find all relative extrema and saddle points of the function. use the second partials test where applicable…

Explanation:

Step1: Find first - order partial derivatives

The function is (z = e^{-x}\sin(y)). The first - order partial derivative with respect to (x) is: [ \begin{align*} z_{x}&=\frac{\partial}{\partial x}(e^{-x}\sin(y))\ &=-\sin(y)e^{-x} \end{align*} ] The first - order partial derivative with respect to (y) is: [ \begin{align*} z_{y}&=\frac{\partial}{\partial y}(e^{-x}\sin(y))\ &=e^{-x}\cos(y) \end{align*} ]

Step2: Find critical points

Set (z_{x}=0) and (z_{y}=0). From (z_{x}=-\sin(y)e^{-x} = 0), we have (\sin(y)=0), so (y = n\pi), (n\in\mathbb{Z}). From (z_{y}=e^{-x}\cos(y)=0), when (\sin(y) = 0), (\cos(y)=\pm1). Since (e^{-x}\gt0) for all (x\in\mathbb{R}), the equations (z_{x} = 0) and (z_{y}=0) are satisfied when (y=n\pi), (n\in\mathbb{Z}).

Step3: Find second - order partial derivatives

(z_{xx}=\frac{\partial}{\partial x}(-\sin(y)e^{-x})=\sin(y)e^{-x}) (z_{yy}=\frac{\partial}{\partial y}(e^{-x}\cos(y))=-e^{-x}\sin(y)) (z_{xy}=\frac{\partial}{\partial y}(-\sin(y)e^{-x})=-\cos(y)e^{-x}) The discriminant (D=z_{xx}z_{yy}-(z_{xy})^{2}) [ \begin{align*} D&=(\sin(y)e^{-x})(-e^{-x}\sin(y))-(-\cos(y)e^{-x})^{2}\ &=-e^{-2x}(\sin^{2}(y)+\cos^{2}(y))\ &=-e^{-2x}\lt0 \end{align*} ]

Answer:

Relative minimum: (\text{DNE}) Relative maximum: (\text{DNE}) Saddle point: ((x,y,z)=(x,n\pi,0)) for all (x\in\mathbb{R}) and (n\in\mathbb{Z})