a) Find the equation of the line tangent to the given curve at a

b) Use a graphing utility to graph the curve and the tangent line on the same set of axes

\(y = e^x ; a = ln 3\)

The answer for a) is: \(y = 3x + 3 - 3ln3\)

But i have no idea how they got that.

Thanks for the help

StayCurly Mar 12, 2018

#1**0 **

First, realize that the drivative of e^x IS e^x

We can find the slope at point a = ln3 by substituting this in to

e^x e^(ln3) = 3 = slope

AT point ln 3 , e^(ln3) or ln 3 , 3

so using y = mx + b form

3 = (3) (ln3) + b solve for b

3-3ln3 = b

equation of the line then becomes

y = 3 x + 3- 3ln3

ElectricPavlov Mar 13, 2018