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

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Answer 1 · 2017-01-20T03:30:03

#y=7x-9#

#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#