The function f: R^2 to R , defined by f(x,y)= 1-y^2+x^2, has an extremum at (0,0).
True or false with full explanation
ANSWER. The function f(x,y)=1−y2+x2f(x,y)=1-{ y }^{ 2 }+{ x }^{ 2 }f(x,y)=1−y2+x2 has no an extremum at the point (0,0).
EXPLANATION. Since f(0,y)=1−y2≤1=f(0,0)f(0,y)=1-{ y }^{ 2 }\le 1=f(0,0)f(0,y)=1−y2≤1=f(0,0) and f(x,0)=1+x2≥1=f(0,0)f(x,0)=1+{ x }^{ 2 }\ge 1=f(0,0)f(x,0)=1+x2≥1=f(0,0) then by the definition of the extremum of the function fff has no extremum at the point (0,0).