Respuesta :

[tex]\bf \begin{cases} f(x)=\cfrac{2}{x}\\\\ g(x)=\cfrac{x}{x+2} \end{cases}\qquad g[\quad f(x)\quad ]=\cfrac{[f(x)]}{[f(x)]+2}[/tex]

[tex]\bf (g\circ f)(x)\iff g[\quad f(x)\quad ]=\cfrac{\boxed{\frac{2}{x}}}{\boxed{\frac{2}{x}}+2} \\\\\\ g[\quad f(x)\quad ]=\cfrac{\frac{2}{x}}{\frac{2x+2}{x}}\implies g[\quad f(x)\quad ]=\cfrac{2}{x}\cdot \cfrac{x}{2x+2} \\\\\\ g[\quad f(x)\quad ]=\cfrac{2x}{2x^2+2x}\implies g[\quad f(x)\quad ]=\cfrac{2x}{2x(x+1)} \\\\\\ g[\quad f(x)\quad ]=\cfrac{1}{x+1}[/tex]

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