Answer to Question #180846 in Calculus for Anuj

Question

Check whether the function: f: R^2→R, defined by

f(x,y)= x+ysinx has extremum at any point in the domain of f.

Answer 1

$f(x, y)=x+y\sin{x}$

$f'_y(x, y)=\sin{x}$

derivative equals 0 when $x=\pi*n$ , $n=0,\pm1,\pm2,...$

$f'_x(x, y)=1+y\cos{x}$

derivative equals 0 when $y=-\frac{1}{\cos{x}}$ .

$y=-1$ when $n$ is even and $y=1$ when $n$ is odd.

$f''_{xx}(x, y)=-y\sin{x}$

$A=f''_{xx}(\pi*n,\pm1)=0$

$f''_{yy}(x,y)=0$

$B=f''_{yy}(\pi*n,\pm1)=0$

$f''_{xy}(x, y)=\cos{x}$

$D=f''_{xy}(\pi*n,\pm1)=\pm1$

$AB-D^2=-1<0$

so the function has no extremum