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

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Answer 1 · 2017-01-20T03:30:03

$y = 7 x - 9$ $y=7x-9$

$y = k x + n$ $y=kx+n$ , 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,

$y=7x-9$

What is the equation of the tangent line of f(x) =(4x) / (3-4x^2-x) f(x)=4x3−4x2−xf(x) =(4x) / (3-4x^2-x) at x = 1x=1 x = 1?