Question

Find the maximum and minimum values of x3-2x2+x+6

Answer 1

f(x) = x³ - 2x² + x + 6

Find the first derivative:

f(x) = x³ - 2x² + x + 6

f'(x) = (3)x² - (2)2x  + 1 + 0

f'(x) = 3x² - 4x  + 1

f'(x) = (x - 1)(3x - 1)

(x -1) (3x - 1) = 0

x = 1 or x = 1/3

Find the second derivative:

f'(x) = 3x² - 4x  + 1

f''(x) = (2)3x - 4 + 0

f''(x) = 6x - 4

Find max / min point:

When x = 1,

f''(x) = 6(2) - 4

f''(x) = 8 > 0 ⇒ minimum point

When x = 1/3

f''(x) = 6(1/3) - 4

f''(x) = - 2 < 0 ⇒ maximum point

Find minimum point:

When x = 1,

f(x) = (1)³ - 2(1)² + (1) + 6

f(x) = 6

⇒ minimum point = 6

Find the maximum point:

when x = 1/3

f(x) = (1/3)³ - 2(1/3)² + (1/3) + 6

f(x) = 166/27

⇒ maximum point = 166/27

Answer:  minimum point = 6, maximum point = 166/27

Answer 2

Answer116/7

Step-by-step explanation: