 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-2/12/e/solution

Along either coordinate axis, the restriction is

h (t, 0) = h (0, t) = t^{6}

, which is positive for

t \neq = 0

. Along either diagonal it is

h (t, t) = h (t, - t) = 2 t^{6} - 24 t^{4} = 2 t^{4} (t^{2} - 12),

which is negative for sufficiently small nonzero

t

. Therefore every neighbourhood of the origin contains both positive and negative values, whereas

h (0, 0) = 0

. Consequently

(0, 0) is a saddle point, neither a local minimum nor a local maximum .

The Hessian matrix vanishes there, so the ordinary second-derivative test is inconclusive; the restrictions settle the question.

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