Let F'(x)=0, then x= 1, or x=- 1 (excluding, not in the target interval).

When x ∈ (0, 1), f' (x)

When x ∈ (1, 2), f' (x) >; 0, F(x) increasing,

In this way, the function gets the minimum value at x= 1, that is, the minimum value F( 1)=4.

The maximum value may be obtained where x=0 and x=2, just compare the function values of these two places.

F(0)=6

F(2)=8

So the maximum value is 8.

To sum up, the minimum value of the function is 4 and the maximum value is 8.

The landlord's d should be 3, right?