Question

Sejam f e g duas funções reais fais que g é a inversa de f. Se f é definida como

Answer 1

Resposta:

letra d

Explicação passo-a-passo:

meu amor acabei de fazer e já corrigiram

Answer 2

$\displaystyle \text {f(x)}=\frac{\text e^{\text x}-\text e^{-\text x}}{\text e^{\text x}+\text e^{-\text x}} \ ; \ \text{g}=\text{f}^{-1} \ ; \ \text g(\frac{1}{2}) = \ ? \\\\\\ \text{Primeiro vamos deixar a f mais arrumada}:$

$\displaystyle \text{f(x)}=\frac{\displaystyle \text e^{\text x}-\frac{1}{\text e^{\text x}}}{\displaystyle \text e^{\text x}+\frac{1}{\text e^{\text x}}} \to \text{f(x)}=\frac{\displaystyle \frac{\text e^{2\text x}-1}{\text e^{\text x}}}{\displaystyle \frac{\text{e}^{2.\text x}+1}{\text e^{\text x}}} \\\\\\ \text{f(x)} = \frac{\text e^{2\text x}-1}{\text e^{2\text x}+1}$

Achando a inversa : Troca x por y e isola o y na f(x)

$\displaystyle \text y =\frac{\text{e}^{2\text x}-1}{\text e^{2\text x}+1} \ \to\ \text x = \frac{\text e^{2\text y}-1}{\text e^{2\text y}+1} \\\\\\ \text{e}^{2\text y}.\text x+\text x=\text e^{2\text y}-1 \ \to \ \text e^{2\text y}.\ [1-\text x]=1+\text x \\\\\\ \text e^{2\text y} = \frac{1+\text x}{1-\text x} \ \to \ \text{ln } \text e^{2\text y}=\text{ln }[\ \frac{1+\text x}{1-\text x} \ ]$

$\displaystyle \text y =\frac{1}{2}.\text{ln }[\ \frac{1+\text x}{1-\text x} \ ] \\\\\\ \underline{\text{Portanto}}: \\\\ \text {g(x)}=\text{ln }\sqrt{\frac{1+\text x}{1-\text x} }$

A questão pede :

$\displaystyle \text e^{\displaystyle \text g(\frac{1}{2})}$

$\displaystyle \text x = \frac{1}{2} \to \text{g}(\frac{1}{2}) =\text{ln }\sqrt{\frac{\displaystyle 1+\frac{1}{2}}{\displaystyle 1-\frac{1}{2}}} \\\\\\ \text{g}(\frac{1}{2})=\text{ln }\sqrt{3}$

Então :

$\displaystyle \text e^{\displaystyle \text g(\frac{1}{2})} \\\\ \text e^{\text{ln }\sqrt{3}}} \\\\ \huge\boxed{\sqrt{3}\ }\checkmark$

Letra d