What is the extreme p(a,x)=a(x+1)+x(x+1)+2(x+1) ?

The extreme p(a,x)=a(x+1)+x(x+1)+2(x+1) is Saddle(-1,-1)

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extreme p(a,x)=a(x+1)+x(x+1)+2(x+1)

extreme points

p (a, x) = a (x + 1) + x (x + 1) + 2 (x + 1)

Saddle (- 1, - 1)

Solution steps

Find the critical points:

(- 1, - 1)

\frac{\partial ^{2} f}{\partial a ^{2}} = 2

D (a, x) = - 1

Check the sign of at each critical point

Check critical point

(- 1, - 1) :

Saddle

Saddle (- 1, - 1)

\frac{4 \frac{x}{5 - 3}}{2 x ^{5} + 9 x}

f (x) = - 3 x^{2} + 6 x + 9

f (x) = x^{2} - x y + y^{2} - 9 x + 6 y + 8

x^{3} - 4 x^{2}

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

What is the extreme p(a,x)=a(x+1)+x(x+1)+2(x+1) ?
The extreme p(a,x)=a(x+1)+x(x+1)+2(x+1) is Saddle(-1,-1)