Calculus

Consider the function h(x)=e^x. Determine the equation for the tangent line at x = 1. All values must be exact (no decimal values).

i know the answer is just e but how do i prove it? by just subbing 1 into x? it cant be that easy..

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  1. h(x) = y = e^x
    you know when x = 1, y = e
    So the tangent makes contact at (1,e)
    But you don't know the slope of the tangent!
    dy/dx = e^x
    when x = 1, dy/dx = e
    So the slope of the tangent is e

    tangent equation:
    y -e = e(x - 1)
    y = ex - e + e

    y = ex is the equation of the tangent.

    proof:
    https://www.wolframalpha.com/input/?i=plot+y+%3D+e%5Ex,+y+%3D+ex

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  2. i don't understand how you got y-e=e(x-1)??

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  3. The point-slope forms straight line equation :

    y − y1 = m ( x − x1 )

    In this case x1 = 1

    y1= eˣ¹ = e¹ = e

    m = slope = first derivation in point x = 1

    m = eˣ¹ = e¹ = e

    y − y1 = m ( x − x1 )

    y − e = e ( x − 1 )

    y − e = e ∙ x − e ∙ 1

    y − e = e ∙ x − e

    add e to both sides

    y − e + e = e ∙ x − e + e

    y = e ∙ x

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