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

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

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

extreme points

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

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

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = 0

into

3 x^{4} - 4 x^{3} + 1 : 1

Saddle (0, 1)

Plug

x = 1

into

3 x^{4} - 4 x^{3} + 1 : 0

Minimum (1, 0)

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

f (x) = 1 - x^{2} - y^{2}

f (x) = sin (x) - cos (x)

f (x, y) = x^{2} - y^{2} 1 - x^{2} - y^{2}

f (x) = x^{2} + y^{2} + 5

x^{3} - 3 x^{2} + 1

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