Question

What is the equation of the line normal to `f(x)=1/(e^x+4)` at `x=6`?

Answer 1

`y=((e^6+4)^2/e^6)(x-6)+(1/(e^6+4))`

Explanation:

To find the equation of Normal to the line `f(x)=1/(e^x+4` we need to find the slope of the normal and the `y` coordinate at `x_1=6`

We know the slope of normal is `(-1)/(dy/dx)` so ,

`y=1/(e^x+4)`

Differentiate both sides with respect to `x` using the chain rule.

`dy/dx=(-e^x)/(e^x+4)^2`

So the slope of the normal ( `m_n`) is `(-1)/(dy/dx)=(-1)/((-e^x)/(e^x+4)^2)`

Therefore `m_n=(e^x+4)^2/e^x`

Now to find the `y_1` coordinate when `x_1=6`

We know `f(x)=1/(e^x+4)`

`f(6)=1/(e^6+4)`

Therefore the `y_1` coordinate is `1/(e^6+4)`

We know the equation of a line is

`(y-y_1)=m_n(x-x_1)`

`(y-1/(e^6+4))=((e^6+4)^2/e^6)(x-6)`

Therefore `y=((e^6+4)^2/e^6)(x-6)+(1/(e^6+4))`