From the given inverse function, we have
[tex]f^{-1}(x-1) = 2x - \dfrac52 \\\\ \implies f^{-1}(x) = f^{-1}((x+1)-1) = 2(x+1) - \dfrac52 = 2x-\dfrac12[/tex]
By definition of inverse function,
[tex]f\left(f^{-1}(x)\right) = x[/tex]
so that
[tex]f\left(2x-\dfrac12\right) = x[/tex]
[tex]a\left(2x-\dfrac12\right) + \dfrac b2 = x[/tex]
[tex]2ax - \dfrac a2 + \dfrac b2 = x[/tex]
[tex]\implies\begin{cases}2a=1 \\\\ -\dfrac a2+\dfrac b2 = 0\end{cases}[/tex]
[tex]\implies \boxed{a=\dfrac12}[/tex]
[tex]\implies \dfrac b2 = \dfrac a2 \implies \boxed{b=\dfrac12}[/tex]
