Question

What is the equation of the normal line of `f(x)=2x+3/x` at `x=1`?

Answer 1

`:. y=x+4`

Explanation:

`f(x) = 2x+3/x`
`:. f(x) = 2x+3x^-1`

Differentiating wrt `x` gives us:
`f'(x) = 2+3(-x^-2)`
`:. f'(x) = 2-3/x^2`

When `x=1 => f(1)=2+3 = 5`
and, `f'(1) = 2-3=-1`

So the gradient of the tangent at `x=1` is `m'=-1`, and the tangent and normal are perpendicular so the product of their gradients is `-1`
Hence, gradient of Normal is `m=-1/(-1)=1`

So the Normal has gradient `m=1` and it passes through `(1,5)`, so using `y-y_1=m(x-x_1)` the Normal equation is given by:

`y-5=(1)(x-1)`
`:. y=x+4`