Show that if 𝑎𝑑−𝑏𝑐 ≠ 0, then the function 𝑓(𝑥)=𝑎𝑥+𝑏/𝑐𝑥+𝑑 is one-to-one and find its inverse.

$f(x)=\frac{ax+b}{cx+d}$

$cx+d\not=0=>x\not=-d/c$

Domain: $(-\infin, -d/c)\cup(-d/c, \infin)$

Replace $f(x)$ with $y$

$y=\dfrac{ax+b}{cx+d}$

Switch $x$ and $y$

$x=\dfrac{ay+b}{cy+d}$

Solve for $y$

$cxy+dx=ay+b$

$y=\dfrac{b-dx}{cx-a}$

$cx-a\not=0=>x\not=c/a$

$f^{-1}\circ f=\dfrac{b-d(\dfrac{ax+b}{cx+d})}{c(\dfrac{ax+b}{cx+d})-a}$

$=\dfrac{bcx+bd-adx-bd}{acx+bc-acx-ad}$

$=\dfrac{bcx-adx}{bc-ad}=x, bc\not=ad=>ad-bc\not=0$

Therefore  if $ad-bc\not=0,$ then the function $f(x)=\dfrac{ax+b}{cx+d}$ is one-to-one and find its inverse.