# Solution of Question2

In this page solution of question2 we are going to see detailed solution of first question in the topic maximum and minimum.

Question 2:

Find the maximum and minimum value of the function x³ - 6 x² + 9 x + 1

Solution:

Let y = f (x) = x³ - 6 x² + 9 x + 1

f ' (x) = 3x² - 6 (2x) + 9 (1) + 0

f ' (x) = 3x² - 12x + 9

set f ' (x) = 0

3x² - 12x + 9 = 0

÷ by 3 => x² - 4 x + 3 = 0

(x - 1) (x - 3) = 0

x - 1 = 0        x - 3 = 0

x = 1              x =  3

f ' (x) = 3x² - 12x + 9

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

f '' (x) = 6 x - 12

Put  x = 1

f '' (1) = 6(1) - 12

= 6 - 12

f '' (1) = -6 < 0 Maximum

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

f (x) = x³ - 6 x² + 9 x + 1

f (1) = (1)³ - 6 (1)² + 9 (1) + 1

= 1 - 6(1) + 9 + 1

= 1 - 6 + 10

= 11 - 6

= 5

Put  x = 3

f '' (3) = 6(3) - 12

= 18 - 12

f '' (3) = 6 > 0 Minimum

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

f (x) = x³ - 6 x² + 9 x + 1

f (3) = (3)³ - 6 (3)² + 9 (3) + 1

= 27 - 6(9) + 27 + 1

= 54 + 1 - 54

= 1

Therefore the maximum value = 5 and

The minimum value = 1

Solution of Question2 to Examples