What is the extreme p(x)=x^2(a+3)x^3+2x+a^2+5 ?

The extreme p(x)=x^2(a+3)x^3+2x+a^2+5 is Saddle(1.43974…,-3.09309…)

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extreme p(x)=x^2(a+3)x^3+2x+a^2+5

extreme points

p (x) = x^{2} (a + 3) x^{3} + 2 x + a^{2} + 5

Saddle (1.43974 \dots, - 3.09309 \dots)

Solution steps

Find the critical points:

(1.43974 \dots, - 3.09309 \dots)

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

D (x, a) = 40 x^{3} (a + 3) - 25 x^{8}

Check the sign of at each critical point

Check critical point

(1.43974 \dots, - 3.09309 \dots) :

Saddle

Saddle (1.43974 \dots, - 3.09309 \dots)

R (x, y) = x^{2} + 5 x y + 6 y^{2} + 7 x + 17 y + 12

f (x) = - 10 x^{2} + 1660 x - 49000

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

f (x) = 4 - (8 + 7 x)^{\frac{2}{7}}

\frac{3}{( x - 1 ) ( x ^{2} - 4 )}

What is the extreme p(x)=x^2(a+3)x^3+2x+a^2+5 ?
The extreme p(x)=x^2(a+3)x^3+2x+a^2+5 is Saddle(1.43974…,-3.09309…)