So f (x) = (1/3) t, t = x 2-2x.

The function f (x) = (1/3) t is a subtraction function.

So when the function t = x 2-2x is in the monotonic increasing interval [1, positive infinity],

The function f (x) = (1/3) (x 2-2x) is a subtraction function.

Because monotonicity is decreasing and increasing, increasing and increasing, decreasing and decreasing, increasing and decreasing.

That is, when x is in the interval [1, positive infinity], the function f (x) = (1/3) (x 2-2x) is a decreasing function.

Similarly, when the function t = x 2-2x is in the monotonically decreasing interval (-infinity, 1),

The function f (x) = (1/3) (x 2-2x) is increasing function.

2, t = x 2-2x can be calculated as: t >；; =- 1.

Because t = x 2-2x,

t=(x- 1)^2- 1.

(x- 1)^2>； =0

So t & gt=- 1.

According to the monotonicity of the function f (x) = (1/3) t,

It is known that the function f (x) = (1/3) t monotonically decreases on [- 1,+infinity].

So f (x) < = f (-1) = (1/3) (-1) = 3.

So the range is (-infinity, 3).