Question

What is the equation of the tangent line of `f(x)=sqrt(1/(x^2-3x+2)` at `x=0`?

Answer 1

`y = 3 / 2^(5/2)x + 1/2^(1/2)`

Explanation:

`p(x) = x^2 - 3x + 2 => f = p^(-1/2)`

`f'(x) = -1/2 p^(-3/2) (2x - 3)`

`f'(0) = -1/2 2^(-3/2) (-3) = + 3 * 2^(-5/2) = a`

The desired line is `g(x) = ax + b`

`g(0) = f(0) = sqrt{1/2} = b`