Find the equation of the tangent to the curve y = 2 - e^-x at the point where x =1.
I know that you have to find the derivative of that function, which is dy/dx = e^-x
I don't really know how to get the answer beyond that though, because the answer is: x - ey + 2e - 2 = 0
The equation is not in the form of "y=mx+b" and I'm not sure how to find the above answer. If any suggestions or tests can be provided, it would be much appreciated. Please and thank you!
Firstly, you find dy/dx. which equals e^-x. Plug in 1 to find that the slope is e^-1, aka 1/e. Now we have the slope, we need a point. you have 1, so plug 1 back into the original equation to get the point (1, 2- 1/e). You now have the point and the slope, and can make the line from that. I doubt i need to tell you how to do this.