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Web skærmformat. - Aarhus Universitet

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IX. Opgaver – 5. Januar 2005 600<br />

Gradienterne ∇f = (fx,fy) beregnes<br />

Så (0,0),(−1, −1) er kritiske punkter.<br />

2) De andenordens partielle afledede er<br />

Hessematricen er<br />

I de kritiske punkter fås:<br />

∇f(1,1) = (6,6)<br />

∇f(0,0) = (0,0)<br />

∇f(−1, −1) = (0,0) .<br />

fxx = 6x, fxy = 3, fyy = 6y .<br />

H(f) =<br />

H (0,0) (f) =<br />

6x 3<br />

3 6y<br />

0 3<br />

3 0<br />

<br />

.<br />

<br />

,<br />

som har determinant = −9. Punktet (0,0) er derfor et saddelpunkt.<br />

<br />

−6 3<br />

H (f) = ,<br />

(−1,−1) 3 −6<br />

som har karakteristisk polynomium<br />

<br />

<br />

<br />

−6 − λ 3 <br />

<br />

3 −6 − λ = λ2 + 12λ + 25

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