MAXIMA AND MINIMA PROBLEMS IN DIFFERENTIATION

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.

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.

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