Solving for Real Numbers $a$ and $b$ in $f(x)=\dfrac{1}{ax+b}$

[anemone]

Hi MHB,


This problem has given me a very hard time because I have exhausted all the methods that I know to figure out a way to find for the values for both a and b but no, there must be a trick to this problem and I admit that it is a question that is out of my reach...

Could you show me how to attack this problem, please?

Thanks in advance.:)

Problem:

For real numbers $a$ and $b$ define $f(x)=\dfrac{1}{ax+b}$. For which $a$ and $b$ are there three distinct real numbers $x_1, x_2, x_3$ such that $f(x_1)=x_2$, $f(x_2)=x_3$ and $f(x_3)=x_1$?

 

[kaliprasad]

Hi MHB,


This problem has given me a very hard time because I have exhausted all the methods that I know to figure out a way to find for the values for both a and b but no, there must be a trick to this problem and I admit that it is a question that is out of my reach...

Could you show me how to attack this problem, please?

Thanks in advance.:)

Problem:

For real numbers $a$ and $b$ define $f(x)=\dfrac{1}{ax+b}$. For which $a$ and $b$ are there three distinct real numbers $x_1, x_2, x_3$ such that $f(x_1)=x_2$, $f(x_2)=x_3$ and $f(x_3)=x_1$?

some one may come with a neat method but one solution I suggest is

f(f(f(x) = x

now f(f(1/(ax+b)) = x

 

[chisigma]

Hi MHB,


This problem has given me a very hard time because I have exhausted all the methods that I know to figure out a way to find for the values for both a and b but no, there must be a trick to this problem and I admit that it is a question that is out of my reach...

Could you show me how to attack this problem, please?

Thanks in advance.:)

Problem:

For real numbers $a$ and $b$ define $f(x)=\dfrac{1}{ax+b}$. For which $a$ and $b$ are there three distinct real numbers $x_1, x_2, x_3$ such that $f(x_1)=x_2$, $f(x_2)=x_3$ and $f(x_3)=x_1$?

With trivial substitions You should arrive to the second order equation...

$\displaystyle a\ (a+b)\ x_{1}^{2} + (a + 2\ b - a\ b)\ x_{1} + b - 1 =0\ (1)$

... and the condition for real solution of Your system is to be $\ge 0$ the discriminant of (1)... if the solutions are also 'distinct' is of course a different problem...

Kind regards

$\chi$ $\sigma$

 

[Opalg]

For real numbers $a$ and $b$ define $f(x)=\dfrac{1}{ax+b}$. For which $a$ and $b$ are there three distinct real numbers $x_1, x_2, x_3$ such that $f(x_1)=x_2$, $f(x_2)=x_3$ and $f(x_3)=x_1$?

I prefer to call the numbers $x,\ y,\ z$, so that $f(x)=y$, $f(y) = z$ and $f(z)= x$. Then $y = \dfrac1{ax+b}$, $$z = \frac1{\frac a{ax+b} + b} = \frac{ax+b}{abx+b+b^2},$$ $$ x = \frac1{\frac{a^2x+ab}{abx+b+b^2} + b} = \frac{abx+b+b^2}{(a^2+ab^2)x + 2ab + b^2}.$$ Therefore \(\displaystyle (a^2+ab^2)x^2 + (2ab + b^2)x = abx+b+b^2\), which simplifies to $a(a+b^2)x^2 + b(a+b^2)x - (a+b^2) = 0.$ Divide through by $a+b^2$ to get $ax^2+bx-1 = 0.$ But now we are in trouble, because if $ax^2+bx-1 = 0$ then (dividing through by $x$) $\frac1x = ax+b$, which means that $x = \frac1{ax+b} = y$, contradicting the requirement that $x$ and $y$ should be distinct.

Looking back to see what went wrong, you notice (if you did not do so at the time) that we divided that quadratic equation by the constant $a+b^2$ without considering the possibility that $a+b^2=0$. That is the only thing that can rescue the problem from having no solution, so it looks as though the answer to the question is that $a$ and $b$ must satisfy the condition $a+b^2=0$. That is certainly a necessary condition, and I leave it to you to check that it is also sufficient for the problem to have a solution.

 

[anemone]

With trivial substitions You should arrive to the second order equation...

$\displaystyle a\ (a+b)\ x_{1}^{2} + (a + 2\ b - a\ b)\ x_{1} + b - 1 =0\ (1)$

... and the condition for real solution of Your system is to be $\ge 0$ the discriminant of (1)... if the solutions are also 'distinct' is of course a different problem...

Kind regards

$\chi$ $\sigma$

some one may come with a neat method but one solution I suggest is

f(f(f(x) = x

now f(f(1/(ax+b)) = x

Thank you to both of you, kaliprasad and chisigma for the help!

I prefer to call the numbers $x,\ y,\ z$, so that $f(x)=y$, $f(y) = z$ and $f(z)= x$. Then $y = \dfrac1{ax+b}$, $$z = \frac1{\frac a{ax+b} + b} = \frac{ax+b}{abx+b+b^2},$$ $$ x = \frac1{\frac{a^2x+ab}{abx+b+b^2} + b} = \frac{abx+b+b^2}{(a^2+ab^2)x + 2ab + b^2}.$$ Therefore \(\displaystyle (a^2+ab^2)x^2 + (2ab + b^2)x = abx+b+b^2\), which simplifies to $a(a+b^2)x^2 + b(a+b^2)x - (a+b^2) = 0.$ Divide through by $a+b^2$ to get $ax^2+bx-1 = 0.$ But now we are in trouble, because if $ax^2+bx-1 = 0$ then (dividing through by $x$) $\frac1x = ax+b$, which means that $x = \frac1{ax+b} = y$, contradicting the requirement that $x$ and $y$ should be distinct.

Looking back to see what went wrong, you notice (if you did not do so at the time) that we divided that quadratic equation by the constant $a+b^2$ without considering the possibility that $a+b^2=0$. That is the only thing that can rescue the problem from having no solution, so it looks as though the answer to the question is that $a$ and $b$ must satisfy the condition $a+b^2=0$. That is certainly a necessary condition, and I leave it to you to check that it is also sufficient for the problem to have a solution.

Good answer, with an excellent explanation! Mr. Opalg, I can't tell you how thankful that I am for all your help!

I realized now that for any given equation, in order to keep it balance, not only we can add/subtract the same quantity from both sides of the equation, we can also take square root to it, square it, divide the whole equation by some quantity which isn't zero (such as the situation in this instance), I am so happy that I learned something so valuable today! Thank you so so much, Opalg!

 

[anemone]

That is the only thing that can rescue the problem from having no solution, so it looks as though the answer to the question is that $a$ and $b$ must satisfy the condition $a+b^2=0$. That is certainly a necessary condition, and I leave it to you to check that it is also sufficient for the problem to have a solution.

Btw,:) I used the graph of the function $f(x)=\dfrac{1}{ax+b}=\dfrac{1}{-b^2x+b}$ below to check that the condition $a+b^2=0$ suggests that there are three distinct real numbers $x_1, x_2, x_3$ such that $f(x_1)=x_2$, $f(x_2)=x_3$ and $f(x_3)=x_1$:

View attachment 1467