Question

What is the equation of the normal line of `f(x)=x^2-x+1/x` at `x = 2`?

Answer 1

The equation is:

`y(x) =-4/11x+71/22`

Explanation:

The general formula of the normal line to a curve is:

`y(x) = f(bar x) -1/(f'(bar x)) (x - bar x)`

Given:

`f(x) = x^2-x+1/x`
`f'(x) =2x-1-1/x^2`

For `x=2`:

`f(x)|_(x=2) =5/2`

`f'(x)|_(x=2) =11/4`

So:

`y(x) =5/2-4/11(x-2)=-4/11x+71/22`