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[f(x+∆x)-f(x)] / ∆x
entonces
f(x+∆x) = (x+∆x)^2 - 4(x+∆x) + 6 = x^2 + 2x∆x + (∆x)^2 - 4x -4∆x + 6 = x^2 - 4x + 6 + 2x∆x -4∆x + (∆x)^2
f(x+∆x)-f(x) = x^2 - 4x + 6 + 2x∆x - 4∆x + (∆x)^2 - (x^2-4x+6) = 2x∆x -4∆x + (∆x)^2
[f(x+∆x)-f(x)] / ∆x = [2x∆x -4∆x + (∆x)^2] / ∆x = 2x - 4 + ∆x
2x - 4 + ∆x esta es la razón de cambio