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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

 Mar 12, 2018
 #1
avatar+17330 
+2

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

 Mar 13, 2018
 #2
avatar+17330 
+2

Here is a graph:

 

 Mar 13, 2018

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