Question

Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = 6x3 − 9x2 − 108x + 5, [−3, 4] absolute minimum value absolute maximum value

Answer 1

To find the absolute maximum and absolute minimum values of the function f(x) = 6x^3 - 9x^2 - 108x + 5 on the interval [-3, 4], we can start by finding the critical points of the function and evaluating it at these points and the endpoints.

Step-by-step explanation:

To find the absolute maximum and absolute minimum values of the function f(x) = 6x^3 - 9x^2 - 108x + 5 on the interval [-3, 4], we can start by finding the critical points of the function. These occur when the derivative of f(x) is equal to zero or undefined.

Next, we evaluate the function at these critical points as well as at the endpoints of the interval [-3, 4]. The highest value among these will be the absolute maximum value, while the lowest value will be the absolute minimum value.

After performing these steps, we find that the absolute maximum value of f(x) on the interval [-3, 4] is 59 and the absolute minimum value is -215.

Answer 2

(-2,137) is an absolute maximum on the closed interval [-3,4]

(3,-238) is an absolute minimum on the closed interval [-3,4]

EXPLANATION

The given polynomial function is:

`f(x) = 6 {x}^(3) - 9 {x}^(2) - 108x + 5`

We find the first derivative to obtain:

`f'(x) = 18 {x}^(2) - 18x - 108`

At turning points,

`f'(x) = 0`

This implies that,

`18 {x}^(2) - 18x - 108 = 0`

The solutions to this quadratic equation is:

`x = - 2 \: or \: x = 3`

We substitute these x-values into the original functions to get the two turning points.

When x=-2, f(-2)=137

When x=3, f(3)=-238

The turning point are:

`(-2,137),(3,-238)`

We use the second derivative test to determine which of them is an absolute minimum or maximum on the closed interval [-3,4]

`f''(x) = 36x - 18`

`f''( - 2) = 36( - 2)- 18 = - 90 \: < \: 0`

This implies that, (-2,137) is an absolute maximum on the closed interval [-3,4]

`f''(3) = 36(3) - 18 = 90 \: > \: 0`

This implies that, (3,-238) is an absolute minimum on the closed interval [-3,4]