+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
+36916 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
