What is the equation of the tangent line of f(x) =(4x) / (3-4x^2-x) at x = 1?

1 Answer
Jan 20, 2017

y=7x-9

Explanation:

y=kx+n, where k=f'(x_0), x_0=1

Lets find the first derivative (using quotient rule):

f'(x)=(4(3-4x^2-x)-4x(-8x-1))/(3-4x^2-x)^2

f'(x)=(12-16x^2-4x+32x^2+4x)/(3-4x^2-x)^2

f'(x)=(16x^2+12)/(3-4x^2-x)^2

k=f'(x_0)=f'(1)=(16+12)/(3-4-1)^2=28/(-2)^2=7

For x_0=1 => y_0=f(1)=4/(3-4-1)=4/(-2)=-2

y_0=kx_0+n => -2=7*1+n => n=-9

Finally,

y=7x-9