What is the extreme p(x)=2x^3-11x^2+ax-40 ?

The extreme p(x)=2x^3-11x^2+ax-40 is Saddle(0,0)

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extreme p(x)=2x^3-11x^2+ax-40

extreme points

p (x) = 2 x^{3} - 11 x^{2} + a x - 40

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, a) = - 1

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

\frac{x ^{4} - 98 x ^{2}}{8}

2 x e^{3 x}

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

f (x, y) = 11 - x^{2} - y^{2}

f (x) = x^{2} - 25

What is the extreme p(x)=2x^3-11x^2+ax-40 ?
The extreme p(x)=2x^3-11x^2+ax-40 is Saddle(0,0)