2 years ago

I need help plsssssssss

1 answer:

True [87]2 years ago

6 0

Answer:

d. 18

Step-by-step explanation:

You might be interested in

san4es73 [151]

16 is the square
root of 4 so i guess 16 is x

4 0

2 years ago

lara31 [8.8K]

$f(x,y)=x^4+y^4-4xy+1$

has critical points wherever the partial derivatives vanish:

$f_x=4x^3-4y=0\implies x^3=y$

$f_y=4y^3-4x=0\implies y^3=x$

Then

$x^3=y\implies x^9=x\implies x(x^8-1)=0\implies x=0\text{ or }x=\pm1$

$f(x,y)$ has Hessian matrix

$H(x,y)=\begin{bmatrix}12x^2&-4\\-4&12y^2\end{bmatrix}$

with determinant

$\det H(x,y)=144x^2y^2-16$

At (0, 0), the Hessian determinant is -16, which indicates a saddle point.
At (1, 1), the determinant is 128, and $f_{xx}(1,1)=12$ , which indicates a local minimum.
At (-1, -1), the determinant is again 128, and $f_{xx}(-1,-1)=12$ , which indicates another local minimum.