What is the extreme f(x)=-4x^3+3x^2+18x ?

The extreme f(x)=-4x^3+3x^2+18x is Minimum(-1,-11),Maximum(3/2 , 81/4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=-4x^3+3x^2+18x

extreme points

f (x) = - 4 x^{3} + 3 x^{2} + 18 x

Minimum (- 1, - 11), Maximum (\frac{3}{2}, \frac{81}{4})

Solution steps

f^{'} (x) = - 12 x^{2} + 6 x + 18

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

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

Decreasing

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

Plug

x = - 1

into

- 4 x^{3} + 3 x^{2} + 18 x : - 11

Minimum (- 1, - 11)

Plug

x = \frac{3}{2}

into

- 4 x^{3} + 3 x^{2} + 18 x : \frac{81}{4}

Maximum (\frac{3}{2}, \frac{81}{4})

Minimum (- 1, - 11), Maximum (\frac{3}{2}, \frac{81}{4})

What is the extreme f(x)=-4x^3+3x^2+18x ?
The extreme f(x)=-4x^3+3x^2+18x is Minimum(-1,-11),Maximum(3/2 , 81/4)