=−(1/x^2 )f′((1/x))=−(1/x^2 )f(x)  ⇔x^2 f′′(x)+f(x)=0  f(x)=x^r   ⇒(r(r−1)+1)x^r =0  ⇒r^2 −r+1=0  r=((1+i(√3))/2),((1−i(√3))/2)  f(x)=x^(1/2) (ax^(i/( 2(√3))) +bx^(−(i/(2 (√3)))) )  =(√x)(asin(((ln(x))/(2 ))(√3))+bcos(((ln(x))/2)(√3))    f′(x)=f((1/x))  ⇒  (1/( (√x)))(((a/2)−(b/( 2))(√3))sin(((ln(x))/(2 ))](√3))+((b/2)+(a/( 2))(√3))cos(((ln(x))/(2 ))(√3)))  =(1/( (√x)))(asin(((ln((1/x)))/( 2))(√3))+bcos(((ln((1/x)))/2)(√3))),∀x>0  ⇒(a/2)−(b/( 2))(√3)=−a  (b/2)+(a/( 2))(√3)=b  b=a(√3)  f(x)=a(√x)(sin(((ln(x)(√3))/2))+(√3)cos(((ln(x)(√3))/2)))  =2a(√x)(sin(((ln(x)(√3))/2)+(π/3))),a∈R
$$\mathrm{f}'\left(\mathrm{x}\right)=\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}}\right) \\ $$$$\mathrm{f}''\left(\mathrm{x}\right)=−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\mathrm{f}'\left(\frac{\mathrm{1}}{\mathrm{x}}\right)=−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\mathrm{f}\left(\mathrm{x}\right) \\ $$$$\Leftrightarrow\mathrm{x}^{\mathrm{2}} \mathrm{f}''\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{x}\right)=\mathrm{0} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{r}} \\ $$$$\Rightarrow\left(\mathrm{r}\left(\mathrm{r}−\mathrm{1}\right)+\mathrm{1}\right)\mathrm{x}^{\mathrm{r}} =\mathrm{0} \\ $$$$\Rightarrow\mathrm{r}^{\mathrm{2}} −\mathrm{r}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{r}=\frac{\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}},\frac{\mathrm{1}−\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{ax}^{\frac{\mathrm{i}}{\:\mathrm{2}\sqrt{\mathrm{3}}}} +\mathrm{bx}^{−\frac{\mathrm{i}}{\mathrm{2}\:\sqrt{\mathrm{3}}}} \right) \\ $$$$=\sqrt{\mathrm{x}}\left(\mathrm{asin}\left(\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\mathrm{2}\:}\sqrt{\mathrm{3}}\right)+\mathrm{bcos}\left(\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\mathrm{2}}\sqrt{\mathrm{3}}\right)\right. \\ $$$$ \\ $$$$\mathrm{f}'\left(\mathrm{x}\right)=\mathrm{f}\left(\frac{\mathrm{1}}{\mathrm{x}}\right) \\ $$$$\Rightarrow \\ $$$$\left.\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}}}\left(\left(\frac{\mathrm{a}}{\mathrm{2}}−\frac{\mathrm{b}}{\:\mathrm{2}}\sqrt{\mathrm{3}}\right)\mathrm{sin}\left(\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\mathrm{2}\:}\right]\sqrt{\mathrm{3}}\right)+\left(\frac{\mathrm{b}}{\mathrm{2}}+\frac{\mathrm{a}}{\:\mathrm{2}}\sqrt{\mathrm{3}}\right)\mathrm{cos}\left(\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\mathrm{2}\:}\sqrt{\mathrm{3}}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}}}\left(\mathrm{asin}\left(\frac{\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)}{\:\mathrm{2}}\sqrt{\mathrm{3}}\right)+\mathrm{bcos}\left(\frac{\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)}{\mathrm{2}}\sqrt{\mathrm{3}}\right)\right),\forall\mathrm{x}>\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{a}}{\mathrm{2}}−\frac{\mathrm{b}}{\:\mathrm{2}}\sqrt{\mathrm{3}}=−\mathrm{a} \\ $$$$\frac{\mathrm{b}}{\mathrm{2}}+\frac{\mathrm{a}}{\:\mathrm{2}}\sqrt{\mathrm{3}}=\mathrm{b} \\ $$$$\mathrm{b}=\mathrm{a}\sqrt{\mathrm{3}} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{a}\sqrt{\mathrm{x}}\left(\mathrm{sin}\left(\frac{\mathrm{ln}\left(\mathrm{x}\right)\sqrt{\mathrm{3}}}{\mathrm{2}}\right)+\sqrt{\mathrm{3}}\mathrm{cos}\left(\frac{\mathrm{ln}\left(\mathrm{x}\right)\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\right) \\ $$$$=\mathrm{2a}\sqrt{\mathrm{x}}\left(\mathrm{sin}\left(\frac{\mathrm{ln}\left(\mathrm{x}\right)\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\pi}{\mathrm{3}}\right)\right),\mathrm{a}\in\mathbb{R} \\ $$$$ \\ $$$$ \\ $$

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