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

The extreme f(x)=(x^3)/3-2x^2+4x+3 is Saddle(2, 17/3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

extreme

Saddle (2, \frac{17}{3})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 2

into

\frac{x ^{3}}{3} - 2 x^{2} + 4 x + 3 : \frac{17}{3}

Saddle (2, \frac{17}{3})

What is the extreme f(x)=(x^3)/3-2x^2+4x+3 ?
The extreme f(x)=(x^3)/3-2x^2+4x+3 is Saddle(2, 17/3)