Calculus

Find the equation of the tangent to the curve y=2-e^-x at the point where x=1.

Can someone please do this step by step and explain?

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  1. Point of contact :
    when x=1 , y = 2 - e^-1 = 2 - 1/e
    = (2e-1)/e

    dy/dx = e^-x
    so when x=1 , slope = e^-1 = 1/e

    using y = mx + b
    (2e-1)/e = (1/e)(1) + b
    b = (1-e)/e

    so y = (1/e)x + (1-e)/e

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