What is the extreme h(x)=(θx^4+3θx^3+3θx^2+θx)/(x^3+2x^2+x) ?

The extreme h(x)=(θx^4+3θx^3+3θx^2+θx)/(x^3+2x^2+x) is Saddle(-1,0)

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extreme h(x)=(θx^4+3θx^3+3θx^2+θx)/(x^3+2x^2+x)

extreme points

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

Saddle (- 1, 0)

Solution steps

Find the critical points:

(- 1, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 0

D (x, θ) = - 1

Check the sign of at each critical point

Check critical point

(- 1, 0) :

Saddle

Saddle (- 1, 0)

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

f (x) = (16 - 2 x) (12 - 2 x) x

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

V (A, t) = A^{t}

f (x) = x^{3} + 32 x^{2} - 6 x + 8, - 4 \leq x \leq 2

What is the extreme h(x)=(θx^4+3θx^3+3θx^2+θx)/(x^3+2x^2+x) ?
The extreme h(x)=(θx^4+3θx^3+3θx^2+θx)/(x^3+2x^2+x) is Saddle(-1,0)