The maximum value of f(x) = x3 – 3x2 – 9x + 2 on the interval [0, 6] is 2 56 135

f(x) = x^3 – 3x^2 – 9x + 2

df/dx  = 3x^2 -6x -9

0 = 3(x^2 -2x-3)

  = 3(x-3)(x+1)

   the maximum or minimum is at 3, -1

since we are in the interval from [0,6]  we will  use 3

f(3) =3^3 -3*3^2 -9(3) +2  = -25  (this is a minimum)

f(0) = 0-0-0+2 = 2

f(6) = 6^3 -3*6^2 -9(6) +2 = 56  ( this will be our max)

hi,

3. 12.15, you are correct.

4.2.01, your first digit is correct,

hope this !

the answer is 15% decrease