Question

Determine f de modo que g(f(x))=x , sendo g(x)=x^2-2x

Answer 1

Chame f(x) de u:

$g(x)=x^{2}-2x\\g(f(x))=[f(x)]^{2}-2[f(x)]=x~~~\therefore~~~[f(x)]^{2}-2[f(x)]-x=0\\\\\\\Delta=b^{2}-4ac\\\Delta=(-2)^{2}-4\cdot1\cdot(-x)\\\Delta=4+4x\\\Delta=4(1+x)\\\sqrt{\Delta}=\sqrt{4}\sqrt{1+x}=2\sqrt{1+x}\\\\\\f(x)=\dfrac{-b\pm\sqrt{\Delta}}{2a}=\dfrac{-(-2)\pm2\sqrt{x+1}}{2\cdot1}=\dfrac{2\pm2\sqrt{x+1}}{2}=1\pm\sqrt{x+1}$

Portanto:

$\boxed{\boxed{f(x)=1+\sqrt{1+x}~~~~~~ou~~~~~~f(x)=1-\sqrt{1+x}}}$