## Calculus: Early Transcendentals 8th Edition

$f^{-1}(x)=\dfrac{-1-3x}{2x-4}$
$f(x)=\dfrac{4x-1}{2x+3}$ 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 $f^{-1}(x)$: $f^{-1}(x)=\dfrac{-1-3x}{2x-4}$.