Question

Podem me ajudar nessa questão? Não consigo chegar na resposta do livro.

f: R→R é definida por $f(x)=2/ x^{2} +1$ , calcule

$f(1+ \sqrt{2})$



Tentei responder e cheguei nessa resposta:
$f(1+ \sqrt{2})=2/(1+ \sqrt{2} )^{2}+1 \\ f(1+ \sqrt{2})=2/( 1^{2}+2 \sqrt{2} +2)+1 \\ f(1+ \sqrt{2})=2/2 \sqrt{2}+4 \\ f(1+ \sqrt{2})= \sqrt{2}*2/ \sqrt{2}*2 \sqrt{2}+4 \\ f(1+ \sqrt{2})= 2* \sqrt{2} /8 \\ f(1+ \sqrt{2})= \sqrt{2}/4$

Porém a resposta do livro é
$2- \sqrt{2}/2$

Answer 1

Olá



$\displaystyle\mathsf{f(x)~=~ \frac{2}{x^2+1} }\\\\\\\mathsf{f(1+ \sqrt{2})=? }\\\\\\\\\\\mathsf{f(1+ \sqrt{2} )~=~ \frac{2}{(1+ \sqrt{2} )^2+1} }\\\\\\\mathsf{f(1+ \sqrt{2} )~=~ \frac{2}{(1+ \sqrt{2} )\cdot(1+ \sqrt{2} )+1} }\\\\\\\mathsf{f(1+ \sqrt{2} )~=~ \frac{2}{(1+ \sqrt{2} )\cdot(1+ \sqrt{2} )+1} }\\\\\\\mathsf{f(1+ \sqrt{2} )~=~ \frac{2}{[1+ \sqrt{2}+ \sqrt{2}+ (\sqrt{2})^2 ]+1} }\\\\\\\mathsf{f(1+ \sqrt{2} )~=~ \frac{2}{[1+ 2\sqrt{2}+2 ]+1} }$

$\displaystyle\mathsf{f(1+ \sqrt{2} )~=~ \frac{2}{1+ 2\sqrt{2}+2 +1} }\\\\\\\\\mathsf{f(1+ \sqrt{2} )~=~ \frac{2}{4+ 2\sqrt{2}} }$


Multiplica pelo conjugado


$\displaystyle\mathsf{f(1+ \sqrt{2} )~=~ \frac{2}{4+ 2\sqrt{2}} ~\cdot ~ \frac{4-2 \sqrt{2} }{4- 2\sqrt{2} } }\\\\\\\mathsf{f(1+ \sqrt{2} )~=~ \frac{2(4- 2\sqrt{2} )}{(4)^2- (2\sqrt{2})^2}}\\\\\\\mathsf{f(1+ \sqrt{2} )~=~ \frac{8-4 \sqrt{2} }{16- 8}}\\\\\\\mathsf{f(1+ \sqrt{2} )~=~ \frac{8-4 \sqrt{2} }{8}}\\\\\\\mathsf{f(1+ \sqrt{2} )~=~ \frac{\diagup\!\!\!\!4(2- \sqrt{2}) }{\diagup\!\!\!\!8}}\\\\\\\boxed{\mathsf{f(1+ \sqrt{2} )~=~ \frac{2- \sqrt{2} }{2}}}$