Please, I need help!

Find the absolute maximum and minimum values of the function below.

f(x) =x³9x²+2

3

2

x12

SOLUTIONSincefis continuous on

3

2

,12

,

we can use the Closed Interval Method:

f(x)	=	x39x2+2
f'(x)	=	.

Since

f'(x)

exists for allx, the only critical numbers offoccur when

f'(x) = ,

that is,

x= 0

or

x= .

Notice that each of these critical numbers lies in

3

2

,12

.

The values offat these critical numbers are

f(0) = and f(6) = .

The values offat the endpoints of the interval are

f

3

2

= and f(12) = .

Comparing these four numbers, we see that the absolute maximum value is

f(12) =

and the absolute minimum value is

f(6) = .

Note that in this example the absolute maximum occurs at an endpoint, whereas the absolute minimum occurs at a critical number. The graph offis sketched in the figure.