What is the extreme f(x)=3x^4-8x^3+6x^2+1 ?

The extreme f(x)=3x^4-8x^3+6x^2+1 is Minimum(0,1),Saddle(1,2)

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extreme f(x)=3x^4-8x^3+6x^2+1

extreme points

f (x) = 3 x^{4} - 8 x^{3} + 6 x^{2} + 1

Minimum (0, 1), Saddle (1, 2)

Solution steps

f^{'} (x) = 12 x^{3} - 24 x^{2} + 12 x

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = 0

into

3 x^{4} - 8 x^{3} + 6 x^{2} + 1 : 1

Minimum (0, 1)

Plug

x = 1

into

3 x^{4} - 8 x^{3} + 6 x^{2} + 1 : 2

Saddle (1, 2)

Minimum (0, 1), Saddle (1, 2)

f (x) = 3 x^{4} - 8 x^{3} + 6 x^{2} + 2

\frac{x ^{2} - 1}{x ^{2} - 9}

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

x^{2} + 6 x - 4

x^{2} + x y + y^{2} - 19 y + 120

What is the extreme f(x)=3x^4-8x^3+6x^2+1 ?
The extreme f(x)=3x^4-8x^3+6x^2+1 is Minimum(0,1),Saddle(1,2)