Suppose f(x,y) represents a polynomial function in x and y in which the coefficients of the highest powers of x and y are positive. Prove that a point P(a,b) lying inside a closed curve f(x,y) must satisfy f(a,b)<0.

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TopNewestI don't get this one. What if \(f(x,y) \equiv -x^2-y^2+1\)? \(f(x,y) = 0\) is the same curve as \(x^2+y^2-1 = 0\); that is, the unit circle. But for the origin \(P(0,0)\) which is inside the curve, we have \(f(P) = f(0,0) = 1 > 0\). Am I missing something? – Ivan Koswara · 3 years, 9 months ago

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Sorry to have missed out on that detail. – Sambit Senapati · 3 years, 9 months ago

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– Mark Hennings · 3 years, 9 months ago

If the highest powers have different signs, the curve \(f(x,y)=0\) is most unlikely to be closed.Log in to reply

And you are going to need to restrict your attention to polynomial functions (or at least ones for which \(f(x,y) \to \infty\) as \((x,y) \to \infty\), since otherwise curves like \[ e^{-x^2} + e^{-y^2} = 1.5 \] will cause you problems as well. – Mark Hennings · 3 years, 9 months ago

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– Sambit Senapati · 3 years, 9 months ago

I have modified the problem.Log in to reply