If x = -1, then x+1 = 0, and f(0) = = 0*f(f(0)) = 0.
If x =0, then x+1 = 1, and f(1) = 1*f(f(0)) = 0 since 2x - x2 = 2*0 -0*0 = 0.
If x = 1, then x+1 = 2, and f(2) = 2*f(f(2×1-1*1=1)) = 2×f(f(1)) =2*0 =0.
Jeff H.
asked 12/02/23"If f is a real function and f(x+1) = (x+1) f(f(2x-x2)), find f(2)".
This was a multiple choice question, but is there a way to mathematically prove that the answer is 0?
If x = -1, then x+1 = 0, and f(0) = = 0*f(f(0)) = 0.
If x =0, then x+1 = 1, and f(1) = 1*f(f(0)) = 0 since 2x - x2 = 2*0 -0*0 = 0.
If x = 1, then x+1 = 2, and f(2) = 2*f(f(2×1-1*1=1)) = 2×f(f(1)) =2*0 =0.
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