What is the extreme y=x^3-8x^2-12x+2 ?

The extreme y=x^3-8x^2-12x+2 is Maximum(-2/3 , 166/27),Minimum(6,-142)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme y=x^3-8x^2-12x+2

extreme points

y = x^{3} - 8 x^{2} - 12 x + 2

Maximum (- \frac{2}{3}, \frac{166}{27}), Minimum (6, - 142)

Solution steps

f^{'} (x) = 3 x^{2} - 16 x - 12

Find intervals:

Increasing

: - \infty < x < - \frac{2}{3},

Decreasing

: - \frac{2}{3} < x < 6,

Increasing

: 6 < x < \infty

Plug

x = - \frac{2}{3}

into

x^{3} - 8 x^{2} - 12 x + 2 : \frac{166}{27}

Maximum (- \frac{2}{3}, \frac{166}{27})

Plug

x = 6

into

x^{3} - 8 x^{2} - 12 x + 2 : - 142

Minimum (6, - 142)

Maximum (- \frac{2}{3}, \frac{166}{27}), Minimum (6, - 142)

y = 3 x + 9

x^{2} + x + 1

f (x) = 3 x^{\frac{2}{3}} - 2 x

f (x) = \frac{x - 7}{x}

f (x) = \frac{x + 5}{x - 2}

What is the extreme y=x^3-8x^2-12x+2 ?
The extreme y=x^3-8x^2-12x+2 is Maximum(-2/3 , 166/27),Minimum(6,-142)