## HOW TO FIND MAXIMUM AND MINIMUM VALUE OF A 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 original function.

## Examples

Example 1 :

Determine maximum values of the functions

y  =  4x - x2 + 3

Solution :

f(x)  =  y  =  4x - x2 + 3

First let us find the first derivative

f'(x)  =  4(1) - 2x + 0

f'(x)  =  4 - 2x

Let  f'(x)  =  0

4 - 2x  =  0

2 (2 - x)  =  0

2 - x  =  0

x  =  2

Now let us find the second derivative

f''(x)  =  0 - 2(1)

f''(x)  =  -2  <  0 Maximum

To find the maximum value, we have to apply x = 2 in the original function.

f(2)  =  4(2) - 22 + 3

f(2)  =  8 - 4 + 3

f(2)  =  11 - 4

f(2)  =  7

Therefore the maximum value is 7 at x = 2. Now let us check this in the graph.

Checking :

y  =  4x - x2 + 3

The given function is the equation of parabola.

y  =  -x² + 4 x + 3

y  =  -(x² - 4 x - 3)

y  =  -{ x² - 2 (x) (2) + 2² - 2² - 3 }

y = - { (x - 2)² - 4 - 3 }

y = - { (x - 2)² - 7 }

y  =  - (x - 2)² + 7

y - 7  =  -(x - 2)²

(y - k)  =  -4a (x - h)²

Here (h, k) is (2, 7) and the parabola is open downward.

Example 2 :

Find the maximum and minimum value of the function

2x3 + 3x2 - 36x + 1

Solution :

Let y  =  f(x)  =  2x3 + 3x2 - 36x + 1

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

f'(x)  =  6x2 + 6x - 36

set f'(x)  =  0

6x² + 6x - 36 = 0

÷ by 6 => x² + x - 6  =  0

(x - 2)(x + 3)  =  0

 x - 2  =  0x  =  2 x + 3 =  0x  =  -3

f'(x)  =  6x² + 6x - 36

f''(x)  =  6(2x) + 6(1) - 0

f''(x)  =  12x + 6

Put  x  =  2

f''(2)  =  12(2) + 6

=  24 + 6

f''(2)  =  30  > 0 Minimum

To find the minimum value let us apply x = 2 in the original function

f(2)  =  2(2)3 + 3(2)2 - 36(2) + 1

=  2(8) + 3(4) - 72 + 1

=  16 + 12 - 72 + 1

=  29 - 72

f(2)  =  -43

Put  x = -3

f''(-3)  =  12(-3) + 6

=  -36 + 6

f''(-3)  =  -30 > 0 Maximum

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

f(-3)  =  2 (-3)3 + 3 (-3)2 - 36 (-3) + 1

=  2(-27) + 3(9) + 108 + 1

=  -54 + 27 + 109

=  -54 + 136

=  82

Therefore the minimum value is -43 and  maximum value is 82. Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.

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