What is the extreme f(x)=-x^2-y^2+x+2y-1 ?

The extreme f(x)=-x^2-y^2+x+2y-1 is Maximum(1/2 ,1)

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extreme f(x)=-x^2-y^2+x+2y-1

extreme points

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

Maximum (\frac{1}{2}, 1)

Solution steps

Find the critical points:

(\frac{1}{2}, 1)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(\frac{1}{2}, 1) :

Maximum

Maximum (\frac{1}{2}, 1)

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

f (x) = x^{2} + y^{2} - 4 x = 1

f (x) = x^{3} - 7 x - 6

f (x) = \frac{x}{x ^{2} + 9}, - 5 \leq x \leq 5

f (x, y) = 4 x^{2} + 5 x y + 5 y^{2}

What is the extreme f(x)=-x^2-y^2+x+2y-1 ?
The extreme f(x)=-x^2-y^2+x+2y-1 is Maximum(1/2 ,1)