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

ANSWER

(-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]