Answer to Question #200116 in Real Analysis for Nikhil

Question

The function f: R^2→R, defined by

f(x,y)= 1-x^2+y^2 has an extremum at (0,0)

True or false with full explanation

Answer 1

Given f(x,y) = 1-x²+y²

differentiating f(x,y) with respect to x and y

f_x(x,y) = -2x

f_y(x,y) = 2y

For critical point, f_x(x,y) = f_y(x,y) = 0

So, x = y = 0

So critical point of f(x,y) = (0,0)

We know that partial derivative of x and y of f(x,y) is (d²f/dx²)(d²f/dy²) - (d²f/dx dy)² = D

So, (d²f/dx²) = -2

(d²f/dy²) = 2

(d²f/dx dy) = 0

D = (-2)(2) - (0)²

D = -4 < 0

If D < 0 , then (0,0) is a saddle point.

So, (0,0) is not an extremum of f(x,y).