# Curve Sketching again~

• Sep 18th 2008, 05:05 AM
ose90
Curve Sketching again~
The function f is defined as $\displaystyle f(t)=(4e^kt -1)/(4e^kt +1)$
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 $\displaystyle when f'(t)=0 , 4e^kt + 1 = 0 but 4e^kt + 1 > 0$ therefore no stationary point
• Sep 18th 2008, 01:55 PM
mr fantastic
Quote:

Originally Posted by ose90
The function f is defined as $\displaystyle f(t)=(4e^kt -1)/(4e^kt +1)$
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 $\displaystyle when f'(t)=0 , 4e^kt + 1 = 0 but 4e^kt + 1 > 0$ therefore no stationary point

You can always substitute a positive value for k to get a feel for things!

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

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

Also, I doubt you've differentiated correctly since the numerator of the derivative is $\displaystyle 8 k e^{kt}$. 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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