If the image of the function f (x) = 1-ex intersects the y-axis at point P, then the equation of the tangent of the curve at point P is______ .
From F (x) = 1-ex, we get that f (0) = 1-e0 = 0. Then f ′ (x) = - ex, f ′ (0) = - E0 = - 1. The tangent equation of F (x) = 1-ex at point P (0, 0) is y-0 = - 1 × (x-0), that is, x + y = 0. So the answer is: x + y = 0
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