What is the extreme f(x)=x^4-8x^3+1 ?

The extreme f(x)=x^4-8x^3+1 is Saddle(0,1),Minimum(6,-431)

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extreme f(x)=x^4-8x^3+1

extreme points

f (x) = x^{4} - 8 x^{3} + 1

Saddle (0, 1), Minimum (6, - 431)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 6,

Increasing

: 6 < x < \infty

Plug

x = 0

into

x^{4} - 8 x^{3} + 1 : 1

Saddle (0, 1)

Plug

x = 6

into

x^{4} - 8 x^{3} + 1 : - 431

Minimum (6, - 431)

Saddle (0, 1), Minimum (6, - 431)

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

y = 2 x - 5 x^{\frac{2}{5}}

f (x) = \frac{8 + x}{8 - x} [4.7]

(1 - 2 x)^{4} (1 - x)^{2} x^{6}

f (x) = \frac{1}{x ^{2} + 3}

What is the extreme f(x)=x^4-8x^3+1 ?
The extreme f(x)=x^4-8x^3+1 is Saddle(0,1),Minimum(6,-431)