Question 976198
 if {{{f(x) = (x+3)/(x+6)}}}, to find {{{f^(-1)(-1)}}} first find  f(x) =  

 {{{f^(-1)(x)}}}

since {{{f(x)=y}}}, we have

{{{y = (x+3)/(x+6)}}}....swap {{{x}}} and {{{y}}}

{{{x = (y+3)/(y+6)}}}.......solve for {{{y}}}

{{{x(y+6) = (y+3)}}}
    

{{{xy+6x = y+3}}}

{{{xy-y = -6x+3}}}

{{{(x-1)y = -3(2x-1)}}}

{{{y = -3(2x-1)/(x-1)}}}

so, {{{f^(-1)(x)=-3(2x-1)/(x-1)}}}

now find {{{f^(-1)(-1)}}}

{{{f^(-1)(-1)=-3(2(-1)-1)/(-1-1)}}}

{{{f^(-1)(-1)=-3(-2-1)/(-2)}}}

{{{f^(-1)(-1)=-3(-3)/(-2)}}}

{{{f^(-1)(-1)=9/(-2)}}}

{{{f^(-1)(-1)=-9/2}}}

or, as decimal

{{{f^(-1)(-1)=-4.5}}}