Verify the linear approximation at (0, 0). 7x + 8 3y + 1 8 + 7x 24y f(x, y) = Let f(x, y) = Then fx(x, y) = and fy(x, y) = so by this theorem, f is differentiable at (0, 0). We have fx(0, 0) = linear approximation of fat (0, 0) is f(x, y) = f(0, 0) + fx(0, 0)(x −0) + fy(0, 0)(y- 0) = y # 7x + 8 3y + 1 I Both fx and fy are continuous functions for , fy(0, 0) = and the

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