Solution: y = (1-x) f' (x)

Let y = 0: x2 = 1 or f'(x) = 0.

As can be seen from the figure, f'(x) = 0 has two solutions: x 1 = -2 x3 = 2.

When x; 0，∴f '(x)& gt； 0

When-2 < x <; y < at 1； At this time it is 0,1-x >; 0∴f '(x)& lt； 0

That is, f(-2) is the maximum value of f(x).

When 1

When x > 2 o'clock, y < 0 1-x at this time.

That is, f(2) is the minimum value of f(x).

Choose d