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Math Help - Curve Sketching again~

  1. #1
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    Curve Sketching again~

    The function f is defined as f(t)=(4e^kt -1)/(4e^kt +1)
    Find lim f(t) when t tends to infinity and sketch the graph of f.

    Thanks, im baffled by the K which is a positive constant.
    I found an asymptote on y = 1, and 2 point of intersection.
    I can't find any stationary point there as  when f'(t)=0 , 4e^kt + 1 = 0 but 4e^kt + 1 > 0 therefore no stationary point
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  2. #2
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    Quote Originally Posted by ose90 View Post
    The function f is defined as f(t)=(4e^kt -1)/(4e^kt +1)
    Find lim f(t) when t tends to infinity and sketch the graph of f.

    Thanks, im baffled by the K which is a positive constant.
    I found an asymptote on y = 1, and 2 point of intersection.
    I can't find any stationary point there as  when f'(t)=0 , 4e^kt + 1 = 0 but 4e^kt + 1 > 0 therefore no stationary point
    You can always substitute a positive value for k to get a feel for things!

     \lim_{t \rightarrow +\infty} \frac{4e^{kt} - 1}{4e^{kt} + 1} = \lim_{t \rightarrow +\infty} \frac{4 - e^{-kt}}{4 + e^{-kt}} = \frac{4 - 0}{4 + 0} = 1.


     \lim_{t \rightarrow -\infty} \frac{4e^{kt} - 1}{4e^{kt} + 1} = \lim_{t \rightarrow +\infty} \frac{0 - 1}{0 + 1} = -1.

    Also, I doubt you've differentiated correctly since the numerator of the derivative is 8 k e^{kt}. Even though this still means that there's no stationary point, you should show your working and the answer you obtained so that your mistake can be pointed out.
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