Solution of Question3





In this page solution of question3 we are going to see detailed solution of first question in the topic maximum and minimum.

Question 3:

Find the maximum and minimum value of the function 2 x³ - 3 x² - 12 x + 5

Solution:

Let y = f (x) = 2 x³ - 3 x² - 12 x + 5

           f ' (x) = 2(3x²) - 3 (2x) - 12 (1) + 0

           f ' (x) = 6x² - 6x - 12

 set f ' (x) = 0

  6x² - 6x - 12 = 0

÷ by 6 => x² - x - 2 = 0

             (x - 2) (x + 1) = 0

             x - 2 = 0        x + 1 = 0

                  x = 2              x =  -1

           f ' (x) = 6x² - 6x - 12

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

           f '' (x) = 12 x - 6

Put  x = 2

           f '' (2) = 12(2) - 6

                     = 24 - 6

           f '' (2) = 18 > 0 Minimum

To find the minimum value let us apply x = 2 in the original function

f (x) = 2 x³ - 3 x² - 12 x + 5

f (2) = 2 (2)³ - 3 (2)² - 12 (2) + 5

       = 2(8) - 3(4) - 24 + 5

       = 16 - 12 - 24 + 5

       = 21 - 36

       = -15

Put  x = -1

          f '' (-1) = 12(-1) -6

                     = -12 - 6

          f '' (-1) = -18 > 0 Maximum

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

f (x) = 2 x³ - 3 x² - 12 x + 5

f (-1) = 2 (-1)³ - 3 (-1)² - 12 (-1) + 5

        = 2(-1) - 3(1) + 12 + 5

        = -2 - 3 + 12 + 5

        = -5 + 17 

        = 12

Therefore themaximum value = 12 and

The minimum value = -15











Solution of Question3 to Examples