What is the extreme f(x)=-8/3 x^3+4x^2+48x+9 ?

The extreme f(x)=-8/3 x^3+4x^2+48x+9 is Minimum(-2,-149/3),Maximum(3,117)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=-8/3 x^3+4x^2+48x+9

extreme points

f (x) = - \frac{8}{3} x^{3} + 4 x^{2} + 48 x + 9

Minimum (- 2, - \frac{149}{3}), Maximum (3, 117)

Solution steps

f^{'} (x) = - 8 x^{2} + 8 x + 48

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 3,

Decreasing

: 3 < x < \infty

Plug

x = - 2

into

- \frac{8}{3} x^{3} + 4 x^{2} + 48 x + 9 : - \frac{149}{3}

Minimum (- 2, - \frac{149}{3})

Plug

x = 3

into

- \frac{8}{3} x^{3} + 4 x^{2} + 48 x + 9 : 117

Maximum (3, 117)

Minimum (- 2, - \frac{149}{3}), Maximum (3, 117)

What is the extreme f(x)=-8/3 x^3+4x^2+48x+9 ?
The extreme f(x)=-8/3 x^3+4x^2+48x+9 is Minimum(-2,-149/3),Maximum(3,117)