Calculus

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

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  1. Equation for the tangent :

    Delta y/Delta x = y'(-1)------>

    [y-exp(-1)]/(x+1) = exp(-1) ------>

    y = exp(-1) + (x+1)exp(-1) =

    exp(-1)(2 + x)

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  2. I really don't understand this. Can you explain it step by step? The answer at the back of the book is y=-ex.

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  3. to find the equation of the tangent you will need the point of contact and the slope at that point of contact.

    when x= -1, y = e^-1 = 1/e

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

    using y = mx + b

    1/e = (1/e)(-1) + b
    2/e = b

    the equation of the tangent is

    y = (1/e)x + 2/e

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