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December 21, 2014

December 21, 2014

Posted by **George** on Monday, November 9, 2009 at 12:48pm.

- Calculus -
**bobpursley**, Monday, November 9, 2009 at 2:14pmHave you graphed it? The graph ought to answer all the questions.

The tangent line will be

y=mx+b at x=2 then y=1/e^.5

if u= -x^-1

d(e^u)= e^u du/dx= e^-1/x * 1/x^2

so all that leads to

y=e^u

y'=1/x^2 * e^-1/x

for x=2, find y' the slope m, put it in the equation, and solve for b, and you have the tangent line.

Graph it all

- Calculus -
**George**, Monday, November 9, 2009 at 4:42pmWhat does it mean asymptotic behavior?

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