What is the extreme f(x)=4x^3+3x^2-6x+1 ?

The extreme f(x)=4x^3+3x^2-6x+1 is Maximum(-1,6),Minimum(1/2 ,-3/4)

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^2-6x+1

extreme points

f (x) = 4 x^{3} + 3 x^{2} - 6 x + 1

Maximum (- 1, 6), Minimum (\frac{1}{2}, - \frac{3}{4})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

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

Increasing

: \frac{1}{2} < x < \infty

Plug

x = - 1

into

4 x^{3} + 3 x^{2} - 6 x + 1 : 6

Maximum (- 1, 6)

Plug

x = \frac{1}{2}

into

4 x^{3} + 3 x^{2} - 6 x + 1 : - \frac{3}{4}

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

Maximum (- 1, 6), Minimum (\frac{1}{2}, - \frac{3}{4})

What is the extreme f(x)=4x^3+3x^2-6x+1 ?
The extreme f(x)=4x^3+3x^2-6x+1 is Maximum(-1,6),Minimum(1/2 ,-3/4)