What are the absolute maximimum and absolute minimum values for: f(x) = x^3 - 12x on the interval [-3, 1]?

f(x) = x^3 - 12x  

Find the maximum
and minimum on the interval [ -3, 1]

First we need to determine the
critical values of f(x) in the interval [ -3, 1]

Then we need to
find the first derivative:

f'(x) = 3x^2 -
12

Now we will calculate the
zeros:

==> 3x^2 -12 =
0

==> 3x^2 = 12

==> x^2 =
4

==> x= +-2

Then there are two
critical values one which is within the interval [ -3, 1]

Then we
will not calculate extreme values at x= 2

Now we will find the
second derivative :

f''(x) =  6x

Then
the function has a maximum if f''(x) < 0

==> 6x
< 0  ==> x < 0

Since the critical values x= -2
is negative then f''(x) < 0 then the function has a maximum value at x=
-2

==> f(-2) = -2^3 -
12*-2

              = -8 + 24 =
16

Then the maximum values if ( -2,
16)