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

The extreme f(x)=x^4-4x^3+1 is Saddle(0,1),Minimum(3,-26)

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

extreme points

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

Saddle (0, 1), Minimum (3, - 26)

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} + 1 : 1

Saddle (0, 1)

Plug

x = 3

into

x^{4} - 4 x^{3} + 1 : - 26

Minimum (3, - 26)

Saddle (0, 1), Minimum (3, - 26)

f (x) = \frac{1}{64} x^{3}

x^{3} + 2 x + 4

f (x) = sin (e^{x} - 1)

f (x) = 8 lo g_{6} (6 x + 8) + 24

\frac{1}{x + 5}

What is the extreme f(x)=x^4-4x^3+1 ?
The extreme f(x)=x^4-4x^3+1 is Saddle(0,1),Minimum(3,-26)