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

The extreme f(x)=x^4-12x^3+4 is Saddle(0,4),Minimum(9,-2183)

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

extreme points

f (x) = x^{4} - 12 x^{3} + 4

Saddle (0, 4), Minimum (9, - 2183)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 9,

Increasing

: 9 < x < \infty

Plug

x = 0

into

x^{4} - 12 x^{3} + 4 : 4

Saddle (0, 4)

Plug

x = 9

into

x^{4} - 12 x^{3} + 4 : - 2183

Minimum (9, - 2183)

Saddle (0, 4), Minimum (9, - 2183)

f (x) = 2 x^{3} + 45 x^{2} - 300 x, 4 \leq x \leq 11

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

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

p (x) = 12 x^{4} + 15 x^{3} + a x^{2} - 69 x - 30

f (x, y) = x + 2 \frac{y}{2} x + y

What is the extreme f(x)=x^4-12x^3+4 ?
The extreme f(x)=x^4-12x^3+4 is Saddle(0,4),Minimum(9,-2183)