A function of two variables is given by, f(x,y) = e^(2x-3y) Find the tangent approximation to f(0.942,0.625) near (0,0).?
回答 (1)
2015-09-27 7:00 pm
Taking partial derivatives,
f_x = 2e^(2x-3y) and f_y = -3e^(2x-3y).
So, the tangent line approximation to f at (0, 0) is given by
f(0,0) + f_x(0,0) (x - 0) + f_y(0,0) (y - 0) = 1 + 2x - 3y.
Therefore,
f(0.942,0.625) ≈ 1 + 2x - 3y {at (0.942,0.625)} = 1.009.
I hope this helps!
f_x = 2e^(2x-3y) and f_y = -3e^(2x-3y).
So, the tangent line approximation to f at (0, 0) is given by
f(0,0) + f_x(0,0) (x - 0) + f_y(0,0) (y - 0) = 1 + 2x - 3y.
Therefore,
f(0.942,0.625) ≈ 1 + 2x - 3y {at (0.942,0.625)} = 1.009.
I hope this helps!
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https://hk.answers.yahoo.com/question/index?qid=20150927082255AAjOVXA