Question

find the maximum and minimum values of function

$y = 2x {}^{3} - 15x {}^{2} + 36x + 1$

Answer 1

Equation: y = 2x³ - 15x² + 36x + 1

Find the first derivative:

f(x) = 2x³ - 15x² + 36x + 1

f'(x) = 6x² - 30x + 36

Find the max and min points

The max and min values happens when the first derivative = 0

6x² - 30x + 36 = 0

x² - 5x + 6 = 0

(x - 2)(x - 3) = 0

x = 2 or x = 3

Find the second derivative:

f'(x) = 6x² - 30x + 36

f''(x) = 12x - 30

When x = 2

f''(2) = 12(2) - 30 = - 6

f''(2) < 0 (It is a maximum point)

When x = 3

f''(3) = 12(3) - 30 = -6

f''(3) > 0 (It is a minimum point)

Answer: The maximum point occurs at x = 2 and minimum points at x = 3

Answer 2

Given Expression is y = 2x^3 - 15x^2 + 36x + 1.

On differentiating, we get

⇒ y' = 6x^2 - 30x + 36

We need to set the derivative equal to 0 in order to find maximum and minima.  

⇒ y' = 0

⇒ 6x^2 - 30x + 36 = 0

⇒ 6(x^2 - 5x + 6) = 0

⇒ x^2 - 5x + 6 = 0

⇒ x^2 - 3x - 2x + 6 = 0

⇒ x(x - 3) - 2(x - 3) = 0

⇒ (x - 2)(x - 3) = 0

⇒ x = 2,3..

Differentiate the given equation again with respect to x.

⇒ y'' = 12x - 30.

Now,

Substitute x = 2 in y'':

⇒ 12x - 30

⇒ 12(2) - 30

⇒ -6.

Substitute x = 3 in y'':

⇒ 12x - 30

⇒ 12(3) - 30

⇒ 36 - 30

⇒ 6.

Hence,

At x = 2, the maximum value is -6.

At x = 3, and the minimum value is 6.

Hope it helps!