Question #180846

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.



1
Expert's answer
2021-04-14T07:36:59-0400

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


fy(x,y)=sinxf'_y(x, y)=\sin{x}

derivative equals 0 when x=πnx=\pi*n , n=0,±1,±2,...n=0,\pm1,\pm2,...


fx(x,y)=1+ycosxf'_x(x, y)=1+y\cos{x}

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

y=1y=-1 when nn is even and y=1y=1 when nn is odd.


fxx(x,y)=ysinxf''_{xx}(x, y)=-y\sin{x}


A=fxx(πn,±1)=0A=f''_{xx}(\pi*n,\pm1)=0


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


B=fyy(πn,±1)=0B=f''_{yy}(\pi*n,\pm1)=0


fxy(x,y)=cosxf''_{xy}(x, y)=\cos{x}


D=fxy(πn,±1)=±1D=f''_{xy}(\pi*n,\pm1)=\pm1


ABD2=1<0AB-D^2=-1<0


so the function has no extremum


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