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

The extreme f(x)=x^4-4x^3+12 is Saddle(0,12),Minimum(3,-15)

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

extreme points

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

Saddle (0, 12), Minimum (3, - 15)

Solution steps

f^{'} (x) = 4 x^{3} - 12 x^{2}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = 0

into

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

Saddle (0, 12)

Plug

x = 3

into

x^{4} - 4 x^{3} + 12 : - 15

Minimum (3, - 15)

Saddle (0, 12), Minimum (3, - 15)

f (x) = 4 x^{3} + 48 x^{2} + 180 x

f (x) = \frac{x ^{2} - 7 x + 26}{x - 5}

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

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

f (x, y) = 18 x^{2} - 32 y^{2} - 36 x - 128 y - 110

What is the extreme f(x)=x^4-4x^3+12 ?
The extreme f(x)=x^4-4x^3+12 is Saddle(0,12),Minimum(3,-15)