Question

A função real f é tal que f(2x -1 ) = x  , sendo f-1(x) a inversa de f , f-1 (x) é igual a:

(por favor, resposta com explicação)​

Answer 1

Podemos fazer de duas formas :

1ª forma

Propriedade da função inversa :

$\text f(\text f^{-1}(\text x))=\text x$

temos :

$\text f(2\text x-1) =\text x$

perceba que está bem parecido com a propriedade da função inversa. Note que ali dentro é a $\text f^{-1}(\text x)$ , pois está igualada a x, igualzinho a propriedade.

logo :

$2\text x-1 = \text f^{-1}(\text x)$

Portanto :

$\huge\boxed{\text{f}^{-1}(\text x)=2\text x-1\ }$

2ª forma de resolver

$\displaystyle \text{f}(2\text x-1)=\text x \ \ ; \ \ \text f^{-1}(\text x) = \ ? \\\\ \underline{\text{Fa{\c c}amos}}: \\\\ \text x = \text a \to \text{f}(2\text a-1) = \text a \\\\ 2\text a-1 = \text x \to \boxed{\text a = \frac{\text x+1}{2}\ } \\\\ \underline{\text{substituindo}}: \\\\\ \text f(\ 2.\frac{(\text x+1)}{2}-1\ ) = \frac{\text x+1}{2} \\\\\\ \text f(\ \text x+1-1\ )=\frac{\text x+1}{2} \\\\\\ \text {f(x)} = \frac{\text x+1}{2} \to \text y = \frac{\text x+1}{2}$

$\displaystyle \underline{\text{Fun{\c c}{\~a}o inversa}}. \\\\\ \text{substituindo x por y e isolando y} : \\\\\\ \text x = \frac{\text y+1}{2} \\\\\ 2.\text x=\text y+1 \\\\ \text y = 2\text x-1 \\\\ \underline{\text{Portanto}}: \\\\ \huge\boxed{\ \text{f}^{-1}(\text x) = 2\text x-1\ }\checkmark$