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?