Hello i am french and I wanted to know if someone could correct (or complete the answers) if possible, I did my best, but for questions 4a) and 4b) I did not find .

A relationship linking the diameter of a tree and its total volume v (including branches) was established towards the end of the 19th century in the form of a function of the third degree.

This relation called "cubage in sylves" is especially used in Switzerland.

We define the function f by: v= f(d) = -10.5d^3 + 26.775d^2 - 9.45d + 0.7 for d belonging to [0.2;1.5] where d is the diameter in meters of the tree at breast height and v the volume in cubic meters.

1) Calculate v for a diameter of 40 cm. Round the result to the nearest thousandth.

2) Determine with the method of your choice, the diameter to the nearest centimeter of a tree of 8 m^3 (cubic meter)

3) Justify that f' (d) = -31.5(d - 0.2) (d - 1.5).

4) a) Determine the sign of f' (d) in the interval [0,2;1,5].

b) Deduce the variations of f on this same interval. .

My answers 40 cm 1 ) V = P ( 0 , 4 ) = = 10 , 5 x ( 0 , 4 ) + 26, 775 x ( 0, 4 ) 39 , 45*0 , 4 to = 1 1876 m ? 2 ) v = f ( d ) = 8 m 3 so, - 10, 5 d' + 26, 7 75 1 2 9 , 45 1 + 0 , 7 = 8 - 10 , 5 d + 26 , 7 75 d 2- 9, 45 1 - 71 3 = 0 Solution ? I don't know how solve it 4 a ) f ( d ) = = 10 , 5 x 3 d2+ 267 7 75 x 2d- 9, 45 = - 31, 5 d2+ 53, 5 5 1- 9, 45. !' 60 , 2 ) = = 1,26 + 101 71 - 9, 45 = 0 101 71 91 45 !' ( 1 1 5 ) = - 70, 875 + 80 , 325 - 9 , 45 = 0