# real functions

• Oct 22nd 2012, 08:08 PM
franios
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
• Oct 22nd 2012, 08:11 PM
Prove It
Re: real functions
Please show us what you've tried.
• Oct 22nd 2012, 08:16 PM
franios
Re: real functions
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?
• Oct 22nd 2012, 08:22 PM
Prove It
Re: real functions
Quote:

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?
• Oct 22nd 2012, 08:24 PM
franios
Re: real functions
x=0
• Oct 22nd 2012, 08:25 PM
Prove It
Re: real functions
No, if x = 0, then the denominator cx + d = c(0) + d = d =/= 0.
• Oct 22nd 2012, 08:31 PM
franios
Re: real functions
cx+d=0 solve for x, xnot=(-d)/c
• Oct 22nd 2012, 08:46 PM
Prove It
Re: real functions
That's better, so what is the domain of your function?
• Oct 23rd 2012, 12:54 PM
franios
Re: real functions
All real numbers except when x=-d/c?
• Oct 23rd 2012, 05:19 PM
Prove It
Re: real functions
Correct, now what is the range? It might help to rewrite this as \displaystyle \begin{align*} \frac{a}{c} + \frac{bc - ad}{c(cx + d)} \end{align*}.
• Oct 24th 2012, 10:26 AM
franios
Re: real functions
the range would still be all real numbers, wouldn't it??