Answer
$f^{-1}(x)=\dfrac{x}{6} +\dfrac{1}{6}$
Work Step by Step
Replace $f(x)$ by $y$ to have:
$y=6x-1$
Interchange $x$ and $y$ then solve for $y$ to have:
$x = 6y-1
\\x+1= 6y
\\\dfrac{x+1}{6}=y
\\\dfrac{x}{6} +\dfrac{1}{6}=y
\\y=\dfrac{x}{6} +\dfrac{1}{6}$
Replace $y$ with $f^{-1}(x)$ to have:
$f^{-1}(x)=\dfrac{x}{6} +\dfrac{1}{6}$
