Por favor, me ajudem. Imploro!! Determine f(x) se f(x/x+1)=x^2.
Soluções para a tarefa
Respondido por
1
Consideremos a função
definida por
![g(x)=\dfrac{x}{x+1}. g(x)=\dfrac{x}{x+1}.](https://tex.z-dn.net/?f=g%28x%29%3D%5Cdfrac%7Bx%7D%7Bx%2B1%7D.)
Pelo enunciado, temos que
![f[g(x)]=x^{2}. f[g(x)]=x^{2}.](https://tex.z-dn.net/?f=f%5Bg%28x%29%5D%3Dx%5E%7B2%7D.)
Isolando
em função de ![g(x): g(x):](https://tex.z-dn.net/?f=g%28x%29%3A)
![g(x)=\dfrac{x}{x+1}\\ \\\\ (x+1)\cdot g(x)=x\\ \\ x\cdot g(x)+g(x)=x\\ \\ x\cdot g(x)-x=-g(x)\\ \\ x\cdot [g(x)-1]=-g(x)\\ \\ \\ x=-\dfrac{g(x)}{g(x)-1}\\ \\ \\ \boxed{x=\dfrac{g(x)}{1-g(x)}}\,,~~~~g(x)\neq 1. g(x)=\dfrac{x}{x+1}\\ \\\\ (x+1)\cdot g(x)=x\\ \\ x\cdot g(x)+g(x)=x\\ \\ x\cdot g(x)-x=-g(x)\\ \\ x\cdot [g(x)-1]=-g(x)\\ \\ \\ x=-\dfrac{g(x)}{g(x)-1}\\ \\ \\ \boxed{x=\dfrac{g(x)}{1-g(x)}}\,,~~~~g(x)\neq 1.](https://tex.z-dn.net/?f=g%28x%29%3D%5Cdfrac%7Bx%7D%7Bx%2B1%7D%5C%5C+%5C%5C%5C%5C+%28x%2B1%29%5Ccdot+g%28x%29%3Dx%5C%5C+%5C%5C+x%5Ccdot+g%28x%29%2Bg%28x%29%3Dx%5C%5C+%5C%5C+x%5Ccdot+g%28x%29-x%3D-g%28x%29%5C%5C+%5C%5C+x%5Ccdot+%5Bg%28x%29-1%5D%3D-g%28x%29%5C%5C+%5C%5C+%5C%5C+x%3D-%5Cdfrac%7Bg%28x%29%7D%7Bg%28x%29-1%7D%5C%5C+%5C%5C+%5C%5C+%5Cboxed%7Bx%3D%5Cdfrac%7Bg%28x%29%7D%7B1-g%28x%29%7D%7D%5C%2C%2C%7E%7E%7E%7Eg%28x%29%5Cneq+1.)
Logo,
![f[g(x)]=x^{2}\\ \\ \\ f[g(x)]=\left[\dfrac{g(x)}{1-g(x)} \right ]^{2} f[g(x)]=x^{2}\\ \\ \\ f[g(x)]=\left[\dfrac{g(x)}{1-g(x)} \right ]^{2}](https://tex.z-dn.net/?f=f%5Bg%28x%29%5D%3Dx%5E%7B2%7D%5C%5C+%5C%5C+%5C%5C+f%5Bg%28x%29%5D%3D%5Cleft%5B%5Cdfrac%7Bg%28x%29%7D%7B1-g%28x%29%7D+%5Cright+%5D%5E%7B2%7D)
Podemos substituir
na igualdade acima:
![f(t)=\left(\dfrac{t}{1-t} \right )^{\!\!2} f(t)=\left(\dfrac{t}{1-t} \right )^{\!\!2}](https://tex.z-dn.net/?f=f%28t%29%3D%5Cleft%28%5Cdfrac%7Bt%7D%7B1-t%7D+%5Cright+%29%5E%7B%5C%21%5C%212%7D)
_____________________
Portanto,
![\boxed{f(x)=\left(\dfrac{x}{1-x} \right )^{\!\!2}} \boxed{f(x)=\left(\dfrac{x}{1-x} \right )^{\!\!2}}](https://tex.z-dn.net/?f=%5Cboxed%7Bf%28x%29%3D%5Cleft%28%5Cdfrac%7Bx%7D%7B1-x%7D+%5Cright+%29%5E%7B%5C%21%5C%212%7D%7D)
Pelo enunciado, temos que
Isolando
Logo,
Podemos substituir
_____________________
Portanto,
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