Question

What is the equation of the normal line of `f(x)=(2+x)e^-x` at `x=0`?

Answer 1

`y=x+2`

Explanation:

The normal line is perpendicular to the tangent line at a given point. Therefore, we must find the slope of the tangent when `x=0`, and to do that, we must find `f'(x)`.

We'll need to use the Product Rule.
`f'(x)=color(red)(d/(dx)[2+x])*e^-x+(2+x)*color(blue)(d/(dx)[e^-x]`

Before we continue, we have to know what the derivatives of these functions are.
`color(red)(d/(dx)[2+x]=1`

The next derivative requires use of the Chain Rule.
`color(blue)(d/(dx)[e^-x])=d/(dx)[-x]*e^-x=-1*e^-x=color(blue)(-e^-x`

Let's plug our derivatives back in.
`f'(x)=color(red)(1)*e^-x+(2+x)*color(blue)(-e^-x)`
`f'(x)=e^-x-(2+x)e^-x`

Factor out an `e^-x`.
`f'(x)=e^-x(1-(2+x))`
`f'(x)=e^-x(-x-1)`
`color(green)(f'(x)=frac(-x-1)(e^x)`

Now, we can compute `f'(0)` to find the slope of the tangent line at `x=0`.
`f'(0)=(-1)/(e^0)=(-1)/1=-1`

Now, since the normal line is perpendicular to the tangent line, its slope will be the opposite reciprocal of `-1`, which is `color(brown)(1)`.

In order to write the equation of the normal line, we should use point-slope form: `y-y_1=m(x-x_1)`

The point through which the normal line passes can be obtained through finding `f(0)`, since that is the point on the original function where we have found the slope of the normal line will be equal to `color(brown)(1)`.

`f(0)=(2+0)e^-0=2e^0=2(1)=2`
We know that the normal line has slope `color(brown)(1)` and passes through the point `(0,2)`.

We can plug this all into point-slope form: `y-2=1(x-0)`
Simplified: `color(purple)(y=x+2`