For which of the following functions f is f(y) = f(1-y) for all of y? Choose the correct choice from the following:
A) f(y) = 3y + 2	B) f(y) = 1-y^2	C) f(y) = 4y^2 * (2-2y)^2	D) f(y) = y/(y-1)	E) f(y) = y/(y+1)
y = 2   a) f(2) = 6*2 + 2= 
<gadget id="calculator">6*2 + 2</gadget>
<output>14</output>
14; f(-1)= -3 + 2 = -1 NO   b) f(2) =1 - 2^2 = 1 - 4 = -3 ; f(-1) = 1 - (-1)^2 = 1 - 1= 
<gadget id="calculator">1 - 1</gadget>
<output>0</output>
0 NO   c) f(2)= 4*2^2 * (2-2*2)^2 = 4*4 * (2-4)^2 = 16 * 4; f(-1) = 4*1 * (2-2(-1))^2 = 4 * (2+2)*2 = 4 * 16 YES   Here we could stop and safely pick the number without doing d &amp; e, but just in case:   d) f(2) = 2/1= 
<gadget id="calculator">2/1</gadget>
<output>2</output>
2 ; f(-1)= -1 / -2 = 1/2 NO   e) f(2) = 2/3; f(-1) = -1/0 = 0 (div by 0) NO   So I proved that in all but f(y) = 4y^2 * (2-2y)^2 case if y=2 the values of f(y) ≠ f(1-y)   Answer: f(y) = 4y^2 * (2-2y)^2
<result>C</result>
For which of the following functions f is f(y) = f(1-y) for all of y?