Answer
Local minimum $f(0,1)=f(\pi,-1)=f(2 \pi,1)=-1$ ;
Saddle points: $(\dfrac{\pi}{2},0),(\dfrac{3\pi}{2},0)$
Work Step by Step
Given: $f(x,y)=y^2-2y \cos x$; $-1 \leq x \leq l$
To solve this problem we will take the help of Second derivative test that suggests the following conditions to determine the local minimum, local maximum and saddle points of $f(x,y)$ or $f(x,y,z)$.
i) If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\gt 0$ , then $f(p,q)$ is a local minimum.
ii) If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\lt 0$ , then $f(p,q)$ is a local maximum.
iii) If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \lt 0$ , then $f(p,q)$ is a not a local minimum and local maximum or, a saddle point.
a)For $(x,y)=(0,0)$; we have $D(0,0)=4 \gt 0$ and $f_{xx}(0,0)=2 \gt 0$
b) For $(x,y)=(0,1)$, we have $D(0,1)=4 \gt 0$ and $f_{xx}=2 \gt 0$
c) For $(x,y)=(\pi,-1)$; we have $D(\pi,-1)=4 \gt 0$ and $f_{xx}=2 \gt 0$
d) For $(x,y)=(2\pi,1)$; we have $D(\pi,-1)=4 \gt 0$ and $f_{xx}=2 \gt 0$
e) For $(x,y)=(\dfrac{\pi}{2},0)$; we have $D(\dfrac{\pi}{2},0)=-2 \lt 0$
f) For $(x,y)=(\dfrac{3\pi}{2},0)$, we have $D(\dfrac{\pi}{2},0)=-2 \lt 0$
From the above conditions we conclude that Local minimum $f(0,1)=f(\pi,-1)=f(2 \pi,1)=-1$ ;
Saddle points : $(\dfrac{\pi}{2},0),(\dfrac{3\pi}{2},0)$
