What is the extreme f(x)=x^4+6x^3+12x^2+8x+10 ?

The extreme f(x)=x^4+6x^3+12x^2+8x+10 is Saddle(-2,10),Minimum(-1/2 , 133/16)

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extreme f(x)=x^4+6x^3+12x^2+8x+10

extreme points

f (x) = x^{4} + 6 x^{3} + 12 x^{2} + 8 x + 10

Saddle (- 2, 10), Minimum (- \frac{1}{2}, \frac{133}{16})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 2,

Decreasing

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

Increasing

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

Plug

x = - 2

into

x^{4} + 6 x^{3} + 12 x^{2} + 8 x + 10 : 10

Saddle (- 2, 10)

Plug

x = - \frac{1}{2}

into

x^{4} + 6 x^{3} + 12 x^{2} + 8 x + 10 : \frac{133}{16}

Minimum (- \frac{1}{2}, \frac{133}{16})

Saddle (- 2, 10), Minimum (- \frac{1}{2}, \frac{133}{16})

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

3 x^{4} - x^{6}

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

f (x) = \frac{( x ^{2} + x + 4 )}{( 3 + x ^{2} )}

(1 - x^{2})^{2}

What is the extreme f(x)=x^4+6x^3+12x^2+8x+10 ?
The extreme f(x)=x^4+6x^3+12x^2+8x+10 is Saddle(-2,10),Minimum(-1/2 , 133/16)