If f(x + y, x - y) = xy,then the arithmetic m | vedclass

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(C) Given $f(x + y, x - y) = xy$.Let $u = x + y$ and $v = x - y$.Solving for $x$ and $y$,we get $x = \frac{u + v}{2}$ and $y = \frac{u - v}{2}$.Substi

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(C) Given $f(x + y, x - y) = xy$.Let $u = x + y$ and $v = x - y$.Solving for $x$ and $y$,we get $x = \frac{u + v}{2}$ and $y = \frac{u - v}{2}$.Substituting these into the function definition,we have $f(u, v) = \left( \frac{u + v}{2} \right) \left( \frac{u - v}{2} \right) = \frac{u^2 - v^2}{4}$.Thus,$f(x, y) = \frac{x^2 - y^2}{4}$ and $f(y, x) = \frac{y^2 - x^2}{4}$.The arithmetic mean of $f(x, y)$ and $f(y, x)$ is $\frac{f(x, y) + f(y, x)}{2} = \frac{1}{2} \left( \frac{x^2 - y^2}{4} + \frac{y^2 - x^2}{4} \right) = \frac{1}{2} (0) = 0$.

If $f(x + y, x - y) = xy$,then the arithmetic mean of $f(x, y)$ and $f(y, x)$ is

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