Question

suppose that f(x,y)is a smooth function and that its partial derivatives have the values, fx(-2,0)=-5 and fy(-2,0)=5 and f(-2,0)= 6 find f(-1,1)

Answer 1

Nice one you got there.

Answer 2

Since fx(-2, 0)=-5, then we can assume that f(x, y) have terms that do not depend on y. And these terms would be 0 on fy, since they are constant in relation to y.

Since fy(-2, 0)=5, we can say that in this point, fx=-fy, and defining Q(x) as the part of f that do not depend on y, and R(x, y) the part that can depend on y, we see that Qx=-Ry on this point.

So f(x, y) = Q(x) + R(x, y) on this point.

Now taking Qx(-2)=-5, we can say that Q=x^2-x

From f(-2, 0)=6, we get that R(-2, 0)=4, and since all the terms that dont depend on y are on Q(x), y must be in the exponent on R. So R(x, y) = 4*exp((5/4)y)\[f(x, y)=4e^{\frac{ 5 }{ 4 }y}+x^2-x\]

Giving fx(-2, 0)=-5, fy(-2, 0)=5, f(-2, 0)=6, and:

f(-1, 1)=4*exp(5/4)

I did that by logic, and I am almost shure there is another way of doing it, but I think better like that, I hope it helps.

Answer 3

its wrong

Answer 4

Crap

Answer 5

Do you have the correct result?

Answer 6

Oh I see it now, the mistake was saying that R(-2, 0)=4, it is actually 0.
Now it gets R=5y*exp(ay).
\[f(x, y)=x^2-x+5ye^{ay}\]And f(-1, 1)=2+5e^a for any a. That is odd but I think it fits.