Solution of Question1





In this page solution of question1 we are going to see detailed solution of first question in the topic maximum and minimum.

Question 1:

Find the maximum and minimum value of the function 2 x³ - 15 x² + 36 x + 18

Solution:

Let y = f (x) = 2 x³ - 15 x² + 36 x + 18

           f ' (x) = 2(3x²) - 15 (2x) + 36 (1) + 0

           f ' (x) = 6x² - 30x + 36

 set f ' (x) = 0

  6x² - 30x + 36 = 0

÷ by 6 => x² - 5 x + 6 = 0

             (x - 2) (x - 3) = 0

             x - 2 = 0        x - 3 = 0

                  x = 2              x =  3

           f ' (x) = 6x² - 30x + 36

           f '' (x) = 6 (2 x) - 30 (1) + 0

           f '' (x) = 12 x - 30

Put  x = 2

           f '' (2) = 12(2) - 30

                     = 24 - 30

           f '' (2) = -6 < 0 Maximum

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

f (2) = 2 (2)³ - 15 (2)² + 36 (2) + 18

       = 2(8) - 15(4) + 72 + 18

       = 16 - 60 + 72 + 18

       = 106 - 60

       = 46

Put  x = 3

           f '' (3) = 12(3) - 30

                     = 36 - 30

           f '' (3) = 6 > 0 Minimum

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

2 x³ - 15 x² + 36 x + 18

f (3) = 2 (3)³ - 15 (3)² + 36 (3) + 18

       = 2(27) - 15(9) + 108 + 18

       = 54 - 135 + 108 + 18

       = 180 - 135

       = 45

Therefore the maximum value = 46 and

The minimum value = 45











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