若f(2x+1)=x^2-4x+2,求f(3-4x)
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若f(2x+1)=x^2-4x+2,求f(3-4x)
令 y = 2x+1,则 x = (y-1)/2
f(y) = (y-1)^2/4 - 4*(y-1)/2 + 2 = (y-1)^2/4 - 2*(y-1) + 2
为什么 f(3-4x) = (2-4x)^2/4 - 2(2-4x)/ + 2
= 4x^2 +4x - 1
令 y = 2x+1,则 x = (y-1)/2
f(y) = (y-1)^2/4 - 4*(y-1)/2 + 2 = (y-1)^2/4 - 2*(y-1) + 2
为什么 f(3-4x) = (2-4x)^2/4 - 2(2-4x)/ + 2
= 4x^2 +4x - 1
令 y = 2x+1,则 x = (y-1)/2
f(y) = (y-1)^2/4 - 4*(y-1)/2 + 2 = (y-1)^2/4 - 2*(y-1) + 2
那么 f(3-4x) = (2-4x)^2/4 - 2(2-4x)/ + 2
= 4x^2 +4x - 1
f(y) = (y-1)^2/4 - 4*(y-1)/2 + 2 = (y-1)^2/4 - 2*(y-1) + 2
那么 f(3-4x) = (2-4x)^2/4 - 2(2-4x)/ + 2
= 4x^2 +4x - 1
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