Question

se considera functia f:R-R,f(x)=xpatrat+4.aratati ca f(-2)+f(2)=4f(0)​

Answer 1

f(-2)=(-2)^2+4

f(-2)=4+4

f(-2)=8

f(2)=2^2+4

f(2)=8

f(0)=0+4

f(0)=4

8+8=4*4

16=16 (adevărat)

Answer 2

$\it f(-2)+f(2) =(-2)^2+4+2^2+4=4+4+4+4=16\ \ \ \ \ (1)\\ \\ 4\cdot f(0) =4\cdot(0+4) =4\cdot4=16\ \ \ \ \ (2)\\ \\ (1),\ (2) \Rightarrow f(-2)+f(2) =4f(0)$