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)

The maximum value of f (x) in the interval [0,6] is 56

Step-by-step explanation:

At maximum value of f (x) we have f' (x) = 0

That is

f' (x) = 3x² - 6x - 9 = 0

x² - 2x - 3 = 0

(x - 3) (x + 1) = 0

x = 3 or x = - 1

Here x = 3 only lies in [0, 6].

Let us find f'' (3) value

f'' (x) = 6x - 6 = 6 x 3 - 6 = 12>0

Sign of f'' (3) is positive, hence at 3 the function is minimum.

Let us find f (0) and f (6)

f (0) = 3 x 0³ - 3 x 0² - 9 x 0 + 2 = 2

f (6) = 6³ - 3 x 6² - 9 x 6 + 2 = 216 - 108 - 54 + 2 = 56

The maximum value of f (x) in the interval [0,6] is 56