Calculus Question

Question

Find the absolute maximum and absolute minimum values of f on the given interval.

f(x) = 6x³ − 18x² − 54x + 9,    [−2, 4]

Answer 1

To find the maxima and minima of any function, you need to calculate its derivative with respect to the variable for which it's a function:

The function in question is

f(x) = 6x³ − 18x² − 54x +9;

it's derivative is

f'(x)=d/dx(  6x³ − 18x² − 54x +9) = 6(3x²) -18(2x) - 54 = 18x² - 36x - 54

The absolute maximum and minimum values will occur where the derivative equals zero.

so find where

f'(x) = 18x² - 36x - 54 = 0

we divide both sides by 18 to simplify this equation and get

x² - 2x - 3 = 0

now we find the roots for this equation by factoring it on the left side:

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

so the two roots are x = 3 and x = -1 which are BOTH in the range  [−2, 4]

Now there are two ways to determine which is the absolute maximum and which is the absolute minimum.

The first way is to simply is set x to either of the two values (3 and -1) in the original function f(x) and check to see which one has the "highest" value and that will be the maximum while the other will be the minimum.

Another way is to take the derivative of the derivative of f(x) (which will be the second derivative of f(x) with respect to x) and see if it is positive or negative for each of the roots.

Since we're asked to find the maximum and minimum values in the first place, we'll use the first way.

f(-1) = 6(-1)³ − 18(-1)² − 54(-1) +9 = 6(-1) - 18(1) -54(-1) + 9 = (-6) -18 +54 + 9 =-24+54+9 =39

f(3) = 6(3)³ − 18(3)² − 54(3) +9 = 162 - 162 - 162 + 9 = -153

so because f(3) is less than f(-1) -153 is the minimum value of f(x) and 39 is the maximum value.

A question I leave for you is what is the value of the SECOND derivative of f(x) for x = -1 and x = 3 and what is the significance of those values?

Answer 2

First find any local max/min on the interval. To do that, take the first derivative and set it equal to zero.

f '(x) = 18x²- 36x - 54

18x²- 36x - 54 = 0 [divide both sides by 18 to get:

x²- 3x - 3 = 0

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

x = 3, x = -1 both of these are on the interval in question.

So now, set up a table with the local max/mins as well as the endpoints and calculate the function values at each point like this:

x | f(x)

-2 | -3

-1 | 39

3 | -153

4 | =111

So, w can see from the table that the absolute minimum occurs at x = 3 and is y = -153 and the absolute maximum occurs at x = -1 and is y = 39.

So we say that the absolute minimum value is -153 and the absolute maximum value is 39 on the interval [−2, 4]