Question

What is the equation of the line tangent to `f(x)=1/(x^2-4)` at `x=-1`?

Answer 1

The equation of the tangent line is:

`y(x) = 2/9x -1/9`

Explanation:

The equation of a line tangent to the curve `y=f(x)` in `x = bar x` is given by:

`y(x) = f(barx)+f'(barx)(x-barx)`

We have:

`f(x) =1/(x^2-4)`

`f'(x) = (-2x)/((x^2-4)^2)`

For `barx =-1`

`f(-1) =1/((-1)^2-4) = -1/3`

`f'(-1) = (-2(-1))/(((-1)^2-4)^2)=2/9`

So that the equation of the tangent line is:

`y(x) = -1/3+2/9(x+1) = 2/9x -1/9`