What is the extreme f(x)=x^4-4x^3+6 ?

The extreme f(x)=x^4-4x^3+6 is Saddle(0,6),Minimum(3,-21)

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extreme f(x)=x^4-4x^3+6

extreme points

f (x) = x^{4} - 4 x^{3} + 6

Saddle (0, 6), Minimum (3, - 21)

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} + 6 : 6

Saddle (0, 6)

Plug

x = 3

into

x^{4} - 4 x^{3} + 6 : - 21

Minimum (3, - 21)

Saddle (0, 6), Minimum (3, - 21)

ln (x + 1)

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

f (x) = ln (5 x)

\frac{cos ( x )}{sin ( x )}

f (x) = 4 - 2 x

What is the extreme f(x)=x^4-4x^3+6 ?
The extreme f(x)=x^4-4x^3+6 is Saddle(0,6),Minimum(3,-21)