Question

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

2
56
135

Answer 1

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

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

Answer 2

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)