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Using the second partial derivative test, I have found (-1,1) to be a saddle point but this option is not available in the MCQ. Have I made a mistake?

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The person who set the question insists that (-1,1) is a minimum, not saddle, because:

"We are talking about the local situation at the 2 points. Since the y coordinate of the 2 points are the same at y=1, that means we pass a vertical plane through the 3-D surface, so it becomes a 2-D curve in terms of x. So using 2nd derivative test (as stated explicitly in the question), it leads to local minimum. "

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2 Answers

up vote 2 down vote accepted

The given function can be written as $$f(x,y)=(1-x^2)^2 -(y-1)^2 -7\ .$$ It follows by inspection that $(0,1)$ is a local maximum. Now when $x=-1$ then $1-x^2=0$, and for all $x$ in a punctured neighborhood of $-1$ one has $(1-x^2)^2>0$. It follows that $f(x,1)>f(-1,1)$ for these $x$; on the other hand we have $f(-1,y)<f(-1,1)$ for all $y\ne 1$. So $(-1,1)$ is definitely a saddle point.

The question remains what is meant by "inconclusive".

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Since $(0,1)$ is a maximum, only option 2. and 5. are left. Since $(-1,1)$ is not a minimum (5.), and further you found it's a saddle point, which is neither a local minimum nor local maximum ($\to$ inconclusive), I would go for option 2.

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