What is the extreme f(z)=3x+2z+4xz ?

The extreme f(z)=3x+2z+4xz is Saddle(-3/4 ,-1/2)

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extreme f(z)=3x+2z+4xz

extreme points

f (z) = 3 x + 2 z + 4 x z

Saddle (- \frac{3}{4}, - \frac{1}{2})

Solution steps

Find the critical points:

(- \frac{3}{4}, - \frac{1}{2})

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

D (z, x) = - 16

Check the sign of at each critical point

Check critical point

(- \frac{3}{4}, - \frac{1}{2}) :

Saddle

Saddle (- \frac{3}{4}, - \frac{1}{2})

f (x) = 12 x^{4} + 3 x^{3} + 5 x^{- 2} - 4

\int sin^{2} (b x) cos (b x) d x

\int 3 x - 2 d x

\frac{d}{d x} (\frac{5}{x + 2})

x^{2} - 1, - x + 2, [0, 1]

What is the extreme f(z)=3x+2z+4xz ?
The extreme f(z)=3x+2z+4xz is Saddle(-3/4 ,-1/2)