What is the extreme f(x)=x^3+x^2-5x-5 ?

The extreme f(x)=x^3+x^2-5x-5 is Maximum(-5/3 , 40/27),Minimum(1,-8)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=x^3+x^2-5x-5

extreme points

f (x) = x^{3} + x^{2} - 5 x - 5

Maximum (- \frac{5}{3}, \frac{40}{27}), Minimum (1, - 8)

Solution steps

f^{'} (x) = 3 x^{2} + 2 x - 5

Find intervals:

Increasing

: - \infty < x < - \frac{5}{3},

Decreasing

: - \frac{5}{3} < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - \frac{5}{3}

into

x^{3} + x^{2} - 5 x - 5 : \frac{40}{27}

Maximum (- \frac{5}{3}, \frac{40}{27})

Plug

x = 1

into

x^{3} + x^{2} - 5 x - 5 : - 8

Minimum (1, - 8)

Maximum (- \frac{5}{3}, \frac{40}{27}), Minimum (1, - 8)

What is the extreme f(x)=x^3+x^2-5x-5 ?
The extreme f(x)=x^3+x^2-5x-5 is Maximum(-5/3 , 40/27),Minimum(1,-8)