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

The extreme f(x)=x^4-4x^3+2 is Saddle(0,2),Minimum(3,-25)

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

extreme points

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

Saddle (0, 2), Minimum (3, - 25)

Solution steps

f^{'} (x) = 4 x^{3} - 12 x^{2}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = 0

into

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

Saddle (0, 2)

Plug

x = 3

into

x^{4} - 4 x^{3} + 2 : - 25

Minimum (3, - 25)

Saddle (0, 2), Minimum (3, - 25)

ln (\frac{1}{x - 1} + 1)

f (x) = \frac{x + 5}{x - 8}

(70.3) (90.2)

\frac{x - 1}{x ^{2}}

(0, 0), (4, 3)

What is the extreme f(x)=x^4-4x^3+2 ?
The extreme f(x)=x^4-4x^3+2 is Saddle(0,2),Minimum(3,-25)