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Question Number 189053 by mathlove last updated on 11/Mar/23

$$findf\left(x\right)$$$$1:f\left(\frac{x+1}{x-1}\right)=x+3;x\ne 1$$$$2:f\left(\frac{2x+1}{x-1}\right)=x^2+2x;x\ne 1$$$$3:f(x+1)+f(x-y)=2f\left(x\right)cosy\mathrm{\forall }x,y$$$$f\left(0\right)=f\left(\frac{\pi }{2}\right)=1$$

Answered by cortano12 last updated on 11/Mar/23

$$\left(1\right)\mathrm{f}\left(\frac{\mathrm{x}+1}{\mathrm{x}-1}\right)=\mathrm{x}+3,\mathrm{x}\ne 1$$$$\mathrm{let}\mathrm{u}=\frac{\mathrm{x}+1}{\mathrm{x}-1}\Rightarrow \mathrm{x}=\frac{\mathrm{u}+1}{\mathrm{u}-1}$$$$\mathrm{f}\left(\mathrm{u}\right)=\frac{\mathrm{u}+1}{\mathrm{u}-1}+3=\frac{4\mathrm{u}-2}{\mathrm{u}-1}$$$$\therefore f\left(x\right)=\frac{4x-2}{x-1},x\ne 1$$

Answered by aleks041103 last updated on 11/Mar/23

$$2.$$$$\frac{2x+1}{x-1}=y$$$$2x+1=yx-y$$$$(2-y)x=-y-1\Rightarrow x=\frac{y+1}{y-2}$$$$f\left(y\right)=x^2+2x+1-1={(x+1)}^2-1=$$$$=\frac{{(y+1)}^2}{{(y-2)}^2}-1=\frac{y^2+2y+1-y^2+4y-4}{y^2-4y+4}=$$$$=\frac{6y-3}{{(y-2)}^2}=\frac{3(2y-1)}{{(y-2)}^2}$$$$\Rightarrow f\left(x\right)=\frac{3(2x-1)}{{(x-2)}^2}$$

Answered by aleks041103 last updated on 11/Mar/23

$$3.$$$$x=-1\Rightarrow f\left(0\right)+f(-y-1)=2f(-1)cos\left(y\right)$$$$1+f\left(z\right)=2f(-1)cos(-z-1)$$$$f\left(z\right)=2f(-1)cos(z+1)-1$$$$\Rightarrow f(-1)=2f(-1)-1\Rightarrow f(-1)=1$$$$\Rightarrow f\left(z\right)=2cos(z+1)-1$$$$butf\left(\pi /2\right)=2cos(\frac{\pi }{2}+1)-1\ne 1$$$$\Rightarrow thereisnosolution$$

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