Solution of Question5





In this page solution of question5 we are going to see detailed solution of first question in the topic maximum and minimum.

Question 5:

Find the maximum and minimum value of the function 4 x³ - 18 x² + 24 x - 7

Solution:

Let y = f (x) = 4 x³ - 18 x² + 24 x - 7

           f ' (x) = 4(3x²) - 18 (2x) + 24 (1) - 0

           f ' (x) = 12 x² - 36 x + 24

 set f ' (x) = 0

  12 x² - 36 x + 24 = 0

÷ by 12 => x² - 3 x + 2 = 0

             (x - 1) (x - 2) = 0

             x - 1 = 0        x - 2 = 0

                  x = 1              x =  2

           f ' (x) = 12 x² - 36 x + 24

           f '' (x) = 12 (2 x) - 36 (1) + 0

           f '' (x) = 24 x - 36

Put  x = 1

           f '' (1) = 24(1) - 36

                     = 24 - 36

           f '' (1) = -12 < 0 Maximum

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

f (x) = 4 x³ - 18 x² + 24 x - 7

f (1) = 4 (1)³ - 18 (1)² + 24 (1) - 7

       = 4(1) - 18(1) + 24 - 7

       = 4 - 18 + 24 - 7

       = 28 - 25

       = 3

Put  x = 2

           f '' (2) = 24(2) - 36

                     = 48 - 36

           f '' (2) = 12 > 0 Minimum

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

f (x) = 4 x³ - 18 x² + 24 x - 7

f (2) = 4 (2)³ - 18 (2)² + 24 (2) - 7

       = 4(8) - 18(4) + 48 - 7

       = 32 - 72 + 48 - 7

       = 80 - 79 

       = 1

Therefore the maximum value = 3 and

The minimum value = 1











Solution of Question5 to Examples