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

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

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extreme f(x)=x^4-12x^3+48x^2-64x

extreme points

f (x) = 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)

\frac{x ^{3} + 2}{5}

\frac{4}{9}

x - e^{x}

\frac{x}{x ^{2} + 4}

- cos (x)

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