Define δ x = x2-x 1, x2 = x 1+δ x, δ x→ 0, δ x > 0,

So f (x 1) =-(x 1)? +2x 1，

f(x2)=-(x 1+δx)？ +2(x 1+δx)=-(x 1)？ -2x 1δx-(δx)？ +2x 1+2δx

∴f(x2)-f(x 1)=-2 x 1δx-(δx)？ +2δx =δx(-2x 1-δx+2)

In the interval (-∞, 0), obviously δ x (-2x 1-δ x+2) > 0.

In the interval (0, 1), there is still δ x (-2x 1-δ x+2) > 0.

∴ f (x2) > f (x 1), ∴ function f (x) =-x 2+2x is the increasing function on (-∞, 1).