Question 739648


the inverse function, if it exists of {{{f(x)=5/(x-4)}}} is {{{f^-1(x)}}}

{{{f(x)=y}}}

{{{y=5/(x-4)}}}......solve for {{{x}}}

{{{(x-4)y=5}}}

{{{xy-4y=5}}}

{{{xy=4y+5}}}

{{{x=(4y+5)/y}}}

...switch {{{y}}} and {{{x}}}
the new "{{{y}}} =" is the inverse:

{{{y=(4x+5)/x}}}