Asked by James
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.
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.
Answers
Answered by
James
Never mind. I figured it out.
L(x,y) = f(2,5) + fx(2,5)(x-2) + fy(2,5)(y-5)
L(x,y) = 6 + 1(x-2) - 1(y-5) = 6+x-2-y+5 = x-y+9
L(2.2,4.9) = 2.2-4.9+9=6.3
L(x,y) = f(2,5) + fx(2,5)(x-2) + fy(2,5)(y-5)
L(x,y) = 6 + 1(x-2) - 1(y-5) = 6+x-2-y+5 = x-y+9
L(2.2,4.9) = 2.2-4.9+9=6.3
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