What is the extreme y(x,t)=x(t-1/4) ?

The extreme y(x,t)=x(t-1/4) is Saddle(0, 1/4)

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extreme y(x,t)=x(t-1/4)

extreme points

y (x, t) = x (t - \frac{1}{4})

Saddle (0, \frac{1}{4})

Solution steps

Find the critical points:

(0, \frac{1}{4})

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

D (x, t) = - 1

Check the sign of at each critical point

Check critical point

(0, \frac{1}{4}) :

Saddle

Saddle (0, \frac{1}{4})

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

f (x) = x^{3} + 9 x^{2} + 27 x + 3

f (x, y) = x^{2} - 12 x y + 2 y^{4}

f (x, y) = - x^{3} + y^{2} + 3 x - 4 y - 1

f (x, y) = x^{3} + y^{3} - 9 x y

What is the extreme y(x,t)=x(t-1/4) ?
The extreme y(x,t)=x(t-1/4) is Saddle(0, 1/4)