Hello,
I'm trying to help my daughter with her math assignment, but this one I can't figure out. Would appreciate help, thanks!
Let f(x)= ax / (2x+3)
Investigate if you can define the 'a' so that f(f(x))=x
+26367 Hello,
I'm trying to help my daughter with her math assignment, but this one I can't figure out. Would appreciate help, thanks!
Let f(x)= ax / (2x+3)
Investigate if you can define the 'a' so that f(f(x))=x
\(\begin{array}{rcl} f(x) &=& \frac{ax}{2x+3} \\\\ f(f(x)) &=& f(\frac{ax}{2x+3}) =x\\\\ f(\frac{ax}{2x+3}) &=& \frac{ a\left( \frac{ax}{2x+3} \right) } { 2\left( \frac{ax}{2x+3} \right) +3} = x \\\\ \frac{ a\left( \frac{ax}{2x+3} \right) } { 2\left( \frac{ax}{2x+3} \right) +3} &=& x \\\\ a\left( \frac{ax}{2x+3} \right) &=& x \left[ 2\left( \frac{ax}{2x+3} \right) +3 \right] \\\\ \frac{a^2}{2x+3} &=& 2\left( \frac{ax}{2x+3} \right) +3 \qquad | \qquad \cdot (2x+3)\\\\ a^2 &=& 2ax +3(2x+3) \\ a^2 &=& 2ax +6x+9 \\ a^2-2x\cdot a -6x-9 &=& 0 \\\\ \boxed{~ \begin{array}{rcl} ax^2+bx+c &=& 0\\ x = {-b \pm \sqrt{b^2-4ac} \over 2a} \end{array} ~}\\ a_{1,2} &=& {2x \pm \sqrt{(2x)^2-4(-6x-9)} \over 2} \\ a_{1,2} &=& {2x \pm \sqrt{4x^2+4(6x+9)} \over 2}\\ a_{1,2} &=& {2x \pm 2\sqrt{x^2+6x+9} \over 2} \\ a_{1,2} &=& x \pm \sqrt{x^2+6x+9} \\ a_{1,2} &=& x \pm \sqrt{(x+3)^2} \\ a_{1,2} &=& x \pm (x+3) \\ a_{1} &=& x + (x+3) \\ \mathbf{a_{1}} & \mathbf{=} & \mathbf{2x+3} \\\\ a_{2} &=& x - (x+3) \\ \mathbf{a_{2}} & \mathbf{=} & \mathbf{ 3 } \end{array}\)
+23246 Since f(x) = ax/(2x+3), to find f( f(x) ), replace the x=term in ax/(2x+3) wwith ax/2x+3):
f( f(x) ) = f( ax/(2x+3) ) = [ a( ax/(2x+3) ) ] / [ 2( ax/(2x+3) ) + 3 ]
= [ a2x / (2x+3) ] / [ 2ax / (2x+3) + 3]
Assume that x isn't -3/2: multiply both the numerator and denominator by 2x + 3 and simplify:
= a2x / ( 2ax + 3(2x+3) ]
= a2x / ( 2ax + 6x + 9 ]
Since f( f(x) ) = x: a2x / ( 2ax + 6x + 9 ] = x
Multiply both sides by 2ax + 6x + 9: a2x = x(2ax + 6x + 9)
Assuming that x isn't 0, divide both sides by x: a2 = 2ax + 6x + 9
---> 2ax + 6x = a2 - 9
---> 2x(a + 3) = (a + 3)(a - 3)
If a is not equal to -3, divide both sides by a + 3:
---> 2x = a - 3
and x = (a - 3) / 2
But, x could be zero (it checks in the original expression).
Also checking -3/2 and -3 -- they don't check.
So, the answer: either 0 or (a - 3)/2.
+26367 Hello,
I'm trying to help my daughter with her math assignment, but this one I can't figure out. Would appreciate help, thanks!
Let f(x)= ax / (2x+3)
Investigate if you can define the 'a' so that f(f(x))=x
\(\begin{array}{rcl} f(x) &=& \frac{ax}{2x+3} \\\\ f(f(x)) &=& f(\frac{ax}{2x+3}) =x\\\\ f(\frac{ax}{2x+3}) &=& \frac{ a\left( \frac{ax}{2x+3} \right) } { 2\left( \frac{ax}{2x+3} \right) +3} = x \\\\ \frac{ a\left( \frac{ax}{2x+3} \right) } { 2\left( \frac{ax}{2x+3} \right) +3} &=& x \\\\ a\left( \frac{ax}{2x+3} \right) &=& x \left[ 2\left( \frac{ax}{2x+3} \right) +3 \right] \\\\ \frac{a^2}{2x+3} &=& 2\left( \frac{ax}{2x+3} \right) +3 \qquad | \qquad \cdot (2x+3)\\\\ a^2 &=& 2ax +3(2x+3) \\ a^2 &=& 2ax +6x+9 \\ a^2-2x\cdot a -6x-9 &=& 0 \\\\ \boxed{~ \begin{array}{rcl} ax^2+bx+c &=& 0\\ x = {-b \pm \sqrt{b^2-4ac} \over 2a} \end{array} ~}\\ a_{1,2} &=& {2x \pm \sqrt{(2x)^2-4(-6x-9)} \over 2} \\ a_{1,2} &=& {2x \pm \sqrt{4x^2+4(6x+9)} \over 2}\\ a_{1,2} &=& {2x \pm 2\sqrt{x^2+6x+9} \over 2} \\ a_{1,2} &=& x \pm \sqrt{x^2+6x+9} \\ a_{1,2} &=& x \pm \sqrt{(x+3)^2} \\ a_{1,2} &=& x \pm (x+3) \\ a_{1} &=& x + (x+3) \\ \mathbf{a_{1}} & \mathbf{=} & \mathbf{2x+3} \\\\ a_{2} &=& x - (x+3) \\ \mathbf{a_{2}} & \mathbf{=} & \mathbf{ 3 } \end{array}\)
Very kind thank you geno3141 and heureka for taking your time to help us!
We are digesting! 
Have a nice day!
+118609 Yes, this one looks reallly interesting. Thanks Geno and Heureka.
I have put it aside for when I have more time. I want to look at both your answers :)
+26367 Hello,
I'm trying to help my daughter with her math assignment, but this one I can't figure out. Would appreciate help, thanks!
Let f(x)= ax / (2x+3)
Investigate if you can define the 'a' so that f(f(x))=x
New edit, without mistake: 
\(\begin{array}{rcl} a_{1} &=& x + (x+3) \\ \mathbf{a_{1}} & \mathbf{=} & \mathbf{2x+3} \\\\ a_{2} &=& x - (x+3) \\ \mathbf{a_{2}} & \mathbf{=} & \mathbf{ -3 } \end{array}\)
