Examine the maxima and minima of the function f(x) = 2x^3 - 21x^2 + 36x - 20 . Also, find the maximum and minimum values of f(x). - Mathematics and Statistics

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Examine the maxima and minima of the function f(x) = 2x3 - 21x2 + 36x - 20 . Also, find the maximum and minimum values of f(x). 

Find the maximum and minimum value of the function:

f(x) = 2x3 - 21x2 + 36x - 20

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Solution 1

`f(x)=2x^3-21x^2+36x-20`

`f^'(x)=6x^2-42x+36`

For finding critical points, we take f'(x)=0

`therefore 6x^2-42x+36=0`

`x^2-7x+6=0`

 (x-6)(x-1)=0

For finding the maxima and minima, find f''(x)

f'(x)=12x-42

for x=6

f''(6)=30>0

Minima

for x=1

f''(1)=-30<0

Maxima

Maximum values of f(x) for x=1

f(1)=-3

minimum values of f(x) for x=6

f(6)=-128

∴ the maximum values of the function is -3 and the minimum value of the function is -128.

Solution 2

f(x) = 2x3 - 21x2 + 36x - 20

∴ f'(x) = `2(3x^2) - 21(2x) + 36(1) - 1`

`= 6x^2 - 42x + 36 = 6(x^2 - 7x + 6)`

= 6(x - 1)(x - 6)

f has a maxima/minima if f'(x) = 0

i.e if 6(x - 1)(x-6) = 0

i.e if x - 1= 0   or x - 6 = 0

i.e if x = 0 or x = 6

Now f"(x) = 6(2x) - 42(1) = 12x - 42

∴ f"(1) = 12(1) - 42 = -30

∴ f"(1) < 0

Hence, f has a maximum at x = 1, by the second derivative test.

Also f"(6) = 12(6)- 42 = 30

∴ f"(6) > 0

Hence, f has a minimum at x = 6, by the second derivative test.

Now, the maximum value of f at 1,

`f(1) = 2(1^3) - 21(1^2) + 36(1) - 20`

= 2-21+36-20 = -3

and minimum value of f at x = 6

`f(6) = 2(6^3) - 21(6^2) + 36(1) - 20`

= 432 - 756 + 216 - 20 = -128

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