**Show that if ππβππ β 0, then the function π(π₯)=ππ₯+π/ππ₯+π is one-to-one and find its inverse.**

The **Answer to the Question**

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**Here's the Solution to this Question**

$f(x)=cx+dax+bβ$

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

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

ReplaceΒ $f(x)$Β withΒ $y$

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

SwitchΒ $x$Β andΒ $y$

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

Solve forΒ $y$

$cxy+dx=ay+b$

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

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

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

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

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

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