f(x)=cx+dax+bβ
cx+dξ =0=>xξ =βd/c
Domain: (ββ,βd/c)βͺ(βd/c,β)
Replace f(x) with y
y=cx+dax+bβ Switch x and y
x=cy+day+bβ Solve for y
cxy+dx=ay+b
y=cxβabβdxβ cxβaξ =0=>xξ =c/a
fβ1βf=c(cx+dax+bβ)βabβd(cx+dax+bβ)β
=acx+bcβacxβadbcx+bdβadxβbdβ
=bcβadbcxβadxβ=x,bcξ =ad=>adβbcξ =0 Therefore if adβbcξ =0, then the function f(x)=cx+dax+bβ is one-to-one and find its inverse.
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