What is the extreme p(x)=(x^4-2x^3+3x^2-ax+3a-7)/(x+2) ?

The extreme p(x)=(x^4-2x^3+3x^2-ax+3a-7)/(x+2) is Saddle(3, 313/5)

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extreme p(x)=(x^4-2x^3+3x^2-ax+3a-7)/(x+2)

extreme points

p (x) = \frac{x ^{4} - 2 x ^{3} + 3 x ^{2} - a x + 3 a - 7}{x + 2}

Saddle (3, \frac{313}{5})

Solution steps

Find the critical points:

(3, \frac{313}{5})

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

D (x, a) = - \frac{25}{( x + 2 ) ^{4}}

Check the sign of at each critical point

Check critical point

(3, \frac{313}{5}) :

Saddle

Saddle (3, \frac{313}{5})

f (x) = 3 3 x^{2} - 2 x

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

f (x) = 2 x + \frac{20}{x}

12 x - 4 x^{2}

f (x, y) = x^{3} + y^{2} - 6 x y - 29 x + 18 y + 20

What is the extreme p(x)=(x^4-2x^3+3x^2-ax+3a-7)/(x+2) ?
The extreme p(x)=(x^4-2x^3+3x^2-ax+3a-7)/(x+2) is Saddle(3, 313/5)