Question

What is the equation of the line tangent to `f(x)=1/(x^2+x+6)` at `x=-1`?

Answer 1

`y\ =\ \frac{1}{36\ }x\ +\ \frac{7}{36}`

Explanation:

start with the derivative, using the quotient rule `(u/v)' = (u' \ v - u \ v')/v^2`

so `f' =- (2x + 1)/(x^2 + x + 6)^2`

`f'(-1) = 1 / 36`

`f(-1) = 1 / 6`

you `y = mx + c`, or here `y = 1/36 x + c` and `1/6 = 1/36 (-1) + c \implies c = 7/36`