Another Functional Equation 3

Let $x=y=0$ . This gives us that $f(0) = 0$ .

Take $x=y$ . So, $f(0) = 2f(x) +x^{2}$ . It implies that, $f(x) = -\dfrac{x^{2}}{2}$ .

So, $f(20.15) = -\dfrac{20.15^{2}}{2} = \boxed{-203.01125}$ .

f(x-y) = f(x) + xy + f(y)

f(y-x) = f(y) + yx + f(x)

Since f:R -> R for all x in R, commutative and associativity holds for addition and multiplication.

Therefore f(x-y) = f(-x+y) = f(x) + xy + f(y) --- [1]

----- Solution Branch 1 - my initial approach, used values for x and y, didn't find f(x) -----

f((x+y)-y) = f(x) = f(x+y) + (x+y)y + f(y) [2]

f(x-(x-y)) = f(y) = f(x) + x(x-y) + f(x-y) [3]

from [2] and [3] f(x) - f(y) = f(x+y) + (x+y)y = -x(x-y) - f(x-y)

f(x+y) + f(x) + f(y) + xy = -(x^2 + y^2) --- apply [1]

f(20+0.15) + f(20) + f(0.15) = -(20^2 + 0.15 + (20)(0.15))

in equation [2], let x = 20, y = 0.15

f(20) = f(20.15) + (20.15)(0.15) + f(0.15)

in equation [3], let x = 0.15, y = 20

f(20) = f(0.15) + (0.15)(-19.85) + f(-19.85)

f(20) = f(0.15) + (0.15)(-19.85) + f(0.15) + (0.15)(20) + f(20) --- apply [1]

2f(0.15) = -(0.15)^2

substitute f(0.15) into the other equations to get a 2x2 linear system. eliminate f(20) to get f(20.15)

----- Solution Branch 2 - much simpler and found f(x) -----

f(x-(x-y)) = f(y) = f(x) + x(x-y) + f(x-y)

f(y) = f(x) + x(x-y) + f(x) + xy + f(y) --- from [1]

2f(x) = -(x^2)

2f(20.15) = -(20.15)^2

f(20.15) = -203.075

since its an even function so f(-x) = f(x) and putting x=2y we get , f(2x)=-2x^2 so f(20.15)= -(20.15)^2/2= -203.01125 = -203.011