How do I find the critical points, and determine whether they are minima, maxima or saddle point, of the following function:
$$f(x,y) = \ln \big( \ (x+y)^2+1 \ \big)$$
For the critical points, I obtained $x = -y$ (which is pretty much the line $x+y=0$)
I tried using the Hessian matrix to do this, but I ended up with all the values in the matrix being the same...which would indicate that no conclusion can be drawn. After I consulted the answers, I found out the the answer is in fact 'all critical points are local minimum'
can anyone please explain how this is done? And can is it possible to solve this using Hessian matrix.
Thank you! :)
It is much easier to study the single variable function $g(t)=\ln(t^2+1)$ first. It has a unique critical point at $t=0$, and the second derivative test shows that it's a minimum.
Now, as you said, $f(x,y)$ has the line $y=-x$ as its critical points. Let $p_0=(x_0,y_0)=(x_0,-x_0)$ be a point on that line. Keeping the expression $x+y$ the same (i.e. following the line $y=-x$) in a neighbourhood of $p_0$ keeps the value of $f(x,y)$ the same. Changing the value of $x+y$ by leaving the line will result in a smaller value for $f(x,y)$ by the analysis done on $g(t)$.
Thus all critical points of $f(x,y)$ are minima points.