Thread: real functions

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

Please show us what you've tried.

a) since c & d are not both zeros, then the Domain is the set of all real numbers, and the range is the set of all real numbers as well.

right?

Originally Posted by franios

a) since c & d are not both zeros, then the Domain is the set of all real numbers, and the range is the set of all real numbers as well.

right?

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?

x=0

No, if x = 0, then the denominator cx + d = c(0) + d = d =/= 0.

cx+d=0 solve for x, xnot=(-d)/c

That's better, so what is the domain of your function?

All real numbers except when x=-d/c?

Correct, now what is the range? It might help to rewrite this as $\displaystyle \displaystyle \begin{align*} \frac{a}{c} + \frac{bc - ad}{c(cx + d)} \end{align*}$.

the range would still be all real numbers, wouldn't it??