f(0)=2f(2)=4

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Solution for f(0)=2f(2)=4 equation:



f(0)=2f(2)=4
We move all terms to the left:
f(0)-(2f(2))=0
determiningTheFunctionDomain f0-2f2=0
We add all the numbers together, and all the variables
-2f^2+f=0
a = -2; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-2)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*-2}=\frac{-2}{-4} =1/2 $
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*-2}=\frac{0}{-4} =0 $

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