Question

Considere a função f(x)=x²+1, em R. Julgue (V ou F) cada igualdade a seguir

A) F(2) + f(0)=6
B) f(-3) - f(3)= 0
c) f(-1).F(5)=12
D) f(4): F(-4)=1

Answer 1

$f(x)=x^2+1\\\\\\\\ a)~~f(2)+f(0) = 6~;~(~~V~~)~,~pois:\\\\ (2^2+1)+(0^2+1)=6\to~ (4+1)+(0+1)=6\to~5+1=6\to~6=6\\\\\\\\ b)~~f(-3)-f(3) = 0~;~(~~V~~)~,~pois:\\\\ ((-3)^2+1)+(3^2+1)=0\to~ (9+1)-(9+1)=0\to~10-10=0\to\\0=0$



$c)~~f(-1)\times f(5) = 12~;~(~~F~~)~,~pois:\\\\ ((-1)^2+1)\times (5^2+1)\to~(1+1)\times(25+1)=12\to~ 2\times26=12\to \\52 \neq 12\\\\\\\\ d)~~f(4):f(-4)=1~;~(~~ V ~~)~,~pois:\\\\ \dfrac{4^2+1}{(-4)^2+1} =1\to~~\dfrac{16+1}{16+1}=1 \to~~\dfrac{17}{17}=1 \to~~1=1$