What is the extreme f(-1.2)=3p^3+4q^3-2p^2q-4p-5q ?

The extreme f(-1.2)=3p^3+4q^3-2p^2q-4p-5q is

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extreme f(-1.2)=3p^3+4q^3-2p^2q-4p-5q

extreme points

f (- 1.2) = 3 p^{3} + 4 q^{3} - 2 p^{2} q - 4 p - 5 q

Minimum \frac{2 ( 59 + 661 )}{235}, \frac{61 + 9 661}{2 470 59 + 661}, Maximum - \frac{2 ( 59 + 661 )}{235}, - \frac{61 + 9 661}{2 470 59 + 661}, Saddle \frac{2 ( 59 - 661 )}{235}, - \frac{9 661 - 61}{2 470 59 - 661}, Saddle - \frac{2 ( 59 - 661 )}{235}, \frac{9 661 - 61}{2 470 59 - 661}

Solution steps

Find the critical points:

\frac{2 ( 59 + 661 )}{235}, \frac{61 + 9 661}{2 470 59 + 661}, - \frac{2 ( 59 + 661 )}{235}, - \frac{61 + 9 661}{2 470 59 + 661}, \frac{2 ( 59 - 661 )}{235}, - \frac{9 661 - 61}{2 470 59 - 661}, - \frac{2 ( 59 - 661 )}{235}, \frac{9 661 - 61}{2 470 59 - 661}

\frac{\partial ^{2} f}{\partial p ^{2}} = 24 q

D (p, q) = 24 q (18 p - 4 q) - 16 p^{2}

Check the sign of at each critical point

Check critical point

\frac{2 ( 59 + 661 )}{235}, \frac{61 + 9 661}{2 470 59 + 661} :

Minimum

Check critical point

- \frac{2 ( 59 + 661 )}{235}, - \frac{61 + 9 661}{2 470 59 + 661} :

Maximum

Check critical point

\frac{2 ( 59 - 661 )}{235}, - \frac{9 661 - 61}{2 470 59 - 661} :

Saddle

Check critical point

- \frac{2 ( 59 - 661 )}{235}, \frac{9 661 - 61}{2 470 59 - 661} :

Saddle

Minimum \frac{2 ( 59 + 661 )}{235}, \frac{61 + 9 661}{2 470 59 + 661}, Maximum - \frac{2 ( 59 + 661 )}{235}, - \frac{61 + 9 661}{2 470 59 + 661}, Saddle \frac{2 ( 59 - 661 )}{235}, - \frac{9 661 - 61}{2 470 59 - 661}, Saddle - \frac{2 ( 59 - 661 )}{235}, \frac{9 661 - 61}{2 470 59 - 661}

f (x) = 8 e x - y - 5 x y + 6 y^{2}

f (x) = - \frac{1}{50} x^{2} + 1.85 x - 8.5

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

f (x, y) = x^{2} - 2 \cdot x + 2 \cdot y^{2} - 4 \cdot y

f (x) = 0.00397 t^{2} - 0.593 t + 33.718

What is the extreme f(-1.2)=3p^3+4q^3-2p^2q-4p-5q ?
The extreme f(-1.2)=3p^3+4q^3-2p^2q-4p-5q is