FIND THE MAXIMUM AND MINIMUM VALUES OF THE FUNCTION EXAMPLES

The value of the function at a maximum point is called the maximum value of the function and the value of the function at a minimum point is called the minimum value of the function.

• Differentiate the given function.
• let f'(x)  =  0 and find critical numbers
• Then find the second derivative f''(x).
• Apply those critical numbers in the second derivative.
• The function f (x) is maximum when f''(x) < 0
• The function f (x) is minimum when f''(x) > 0
• To find the maximum and minimum value we need to apply those x values in the given function.

Question 1 :

Find the maximum and minimum value of the function

2x³ - 15x² + 36x + 18

Solution :

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

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

f'(x)  =  6x² - 30x + 36

f'(x)  =  0

6x² - 30x + 36  =  0

÷ by 6 => x² - 5 x + 6  =  0

f'(x)  =  6x² - 30x + 36

f''(x)  =  6(2x) - 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 given function.

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

  =  2(8) - 15(4) + 72 + 18

  =  16 - 60 + 72 + 18

  =  106 - 60 

f(2)  =  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 given function.

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 is 46 and minimum value  is 45.

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

f'(x)  =  0

  3x² - 12x + 9 = 0

÷ by 3 => x² - 4 x + 3 = 0

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 - 6x² + 9x + 1

f (3)  =  3³ - 6 (3)² + 9 (3) + 1

  =  27 - 6(9) + 27 + 1

  =  54 + 1 - 54

  =  1

Therefore the maximum value is 5 and minimum value is 1.

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