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)