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

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

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extreme y=x^4+6x^3+12x^2+8x+10

extreme points

y = 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} - 2 x y + 3 y^{2}

f (x) = \frac{e ^{x}}{1 + x ^{2}}

f (x, y) = 6 x y - 2 x^{2} y - 3 x y^{2}

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

f (x) = (x^{2} - 1)^{3}, - 1 \leq x \leq 2

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