MAXIMA AND MINIMA PROBLEMS IN DIFFERENTIATION

How to Find Maxima and Minima Points Using Differentiation ?

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 maxima when f''(x) < 0
• The function f (x) is minima 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

Example 1 :

Find the maximum and minimum value of the function

x³ - 3x² - 9x + 12

Solution :

Let y  =  f(x)  =  x³ - 3 x² - 9 x + 12

f'(x)  =  3x² - 3 (2x) - 9 (1) + 0

f'(x)  =  3x² - 6x - 9

f'(x)  =  0

3x² - 6x - 9  =  0

÷ by 3 => x² - 2 x - 3  =  0

f'(x)  =  3x² - 6x - 9

f''(x)  =  3(2x) - 6(1) - 0

f''(x)  =  6x - 6

Put  x = -1

f '' (-1)  =  6(-1) - 6

  =  -6 - 6

f''(-1)  =  -12 < 0 Maximum

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

f (x)  =  x³ - 3 x² - 9 x + 12

f (-1)  =  (-1)³ - 3 (-1)² - 9 (-1) + 12

  =  -1 - 3(1) + 9 + 12

  =  -1 - 3 + 9 + 12

  =  -4 + 21

  =  17

Put  x  =  3

f''(3)  =  6(3) - 6

  =  18 - 6

 f''(3)  =  12  >  0 Minimum

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

f (x)  =  x³ - 3x² - 9 x + 12

f (3)  =  3³ - 3 (3)² - 9 (3) + 12

  =  27 - 3(9) - 27 + 12

  =  27 - 27 - 27 + 12

  =  -27 + 12 

  =  -15

Therefore the maximum value  =  17 and

The minimum value  =  -15

Example 2 :

Find the maximum and minimum value of the function

4x³ - 18x² + 24 x - 7

Solution :

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

f'(x)  =  4(3x²) - 18(2x) + 24(1) - 0

f'(x)  =  12x² - 36x + 24

f'(x)  =  0

12x² - 36 x + 24 = 0

÷ by 12 => x² - 3 x + 2 = 0            

f'(x)  =  12x² - 36 x + 24

f''(x)  =  12 (2x) - 36(1) + 0

f''(x)  =  24x - 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 given function.

f (x)  =  4x³ - 18x² + 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 given function

f(x)  =  4x³ - 18x² + 24x - 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 is 3 and the 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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