 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2013/discrete-mathematics-probability-and-continuous-mathematics/4/d/solution

f_{x} = f_{y} = 0

. Since

y^{4} + y^{2} + 1 > 0

for every real

y

x = 0

. The second then reduces to

3 y^{2} - 3 = 0

, so the only critical points are

(0, 1)

and

(0, - 1)

At these points the Hessian matrix is diagonal:

H_{f} (0, 1) = (6006), H_{f} (0, - 1) = (60 0 - 6) .

f (0, 1) = - 1

and

f (0, - 1) = 3

. Hence

(0, 1) : strict local minimum, value - 1; (0, - 1) : saddle point, value 3; no local maxima .

The minimum is local: along

x = 0

f (0, y) = y^{3} - 3 y + 1

y \to - \infty

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