What is the extreme y=x^4-12x^3+48x^2-64x ?

The extreme y=x^4-12x^3+48x^2-64x is Minimum(1,-27),Saddle(4,0)

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extreme y=x^4-12x^3+48x^2-64x

extreme points

y = x^{4} - 12 x^{3} + 48 x^{2} - 64 x

Minimum (1, - 27), Saddle (4, 0)

Solution steps

f^{'} (x) = 4 x^{3} - 36 x^{2} + 96 x - 64

Find intervals:

Decreasing

: - \infty < x < 1,

Increasing

: 1 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 1

into

x^{4} - 12 x^{3} + 48 x^{2} - 64 x : - 27

Minimum (1, - 27)

Plug

x = 4

into

x^{4} - 12 x^{3} + 48 x^{2} - 64 x : 0

Saddle (4, 0)

Minimum (1, - 27), Saddle (4, 0)

f (x, y) = x^{2} + x y + 2 x + 5 y - 1

0.01 x^{3} - 0.45 x^{2} + 2 x + 300

x e^{\frac{2}{x}}

50000 x + 40000 y - 10 x^{2} - 20 y^{2} - 10 x y

f (x) = \frac{1}{2} x^{2} - 9 x + 7

What is the extreme y=x^4-12x^3+48x^2-64x ?
The extreme y=x^4-12x^3+48x^2-64x is Minimum(1,-27),Saddle(4,0)