Find the absolute minimum and absolute maximum values of y=x^3-12x on the interval [0,3]

Question

I really need to learn how to do this problem. Thanks so much!

Answer 1

Set y' equal to zero.

 

3x² - 12 = 0

 

 

Solving for x,

 

3(x² - 4) = 0

3(x - 2)(x + 2) = 0

 

x = -2        and         x = 2

 

Since x=-2 is not in the interval defined, we only look at x=2.

 

Evaluate the derivative around this point.

 

f'(1) = negative value

f'(2.5) = positive value

 

f(2) is a local minimum.  It could also be absolute minimum.  But, we need to evaluate the other values and compare them with this minimum value.

 

f(2) = 2³ - 12(2) = 8 - 24= -16

 

Now we have to evaluate f(0) and f(3).

 

f(0) = 0

 

f(3) = 3³ - 12(3) = 27 - 36 = -9

 

 

After comparing f(0) , f(2) and f(3), which is the least value and which is the greatest value?  The least value is the absolute minimum, while greatest value is the absolute maximum.

Answer 2

y = x³ - 12x

 

y' = 3x² - 12 = 0 ⇒ x = ±2

 

But -2 is not in [0,3], so x = 2

 

Is it a maximum, a minimum, or neither?  To be a maximum, 

 

f''(2) < 0  [concave down]

 

To be a minimum,

 

f''(2) > 0  [concave up]

 

f'' = 6x

 

f''(2) = 6(2) = 12 > 0 ⇒ concave up

 

The point (2,f(2)) is a minimum.

 

f(2) = 2³ - 12(2) = 8 - 24 = -16

 

The point (2,-16) is a minimum.