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Question Number 155809 by zainaltanjung last updated on 05/Oct/21

$$\mathrm{Given}\mathrm{that}f\circ g=\frac{x^2}{2x^2-x+4}\mathrm{and}$$$$g\left(x\right)=\frac{x}{x-2},\mathrm{find}f\left(x\right)?$$$$$$

Answered by PRITHWISH SEN 2 last updated on 05/Oct/21

$$\mathrm{put}\mathrm{t}=\frac{\mathrm{x}}{\mathrm{x}-2}$$$$\mathrm{x}=\frac{2}{\mathrm{t}-1}+2=\frac{2\mathrm{t}}{\mathrm{t}-1}$$$$\mathrm{f}\left(\mathrm{t}\right)=\frac{{\left(\frac{2\mathrm{t}}{\mathrm{t}-1}\right)}^2}{2{\left(\frac{2\mathrm{t}}{\mathrm{t}-1}\right)}^2-\frac{2\mathrm{t}}{\mathrm{t}-1}+4}$$$$=\frac{{4\mathrm{t}}^2}{{10\mathrm{t}}^2-6\mathrm{t}+4}$$$$\therefore \mathrm{f}\left(\mathrm{x}\right)=\frac{{4\mathrm{x}}^2}{{10\mathrm{x}}^2-6\mathrm{x}+4}$$

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