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)
Answer:
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
