Question

Nice little problem.

Show that \(\dfrac{(ax - b)(dx -c)}{(bx - a)(cx - d)}\) will be capable of all real values if \(a^2 - b^2\) and \(c^2 - d^2\) have the same sign.

Answer 1

@mathslover and @mukushla try this.

Answer 2

capable ??

Answer 3

@amistre64

Answer 4

Can I give the solution off now? Or is it too early?

Answer 5

if you want to give the solution, especially to your own question, then feel free :)

Answer 6

if you want it to sit for a few days, bumping, then thats fine too

Answer 7

Hehe, I'm too impatient. :P

Any ideas about the problem?

Answer 8

i would try \(\iff\) i know what do you mean by capable

Answer 9

for what domain of x?

Answer 10

Real.

Answer 11

you are essentially saying that this maps whole R to R.

Answer 12

@BSwan It means that this function can return any possible real number for a particular \(x\) in reals.

Answer 13

well well ... let y=that function and solve the quadratic equation.

Answer 14

Haha nice.

Answer 15

i can see that there exists a map from R to R, but I am not sure if it's one to one map.

Answer 16

it's because there exists a thing called pole of order 1.

Answer 17

actually two poles that maps to +infinity and -infinity ,,, i am guessing this is not one to one function.

Answer 18

\[f(1) = 1\]\[f(-1)=1\]

Answer 19

interesting

Answer 20

so it's not a one-to-one function. :)

Answer 21

yep!!

Answer 22

anyway, experimentX's solution is right.

1. Suppose that the function is \(y\).
2. Take the denominator to the other side.
3. Rearrange and obtain an expression in the form \(ax^2 + bx + c = 0\). \(a,b,c\) can contain terms in \(y\).
4. You basically have to find the condition for which this quadratic has real roots. Use the discriminant. This will leave you with an expression that is \(\ge 0\).
5. Manipulate that expression until you get \((a^2 - b^2)(c^2 - d^2) \ge 0\).

Answer 23

well well ... nice job there (y)
I suppose I am just too lazy :(