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Math Help - derivative and equation of tangent line of function with base e

  1. #1
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    derivative and equation of tangent line of function with base e

    given f(x)= 4e^x -1x + 4

    a) find derivative
    I got 4ex -1 (which is correct)

    ***b) find the equation of a tangent line:

    I thought that to find the equation of the tangent line you go
    [f(x) - f(x+h)] / h
    so i did this, plugged in
    [(4ex -1 ) - (4e(x+h) -1 )] / h
    cancelled and retrieved -4e for my answer.
    which is wrong..

    help plzz
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  2. #2
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    Quote Originally Posted by williamb View Post
    given f(x)= 4e^x -1x + 4

    a) find derivative
    I got f'(x) = 4e^x -1 (which is correct)

    ***b) find the equation of a tangent line:
    general form for a tangent line to a curve at the point (a, f(a))

    y - f(a) = m(x - a)

    where m = f'(a)
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  3. #3
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    where could you get the values from?
    can you explain in more detail?
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  4. #4
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    y-f(a) = m(x-a)

    so m is f'(a) as in 4ex-1
    and f(a) is 4e^x - x + 4 ?

    y - 4e^x - x +4 = (4ex-1)(x-a)

    ??
    is this right?
    if so, what is a?
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  5. #5
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    f(x) = 4e^x - x + 4

    f'(x) = 4e^x - 1

    the tangent line to the curve f(x) = 4e^x - x + 4 at x = a is ...

    y - f(a) = f'(a) \cdot (x - a)

    y - (4e^a - a + 4) = (4e^a - 1) \cdot (x - a)

    normally, one is given a specific value of x ... I'm using x = a because it seems you were not given an x.
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  6. #6
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    THANKS SO MUCH AWESOMEEEEE
    you rock my world
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