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

1 Answer
May 15, 2017

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