What is the extreme f(x)=x^4-8x^3+18x^2 ?

The extreme f(x)=x^4-8x^3+18x^2 is Minimum(0,0),Saddle(3,27)

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extreme f(x)=x^4-8x^3+18x^2

extreme points

f (x) = x^{4} - 8 x^{3} + 18 x^{2}

Minimum (0, 0), Saddle (3, 27)

Solution steps

f^{'} (x) = 4 x^{3} - 24 x^{2} + 36 x

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = 0

into

x^{4} - 8 x^{3} + 18 x^{2} : 0

Minimum (0, 0)

Plug

x = 3

into

x^{4} - 8 x^{3} + 18 x^{2} : 27

Saddle (3, 27)

Minimum (0, 0), Saddle (3, 27)

f (x) = 4 x^{3} - 3 x^{2} - 6 x + 3, (- 1, 10)

y = 2 x^{3} - 24 x^{2} + 90 x + 8

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

f (x) = 4 x^{3} - 24 x^{2} + 36 x

f (x, y) = x^{3} + 2 x y - 5 x - y^{2}

What is the extreme f(x)=x^4-8x^3+18x^2 ?
The extreme f(x)=x^4-8x^3+18x^2 is Minimum(0,0),Saddle(3,27)