State and prove the second derivative test for z= f(x,y) Solution Proof of the s

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State and prove the second derivative test for z= f(x,y)

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Proof of the second-derivative test. Our goal is to derive the second-derivative test, which determines the nature of a critical point of a function of two variables, that is, whether a critical point is a local minimum, a local maximum, or a saddle point, or none of these. In general for a function of n variables, it is determined by the algebraic sign of a certain quadratic form, which in turn is determined by eigenvalues of the Hessian matrix [Apo, Section 9.11]. This approach however relies on results on eigenvalues, and it may take several lectures to fully develop. Here we focus on the simpler setting when n = 2 and derive a test using the algebraic sign of the second derivative of the function. The statement of the test is in [Apo, Theorem 9.7]. Theorem 1 (The second-derivative test). Let the scalar ?eld f(x1; x2) have continuous second derivatives in an open ball containing a = (a1; a2). Suppose that D1f(a) = D2f(a) = 0. Let A = D11f(a), B = D12f(a) = D21f(a) and C = D22f(a). Let 4 = det A B B C = AC B 2 : Then, we have (a) If 4 < 0, then a is a saddle point. (b) If 4 > 0 and A > 0, then f(a) is a local minimum. (c) If 4 > 0 and A < 0, then f(a) is a local maximum. (d) If 4 = 0, then the test is inconclusive. The proof uses the second-orde Taylor formula, which we will state for general scalar ?elds. Theorem 2 (Second-order Taylor formula). Let f be a scalar ?eld with continuous second-order partial derivatives Dijf in an open ball B(a). Then, for all y 2 R n such that a + y 2 B(a) we have f(a + y) f(a) = rf(a) y + 1 2 Xn i=1 Xn j=1 Dijf(a + y)yiyj ; where y = (y1; : : : ; yn) and 0 < < 1. The statement and its proof is in [Apo, Section 9.10], and hence it is omitted. The coef?cient of the quadratic form in the above expansion are the second-order partial derivatives. The n n matrix of second-order derivatives Dijf(a) is often called the Hessian matrix. In our present setting, the above Taylor expansion leads to f(a + ty) =f(a) + t(D1f(a)h + D2f(a)k) + t 2 2 (D11f(a )h 2 + 2D12f(a )hk + D22f(a )k 2 ); (*) where y = (h; k) and a = a + y for some 0 <

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