Inverse Function Formula

[soroban]

One semester I was asked to find the inverse of \(\displaystyle \,f(x) \:=\:\dfrac{3x - 5}{2x+1}\)
Later, I had to find the inverse of \(\displaystyle \,f(x) \:=\:\dfrac{2x+7}{4x-3}\)


It occurred to me that a general formula would a handy tool.
Especially since I planned to teach Mathematics and I might
be teaching this very topic every semester.

So I solved it for: \(\displaystyle \,f(x) \:=\:\dfrac{ax+b}{cx+d}\)

And arrived at: \(\displaystyle \:f^{\text{-}1}(x) \;=\;\dfrac{dx-b}{\text{-}cx+a}\)

This is easily remembered . . .

(1) Switch the coefficients on the main diagonal (\(\displaystyle a\) and \(\displaystyle d\)).

(2) Change the signs on the minor diagonal (\(\displaystyle b\) and \(\displaystyle c\)).

 

[Steven G]

One semester I was asked to find the inverse of \(\displaystyle \,f(x) \:=\:\dfrac{3x - 5}{2x+1}\)
Later, I had to find the inverse of \(\displaystyle \,f(x) \:=\:\dfrac{2x+7}{4x-3}\)


It occurred to me that a general formula would a handy tool.
Especially since I planned to teach Mathematics and I might
be teaching this very topic every semester.

So I solved it for: \(\displaystyle \,f(x) \:=\:\dfrac{ax+b}{cx+d}\)

And arrived at: \(\displaystyle \:f^{\text{-}1}(x) \;=\;\dfrac{dx-b}{\text{-}cx+a}\)

This is easily remembered . . .

(1) Switch the coefficients on the main diagonal (\(\displaystyle a\) and \(\displaystyle d\)).

(2) Change the signs on the minor diagonal (\(\displaystyle b\) and \(\displaystyle c\)).

Very nice. Now can someone explain why this is similar to the inverse of the matrix (a b c d)? Just the determinant is not there.

I was wondering why Soroban wrote -cx + a instead of a - cx but this may be why (visual aid)