Question

What is the equation of the line normal to `f(x)=e^(2x)/(e^x+4)` at `x=2`?

Answer 1

In point slope form it is
`y-e^4/(e^2+4) = (e^2+4)^2/ (e^6+8e^4) (x-2)`.

Explanation:

For `f(x)=e^(2x)/(e^x+4)` at `x=2`, we get `y=e^4/(e^2+4)`.

The slope of the tangent line is given by:

`f'(x) = ((2e^(2x))(e^x+4)-e^(2x)(e^x))/(e^x+4)^2 = (e^(3x)+8e^2x)/(e^x+4)^2`.

At the point where `x=2`, the slope of the tangent is

`m_(tan) = (e^6+8e^4)/(e^2+4)^2`.

So the slope of the normal line is

`m_(norm) = (e^2+4)^2/ (e^6+8e^4)`.

The equation of the line (in point slope form) through `(2,e^4/(e^2+4))` with slope `m = (e^2+4)^2/ (e^6+8e^4)` is

`y-e^4/(e^2+4) = (e^2+4)^2/ (e^6+8e^4) (x-2)`.

Rewrite using algebra as you see fit. (Or as your grader demands.)