Analyzing functions of max and min

Question

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

Choices:

1. 2
2. 56
3. 135
4. There is no maximum on [0,6]

Answer 1

Set f'(x) = 0 and solve for x in [0,6].

f'(x) = 3x² - 6x - 9 = 0

x² - 2x - 3 = 0

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

Find f''(3).

f''(x) = 6x - 6

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

f'(3) = 0 and f''(3) > 0 ⇒ f(x) is concave upward and therefore you have a minimum at x=3

There is no local maximum on [0,6].

Answer 2

Taking the derivative of the function,

f'(x)=3x²-6x-9

If I set this equal to 0, I can solve it and find where there are potential min and max

0=3(x²-2x-3)

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

x=3,x=-1 at the possibilities

I decided at this point to graph the function since nothing matched the choices, I see by the graph that there is no max between [0,6], there is a max at x=-1 and a min at x=3

The correct choice based on what you have is There is no maximum on [0,6]

Answer 3

f(x) = x³ - 3x² - 9x + 2 on [0,6]

f(0) = 2, f(6) = 56

f’(x) = 3x² - 6x -9 = 0 when x = 3; here is a local extrema (in a smaller neighborhood)

f(3) = -25

so the maximum of f(x) on [0,6] is 56; Answer choice 2