Second Derivative Test

Question

Given f(x, y) = x²y − 2xy + x² − 2x + y² − y.

(a) Fi‎nd the criti‎cal poin‎ts of f.

(b) U‎se the Se‎cond Derivati‎ves T‎est to cl‎assify the c‎ritical poi‎nts obt‎ained in (a).

Answer 1

(a) f_x = 2xy - 2y + 2x - 2 = 0 ; f'_y = x² - 2x + 2y - 1 = 0;

(x - 1)(y + 1) = 0; x = 1; 1 - 2 + 2y - 1 = 0; y = 1 (1, 1); y = - 1; x² - 2x - 3 = 0; (x - 3)(x + 1) = 0; x =3 or x = -1

(3, - 1); (- 1, - 1) and (1, 1) are critical points

(b) f_xx = 2y + 2; f_yy = 2; f_xy = 2x - 2

Point (3, - 1): f_xx = 0, f_yy = 2, f_xy = 4; Δ = 0·2 - 4² = - 16 < 0; at (3, - 1) is saddle point

Point (-1, - 1): f_xx = 0, f_yy = 2, f_xy = - 4; Δ = 0·2 - 16 = - 16 < 0 at (- 1, - 1) is saddle point

Point(1, 1): f_xx = 4 > 0, f_yy = 2, f_xy = 0; Δ = 4·2 - 0² = 8 > 0; at (1, 10 is local minimum