Answer to Question #252422 in Discrete Mathematics for gael

Question #252422

Show that if π‘Žπ‘‘βˆ’π‘π‘ β‰  0, then the function 𝑓(π‘₯)=π‘Žπ‘₯+𝑏/𝑐π‘₯+𝑑 is one-to-one and find its inverse.


1
Expert's answer
2021-10-19T15:44:01-0400
f(x)=ax+bcx+df(x)=\dfrac{ax+b}{cx+d}

cx+d=ΜΈ0=>x=ΜΈβˆ’d/ccx+d\not=0=>x\not=-d/c

Domain: (βˆ’βˆž,βˆ’d/c)βˆͺ(βˆ’d/c,∞)(-\infin, -d/c)\cup(-d/c, \infin)

Replace f(x)f(x) with yy


y=ax+bcx+dy=\dfrac{ax+b}{cx+d}

Switch xx and yy


x=ay+bcy+dx=\dfrac{ay+b}{cy+d}

Solve for yy


cxy+dx=ay+bcxy+dx=ay+b

y=bβˆ’dxcxβˆ’ay=\dfrac{b-dx}{cx-a}

cxβˆ’a=ΜΈ0=>x=ΜΈc/acx-a\not=0=>x\not=c/a


fβˆ’1∘f=bβˆ’d(ax+bcx+d)c(ax+bcx+d)βˆ’af^{-1}\circ f=\dfrac{b-d(\dfrac{ax+b}{cx+d})}{c(\dfrac{ax+b}{cx+d})-a}

=bcx+bdβˆ’adxβˆ’bdacx+bcβˆ’acxβˆ’ad=\dfrac{bcx+bd-adx-bd}{acx+bc-acx-ad}

=bcxβˆ’adxbcβˆ’ad=x,bc=ΜΈad=>adβˆ’bc=ΜΈ0=\dfrac{bcx-adx}{bc-ad}=x, bc\not=ad=>ad-bc\not=0

Therefore  if adβˆ’bc=ΜΈ0,ad-bc\not=0, then the function f(x)=ax+bcx+df(x)=\dfrac{ax+b}{cx+d} is one-to-one and find its inverse.



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