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

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### 2 Answers

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)**

f(x) = x^3-12.

We know that a function f(x) has the maximum or minimum for x = c if f'(x) = 0 and f''(c) < 0 for maximum and f''(c) > 0 for minimum.

f'(x) = 0 gives (x^3-12) = 0. Or (3x^2-12) =0. S0 x^2-4 = 0. Therefore x^2 = 4. Or x= 2 Or x = -2. So x= c1 = 2 Or x= c2 = -2

f"(x) = (4x^2-12)' = 8x.

f"(2) = 8*2 >0 and f"(-2) = 8(-2) = -16 < 0.

So for x = 2, f(2) = 2^3-12*2 = -16 is minimum,

For x = -2, f(-2) = (-2)^3 -12(-2) = 16 is the maximum in (-3,1)