Solution of Question4





In this page solution of question4 we are going to see detailed solution of first question in the topic maximum and minimum.

Question 4:

Find the maximum and minimum value of the function x³ - 3 x² - 9 x + 12

Solution:

Let y = f (x) = x³ - 3 x² - 9 x + 12

           f ' (x) = 3x² - 3 (2x) - 9 (1) + 0

           f ' (x) = 3x² - 6x - 9

 set f ' (x) = 0

  3x² - 6x - 9 = 0

÷ by 3 => x² - 2 x - 3 = 0

             (x + 1) (x - 3) = 0

             x + 1 = 0        x - 3 = 0

                  x = -1             x =  3

           f ' (x) = 3x² - 6x - 9

           f '' (x) = 3 (2 x) - 6 (1) - 0

           f '' (x) = 6 x - 6

Put  x = -1

           f '' (-1) = 6(-1) - 6

                      = -6 - 6

           f '' (-1) = -12 < 0 Maximum

To find the maximum value let us apply x = -1 in the original function

f (x) = x³ - 3 x² - 9 x + 12

f (-1) = (-1)³ - 3 (-1)² - 9 (-1) + 12

        = -1 - 3(1) + 9 + 12

        = -1 - 3 + 9 + 12

        = -4 + 21

        = 17

Put  x = 3

           f '' (3) = 6(3) - 6

                     = 18 - 6

           f '' (3) = 12 > 0 Minimum

To find the minimum value let us apply x = 3 in the original function

f (x) = x³ - 3 x² - 9 x + 12

f (3) = (3)³ - 3 (3)² - 9 (3) + 12

       = 27 - 3(9) - 27 + 12

       = 27 - 27 - 27 + 12

       = -27 + 12 

       = -15

Therefore themaximum value = 17 and

The minimum value = -15











Solution of Question4 to Examples