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

The extreme f(x)=(x^4)/4-(4x^3)/3+2x^2-1 is Minimum(0,-1),Saddle(2, 1/3)

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extreme f(x)=(x^4)/4-(4x^3)/3+2x^2-1

extreme points

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

Minimum (0, - 1), Saddle (2, \frac{1}{3})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 0

into

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

Minimum (0, - 1)

Plug

x = 2

into

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

Saddle (2, \frac{1}{3})

Minimum (0, - 1), Saddle (2, \frac{1}{3})

- 30 x^{2} - 25 y^{2} + 480 x + 300 y + 170

f (x) = \frac{( x - 2 ) ^{2}}{x - 5}

f (x) = - x^{2} - 5 x - 2

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

F (X, Y) = 1 - X^{2} + 2 Y^{2}

What is the extreme f(x)=(x^4)/4-(4x^3)/3+2x^2-1 ?
The extreme f(x)=(x^4)/4-(4x^3)/3+2x^2-1 is Minimum(0,-1),Saddle(2, 1/3)