Answer
$f^{-1}(x)=\dfrac{x+3}{8x+4}$
Work Step by Step
$f(x)=\dfrac{3-4x}{8x-1}$
Rewrite this expression as $y=\dfrac{3-4x}{8x-1}$ and solve for $x$:
$y=\dfrac{3-4x}{8x-1}$
Take $8x-1$ to multiply the left side:
$y(8x-1)=3-4x$
$8xy-y=3-4x$
Take $-4x$ to the left side and $-y$ to the right side:
$8xy+4x=y+3$
Take out common factor $x$ from the left side:
$x(8y+4)=y+3$
Take $8y+4$ to divide the right side:
$x=\dfrac{y+3}{8y+4}$
Interchange $x$ and $y$:
$y=\dfrac{x+3}{8x+4}$
The inverse of the original function is $f^{-1}(x)=\dfrac{x+3}{8x+4}$