Question 2338
ok, inverses of functions...very simple. All you are being asked is basically find the "reverse" function...in "earlier Maths"...what is x equal to, that is all.

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

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

so x = {{{((1/y)+3)/2}}}

this is your answer...{{{f^-1(x) = ((1/x)+3)/2}}}

Note, i have changed the y into an x, because the function asks for {{{f^-1(x)}}}, not {{{f^-1(y)}}}.

The acid test is that f(x) and {{{f^-1(x)}}} should be opposites of each other,

so put a value in f(x) to give an answer.
then put that answer in {{{f^-1(x)}}} to get the original number back as the answer.

eg x=1 into f(x) is -1

so x=-1 into {{{f^-1(x)}}} is 1. CORRECT :-)

PS, this working assumes that i have read your f(x) correctly...or is f(x) = {{{(1/(2x))-3}}}? If so, you re-do the work to find the inverse :-)

jon.