We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
308
2
avatar+31 

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+18957 
+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+18957 
+2

Here is a graph:

 

 Mar 13, 2018

32 Online Users

avatar
avatar
avatar