Help with Partial derivatives please
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Help with Partial derivatives please

[From: ] [author: ] [Date: 12-03-25] [Hit: ]
So, (x, y) = (1, -4) is the only critical point.Classify this with the second derivative test.f_xx = 10x^3 y + 60x^2,......
Consider the function f(x,y) = x^(5)y + 5 x^(4) - 1 y. Find and classify the critical point of the function. is it local minimum, local maximum, saddle point?

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For the critical points:
f_x = 5x^4 y + 20x^3 and f_y = x^5 - 1.

Set these equal to 0:
5x^4 y + 20x^3 = 0
x^5 - 1 = 0 ==> x = 1 (other roots are imaginary).

Since x = 1, we have 5y + 20 = 0 ==> y = -4.

So, (x, y) = (1, -4) is the only critical point.
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Classify this with the second derivative test.
f_xx = 10x^3 y + 60x^2, f_yy = 0, f_xy = 5x^4
==> D = (f_xx)(f_yy) - (f_xy)^2 = 0 - (5x^4)^2 = -25x^8.

Since D(1, -4) = -25 < 0, we have a saddle point at (1, -4).

I hope this helps!
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