Find the absolute maximum and minimum values of f(x) = 9x3 - 54x2 + 81x + 13 on the interval [-6, 2].?

Question

Find the absolute maximum and minimum values of f(x) = 9x3 - 54x2 + 81x + 13 on the interval [-6, 2].

Answer 1

First take the derivative and set equal to zero:

f'(x) = 27x^2 -108x +81 = 0

Divide by 27 to get x^2 - 4x + 3 =0

factor: (x-1)(x-3) = 0

So the critical points are x = 1 and 3. However, x=3 is not in the interval so we reject it.

Now this is a continuous function on a closed, bounded interval. So by the Extreme value Thm, it must have both a max and min but they may occur at the endpoints. So see what f(x) is at the endpoints and at the critical point:

f(-6) = -4361

f(1) = 49

f(2) = 31

Comparing, we see that the max of f occurs at x = 1 and the min of f occurs at x = -6.

[ If you had tested the critical point with the 2nd derivative test you would have gotten:

f''(x) = 54x - 108, and at x=1 this is negative. That tells us that x=1 is at least a  relative max.]

Note: To avoid confusion on these postings, it is best to indicate powers by using the ^ symbol

eg: x-cubed is  typed x^3.

Answer 2

df/dx = 27x2 - 108x + 81

Let df/dx = 0 and simplify:  df/dx = x2 - 4x + 3

x = (1, 3)

x=3 is outside the interval, so only x=1 is relevant and a critical point.

f(1) = 49, which is the maximum value.

f(-6) = -4361, which is the minimum value.