FIND THE MAXIMUM AND MINIMUM VALUES OF THE FUNCTION EXAMPLES

Find the Maximum and Minimum Values of the Function Examples :

In this section, we will see some example problems of finding maximum and minimum values of the function.

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.

Finding the Maximum and Minimum Values of the Function Examples

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

x - 2  =  0

x  =  2

x - 3  =  0

x  =  3

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

x - 1  =  0

x  =  1

x - 3  =  0

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 - 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.

After having gone through the stuff given above, we hope that the students would have understood how to find maximum and minimum value of the function.

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