What is the extreme f(y)=2x^2+3xy+4y^2+5x+6y-7 ?

The extreme f(y)=2x^2+3xy+4y^2+5x+6y-7 is Minimum(-9/23 ,-22/23)

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extreme f(y)=2x^2+3xy+4y^2+5x+6y-7

extreme points

f (y) = 2 x^{2} + 3 x y + 4 y^{2} + 5 x + 6 y - 7

Minimum (- \frac{9}{23}, - \frac{22}{23})

Solution steps

Find the critical points:

(- \frac{9}{23}, - \frac{22}{23})

\frac{\partial ^{2} f}{\partial y ^{2}} = 4

D (y, x) = 23

Check the sign of at each critical point

Check critical point

(- \frac{9}{23}, - \frac{22}{23}) :

Minimum

Minimum (- \frac{9}{23}, - \frac{22}{23})

(x^{2} + y^{2} - 3) d x = (y + x y) d y, y (0) = 1

\frac{d y}{d x} = 1 + e^{y - x + 5}

\frac{d}{d y} (\frac{y}{3})

x \to \infty lim (4 - \frac{1}{x})

\int \frac{sec ^{2} ( x )}{tan ^{2} ( x )} d x

What is the extreme f(y)=2x^2+3xy+4y^2+5x+6y-7 ?
The extreme f(y)=2x^2+3xy+4y^2+5x+6y-7 is Minimum(-9/23 ,-22/23)