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Math Help - Tangent lines

  1. #1
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    Tangent lines

    I have a problem that I do not know how to start. If anyone could give me a hint on how to start it, that would be great.

    Let f be the function given by f(x) = e^{-x}, and let g be the function given by g(x) = kx, where k is the nonzero constant such that the graph of f is tangent to the graph of g.

    Find the x-coordinate of the point of tangency and the value of k.
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  2. #2
    Member arpitagarwal82's Avatar
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    By condition of tangance df/dx x dg/dx = -1

    so  -e^{-x} * k = -1<br />
    e^{-x} = 1/k

    e^x = k
    x = \ln k

    So g(x) is tangent to f(x) at point x = \ln k

    Now for x = \ln k
    you cordinates of both curve should be same

    so x = \ln k satisfies
    e^{-x}= kx

    solve for k
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  3. #3
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    Tangent

    Hello Cursed
    Quote Originally Posted by Cursed View Post
    I have a problem that I do not know how to start. If anyone could give me a hint on how to start it, that would be great.

    Let f be the function given by f(x) = e^{-x}, and let g be the function given by g(x) = kx, where k is the nonzero constant such that the graph of f is tangent to the graph of g.

    Find the x-coordinate of the point of tangency and the value of k.
    First, you need to find the value of x where the two graphs meet: that is where f(x) = g(x), or

     e^{-x}= kx (1)

    Then if g(x) is a tangent to f(x) for this value of x, their gradients are equal at this point. So f'(x) = g'(x) or:

     -e^{-x} = k

    \Rightarrow e^{-x} = -k (2)

    Now you can use (2) to eliminate the e^{-x} term in (1), and solve the resulting equation for x. This is the x-coordinate of the point of tangency, which you can then use to find the value of k, by substituting it back into (1).

    Can you complete it now?

    Grandad
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  4. #4
    Member arpitagarwal82's Avatar
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    Ok
    my solution was wrong.
    Proceed with Granded's solution.

    Thanks Granded
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