Question

What is the equation of the normal line of `f(x)=e^(4+x)/(4-x)` at `x=0`?

Answer 1

`16/5e^(-4)x-y+1/4e^4=0` that gives `0.0586-y+13.65`, nearly. See the normal-inclusive Socratic graph..

Explanation:

`f(x-4)=e^4e^x`. Differentiating,

f'(4-x)-f-e^4e^x=0.

At `P(0, e^4/4),`

`i'=-5/16e^4`

The slope of the normal is

`-1/(f')=16/5e^(-4)`. So, the equation to the normal at `P(0, 1/4e^4)` is

`y-1/4e^4=16/5e^(-4)x`.
graph{(y(4-x)-e^(4+x))(y-13.65)(x^2+(y-13.65-.09x)^2-.1)=0 [-32.76, 32.71, 0, 32]}