Posted by James on .
Given that f is a differentiable function with f(2,5) = 6, fx(2,5) = 1, and fy(2,5) = 1, use a linear approximation to estimate f(2.2,4.9).
The answer is supposed to be 6.3.
Here's what I've done so far:
L(x,y) = f(2,5) + fx(2,5)(x) + fy(2,5)(y)
L(x,y) = 6 + x  y
L(2.2,4.9) = 6 + 2.2  4.9 = 3.3
So I'm three off. Any help would be greatly appreciated.

Calculus 
James,
Never mind. I figured it out.
L(x,y) = f(2,5) + fx(2,5)(x2) + fy(2,5)(y5)
L(x,y) = 6 + 1(x2)  1(y5) = 6+x2y+5 = xy+9
L(2.2,4.9) = 2.24.9+9=6.3