find the critical points and classify them as local maxima, local minima, saddle points, or none of these.

This question is incomplete, the complete question is;

find the critical points and classify them as local maxima, local minima, saddle points, or none of these.

f(x,y) = (x + y)(xy + 1)

(x,y) = (-1, 1), (1, -1) area critical points

f(xx) =2y, fyy =2x,f(xy) =2x + 2y, D = f(xx)fyy - f(xy² )

at (-1, 1)

f(xx) = 2 ,fyy =-2,f(xy) = 0, D = -4 < 0 saddle point

at (1, -1)

f(xx) = -2, fyy =2,f(xy) =0, D = -4 < 0 saddle point

Step-by-step explanation:

Given that;

f(x,y) = (x + y)(xy + 1)

f(x,y) =x²y + xy² + x + y

for critical points fx =0 ,fy =0

fx = 2xy + y² + 1 = 0, fy = x² + 2xy + 1 = 0

2xy + y² + 1 = 0, x²+ 2xy + 1 = 0

2xy + y² + 1 - x² - 2xy - 1 = 0

x² = y²

=> x = y, x = -y

2xy + y² + 1 = 0, x = y

2yy + y² + 1 = 0

3y² = -1 , no solution

2xy + y² + 1 = 0, x = -y

-2yy + y² + 1 = 0

=> -y2 + 1 = 0

=> y = -1, y = 1  

y = -1 => x = 1, y = 1 => x = -1  

(x,y) = (-1, 1), (1, -1) area critical points

f(xx) =2y, fyy =2x,f(xy) =2x + 2y, D = f(xx)fyy - f(xy² )

at (-1, 1)

f(xx) = 2 ,fyy =-2,f(xy) = 0, D = -4 < 0 saddle point

at (1, -1)

f(xx) = -2, fyy =2,f(xy) =0, D = -4 < 0 saddle point

henry ordered 5 large pizzas

step-by-step explanation:

sorry no step by step

since we know that the two functions are equal, f (x) = g

(x), therefore when we equate the two equations the result should be similar to

3x = x – 4:

a. f(x) = 3x, g(x) = x – 4

equating the two functions:

3x = x – 4             (correct)

so we now got our answer (a)