What is the extreme f(x)=x^4-6x^3+12x^2-8x+1 ?

The extreme f(x)=x^4-6x^3+12x^2-8x+1 is Minimum(1/2 ,-11/16),Saddle(2,1)

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extreme f(x)=x^4-6x^3+12x^2-8x+1

extreme points

f (x) = x^{4} - 6 x^{3} + 12 x^{2} - 8 x + 1

Minimum (\frac{1}{2}, - \frac{11}{16}), Saddle (2, 1)

Solution steps

f^{'} (x) = 4 x^{3} - 18 x^{2} + 24 x - 8

Find intervals:

Decreasing

: - \infty < x < \frac{1}{2},

Increasing

: \frac{1}{2} < x < 2,

Increasing

: 2 < x < \infty

Plug

x = \frac{1}{2}

into

x^{4} - 6 x^{3} + 12 x^{2} - 8 x + 1 : - \frac{11}{16}

Minimum (\frac{1}{2}, - \frac{11}{16})

Plug

x = 2

into

x^{4} - 6 x^{3} + 12 x^{2} - 8 x + 1 : 1

Saddle (2, 1)

Minimum (\frac{1}{2}, - \frac{11}{16}), Saddle (2, 1)

f (x, y) = - x^{3} - y^{3} + \frac{3 x ^{2}}{3} + 3 y^{2}

3 x^{5} - 10 x^{3}

f (x) = 8 x^{\frac{1}{3}} - x^{\frac{4}{3}}

f (x, y) = x^{2} + 3 x y + 2 y^{4}

f (x) = x^{2} + 6 x + 15

What is the extreme f(x)=x^4-6x^3+12x^2-8x+1 ?
The extreme f(x)=x^4-6x^3+12x^2-8x+1 is Minimum(1/2 ,-11/16),Saddle(2,1)