Answer
$f^{-1}(x)=\dfrac{-1-3x}{2x-4}$
Work Step by Step
$f(x)=\dfrac{4x-1}{2x+3}$
Substitute $f(x)$ by $y$:
$y=\dfrac{4x-1}{2x+3}$.
Take $2x+3$ to multiply both sides:
$y(2x+3)=4x-1$.
Evaluate the product on the left:
$2xy+3y=4x-1$.
Take $4x$ to the left side and $3y$ to the right side:
$2xy-4x=-1-3y$.
Take out common factor $x$ on the left side:
$x(2y-4)=-1-3y$.
Take $2y-4$ to divide both sides:
$x=\dfrac{-1-3y}{2y-4}$.
Interchange $x$ and $y$:
$y=\dfrac{-1-3x}{2x-4}$.
Substitute $y$ by $f^{-1}(x)$:
$f^{-1}(x)=\dfrac{-1-3x}{2x-4}$.
