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)

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

extreme points

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

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})

f (x) = \frac{6 + x}{8 - x}

x + y = - 3

x^{2} - 2 x - 15

x^{2} - 3 x + 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)