We are given the function:
f(x)=2ax+a−x.
Now, let's calculate f(x+y) and f(x−y):
f(x+y)=2ax+y+a−(x+y),f(x−y)=2ax−y+a−(x−y).
By adding these two expressions together:
f(x+y)+f(x−y)=2ax+y+a−(x+y)+ax−y+a−(x−y).
Next, simplify this using the properties of exponents:
f(x+y)+f(x−y)=2ax(ay+a−y)+a−x(ay+a−y).
Factor the expression:
f(x+y)+f(x−y)=2(ax+a−x)(ay+a−y).
Now, recognizing that f(x)=2ax+a−x and f(y)=2ay+a−y, we substitute these back in:
f(x+y)+f(x−y)=2f(x)f(y).Final Answer:2f(x)f(y)