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Math Help - Need help with some Ln and base e functions

  1. #1
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    Need help with some Ln and base e functions

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

    for this i found the slope of the tangent to curve at x = -1 to be -e (not sure if this is correct) and i dunno what to do after that


    2) Find the equation of the tangent to the curve y = 2 - e^-x at the point where x = 1.
    for this i found the slope of the tangent to curve at x = 1 to be 1/e (not sure if this is correct) and i dunno what to do after that


    3) Find the equation of the tangent to the curve y = ln 2x at the point where x = e/2.
    for this i found the slope of the tangent to curve at x = 1 to be 1/2e (not sure if this is correct) and i dunno what to do after that


    Thanks in advance
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  2. #2
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    1) Find the equation of the tangent to the curve y = e^-x at the point where x = -1.

    for this i found the slope of the tangent to curve at x = -1 to be -e (not sure if this is correct) and i dunno what to do after that

    after that use y - y1 = m(x - x1) to het
    y - e = -e(x + 1)
    y = -ex


    parts 2 and 3 to come soon
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  3. #3
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    2) Find the equation of the tangent to the curve y = 2 - e^-x at the point where x = 1.
    for this i found the slope of the tangent to curve at x = 1 to be 1/e (not sure if this is correct) and i dunno what to do after that

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




    3) Find the equation of the tangent to the curve y = ln 2x at the point where x = e/2.
    for this i found the slope of the tangent to curve at x = 1 to be 1/2e (not sure if this is correct) and i dunno what to do after that

    gradient at x is actually 1/x so at x it is 2/e

    same again

    y - 1 = 2x/e - 1
    y = 2x/e

    hope that helps you
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