Question

Suppose (1,1) is a critical point of a function f with continuous second derivatives. In each case, what can you say about f? (6 Pts) (a) fxx(1,1) = −3, fxy(1,1) = 1, fyy(1,1) = 2 (b) fxx(1,1) = −4, fxy(1,1) = 2, fyy(1,1) = −1 (c) fxx(1,1) = −4, fxy(1,1) = −2, fyy(1,1) = −2 (d) fxx(1,1) = 4, fxy(1,1) = 3, fyy(1,1) = 3

Answer 1

Alote: - let f(x,y) have continuous second order partial derivatives Critical point (a, b) & let D = fx. (a, b) fry (a,b) - (bay (a,b)? 1) If Do & fox (a,b) >0, then of a relatine minimum cet ca, b) has 2) If DO & fxx (a, b) co, then relative maximum cut (a,b) has 3 has If Dco, then (a,b). suddle point cet f 4) If D = 0, then conclusion can be drawn. A Now, here given that has continuous second douvatives with a ceutical point (1, 1). (a) fxx (1,1)=-3 fay(1,1)= 1, ty (1, 1) = 2 So D = (-3) (2) -6 -7 So, Dco So with these values (1, 1) is a suddle paint I? 1

(b) fax (1,1)=-4, bxy (1,1)=2, fyy (1, 1) = -3 So, D =(-4) (-1) - (2)² = 4-4 = 0 5 D = 0 So, no conclusion these values. Coin be drown with (c) bxx(1, 1) = -4 (1,1)=-2, Dry (1, 1)=–2 So, D =(-4) (-2) -(-2)2 8-4 - - 4 t has » Do & faxli, 1) co So, with these ucelues these ucelues, cet (1, 1) relative maximum. (d) foxx (1,1)= 4, bxy (1, 1) = 3, bxo (1, 1) = 3 D (4) (3) - 32 = 12-9 = 3 > 0 & box (1, 1) > 0 So, So, with these values, at (1, 1), t has relative minimum. o Hope this is helpful! og