Question

What is the equation of the line tangent to `f(x)=2/(4 − x^2)` at `x=3`?

Answer 1

`y = 12/25 x - 36/25` or `25y = 12x -36`

Explanation:

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

when `x =3, y = f(3) = 2/(4-3^2) =-2/5`

let say,
`u = 2, u' = 0, v = 4-x^2, v' = -2x`

Gradien of tangent, f'(x),

`f'(x) = (0*(4- x^2) - 2*(-2x))/(4 - x^2)^2`

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

at `x =3, f'(3) = (4*3)/(4 - 3^2)^2 = 12/25`-->m, gradient of tangent at `x =3`

plug in value of `x, y and m` in the equation of `y =mx + c`

`-2/5 = 12/25 (3) + c`

`-10/25 - 36/25 = c`

`-36/25 = c`

therefore the equation of line tangent,
`y = 12/25 x - 36/25` or `25y = 12x -36`