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For any real numbers a,b,c & d such that c & d are not both zero, the formula f(x)=(ax+b)/(cx+d) defines a real function.
a)find the domain & range of f.
b)Show that f is a one-to-one function if and only if ad-bc not=0
c)Show that if f is one-to-one, then f' is given by f-1(x)=(dx-b)/(-cx+a) where c & a are both not zero
d) show that if f is one-to-one and a=-d then f-1=f
No, for any functions which are defined for all the reals, as linear functions like those are, their quotient will be defined for all the reals EXCEPT for where the denominator is 0. What value of x will make the denominator 0?Originally Posted by franios
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