Question

What is the equation of the line tangent to `f(x)= -1/x` at `x=-3`?

Answer 1

`y = 1/9x + 2/3`

Explanation:

To find the full co-ordinates of the point that you're finding the tangent of, substitute `x = -3` into the original equation.

`y = f(-3) = -1/-3 = 1/3`

Which gives the `(x, y)` co-ordinates

`(-3, 1/3)`

Differentiate f(x) by switching around the function to make it more friendly and then differentiating in the normal fashion.

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

This is the gradient of the line, or `m` in the line equation `y = mx + c`. Substituting our value of `x`, `m` more properly is

`1/(-3)^2 = 1/9`, so

`y = 1/9x + c`

Substituting all of the co-ordinates into this to find `c`, we get

`y = 1/9x + c`
`1/3 = -3/9 + c = -1/3 + c`
`1/3 + 1/3 = c`
`c = 2/3`

Putting this back into the form `y = mx + c`,

`y = 1/9x + 2/3`

Which is the equation of the tangent.