recall that for inverse functions, the range of the original is the domain of the inverse and othe other way around.
so f⁻¹(8) is f⁻¹(x) but making x = 8, so the domain value in this case for the inverse is 8, and that's going to give us something, but let's nevermind that.
since the domain value is 8 for the inverse, that means, the same 8, is a range for the original f(x).
meaning, "some value of x" on the original, gives us a range of 8, then let's use that 8 in the original, BUT not on "x", but on "y", or f(x).
[tex]\bf f^{-1}(8)~\hspace{7em}\stackrel{f(x)}{8}=2x+5 \\\\[-0.35em] ~\dotfill\\\\ 8=2x+5\implies 3=2x\implies \boxed{\cfrac{3}{2}=x} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{original f(x)}}{\left(\stackrel{x}{\frac{3}{2}}~~,~~\stackrel{y}{8} \right)}~\hspace{7em} \stackrel{\textit{inverse }f^{-1}(x)}{\left(\stackrel{x}{8}~~,~~\stackrel{y}{\frac{3}{2}} \right)}~\hfill f^{-1}(8)=\cfrac{3}{2}[/tex]
so notice how the values swap places for the inverses, one's domain, is the other's range.
