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

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Answer 1 · 2017-05-15T05:10:25

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

#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#