What is the extreme f(x)=((x^4))/4-2x^3+1 ?

The extreme f(x)=((x^4))/4-2x^3+1 is Saddle(0,1),Minimum(6,-107)

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extreme f(x)=((x^4))/4-2x^3+1

extreme points

f (x) = \frac{( x ^{4} )}{4} - 2 x^{3} + 1

Saddle (0, 1), Minimum (6, - 107)

Solution steps

f^{'} (x) = x^{3} - 6 x^{2}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 6,

Increasing

: 6 < x < \infty

Plug

x = 0

into

\frac{x ^{4}}{4} - 2 x^{3} + 1 : 1

Saddle (0, 1)

Plug

x = 6

into

\frac{x ^{4}}{4} - 2 x^{3} + 1 : - 107

Minimum (6, - 107)

Saddle (0, 1), Minimum (6, - 107)

81 - x^{2}

f (x) = \frac{3 - 4 x}{x ^{2} + 1}, - 2 \leq x \leq 4

y = 4 x + 8 (\frac{20000}{x})

f (x) = (x^{2} + 4)^{2}

f (x) = 9 x^{2} - 72 x + 108

What is the extreme f(x)=((x^4))/4-2x^3+1 ?
The extreme f(x)=((x^4))/4-2x^3+1 is Saddle(0,1),Minimum(6,-107)