What is the extreme f(x)=x^4-4x^3+5 ?

The extreme f(x)=x^4-4x^3+5 is Saddle(0,5),Minimum(3,-22)

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extreme f(x)=x^4-4x^3+5

extreme points

f (x) = x^{4} - 4 x^{3} + 5

Saddle (0, 5), Minimum (3, - 22)

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} + 5 : 5

Saddle (0, 5)

Plug

x = 3

into

x^{4} - 4 x^{3} + 5 : - 22

Minimum (3, - 22)

Saddle (0, 5), Minimum (3, - 22)

x^{4} - 10 x^{2} + 9

P (x, y) = 3 x^{2} + 4 x y + y^{2} + 4 x + 2 y + 1

\frac{( x - 1 ) ^{2}}{x ^{2} + 1}

f (x) = x^{\frac{4}{5}} (x - 4)

\frac{x ^{2} - x}{8 x ^{2} + 1}

What is the extreme f(x)=x^4-4x^3+5 ?
The extreme f(x)=x^4-4x^3+5 is Saddle(0,5),Minimum(3,-22)