Question

Determine the maximum and minimum values of the functions f(x) = 2x³ - 21x² + 36x - 20

Answer 1

Let y = f (x) = 2 x³ + 3 x² - 36 x + 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

             (x - 2) (x + 3) = 0

             x - 2 = 0        x + 3 = 0

                  x = 2              x = - 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 

       = -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

Answer 2

Answer:

Step-by-step explanation:

f(x) = 2x³  - 21x²  + 36x  - 20

f'(x) = 6x² - 42x + 36

put f'(x) = 0

=>  6x² - 42x + 36 = 0

=>  x² - 7x + 6 = 0

=> x² - 6x - x + 6 = 0

=> x(x - 6) - 1(x - 6) =0

=> (x - 1)(x - 6) = 0

x = 1 or x = 6

f''(x) = 12x - 42

f''(1) = 12 - 42 = -30  -ve hence maxima

f(1) = 2 -21 + 36 - 20  = -3

f''(6) = 12*6 - 42 = +30 +ve hence minima

f(6) = 2(216) -21(36) + 36(6) - 20  = -128

Minimum Vale = -128   at x = 6

Maximum Value = -3   at x = 1