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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

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

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.6: The Inverse Trigonometric Functions
Problem 91E
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Verify the linear approximation at (0, 0). 7x + 8 8 + 7x 24y 3y + 1 f(x, y) = = 7x + 8 3y + 1 Let f(x, y): = Then fx(x, y) = and fy(x, y) = 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) = I I Both fx and fy are continuous functions for , fy(0, 0) = I and the
Transcribed Image Text:Verify the linear approximation at (0, 0). 7x + 8 8 + 7x 24y 3y + 1 f(x, y) = = 7x + 8 3y + 1 Let f(x, y): = Then fx(x, y) = and fy(x, y) = 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) = I I Both fx and fy are continuous functions for , fy(0, 0) = I and the
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