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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

Guest Nov 26, 2015

Best Answer 

 #2
avatar+18715 
+30

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

 

\(\begin{array}{rcl} f(x) &=& \frac{ax}{2x+3} \\\\ f(f(x)) &=& f(\frac{ax}{2x+3}) =x\\\\ f(\frac{ax}{2x+3}) &=& \frac{ a\left( \frac{ax}{2x+3} \right) } { 2\left( \frac{ax}{2x+3} \right) +3} = x \\\\ \frac{ a\left( \frac{ax}{2x+3} \right) } { 2\left( \frac{ax}{2x+3} \right) +3} &=& x \\\\ a\left( \frac{ax}{2x+3} \right) &=& x \left[ 2\left( \frac{ax}{2x+3} \right) +3 \right] \\\\ \frac{a^2}{2x+3} &=& 2\left( \frac{ax}{2x+3} \right) +3 \qquad | \qquad \cdot (2x+3)\\\\ a^2 &=& 2ax +3(2x+3) \\ a^2 &=& 2ax +6x+9 \\ a^2-2x\cdot a -6x-9 &=& 0 \\\\ \boxed{~ \begin{array}{rcl} ax^2+bx+c &=& 0\\ x = {-b \pm \sqrt{b^2-4ac} \over 2a} \end{array} ~}\\ a_{1,2} &=& {2x \pm \sqrt{(2x)^2-4(-6x-9)} \over 2} \\ a_{1,2} &=& {2x \pm \sqrt{4x^2+4(6x+9)} \over 2}\\ a_{1,2} &=& {2x \pm 2\sqrt{x^2+6x+9} \over 2} \\ a_{1,2} &=& x \pm \sqrt{x^2+6x+9} \\ a_{1,2} &=& x \pm \sqrt{(x+3)^2} \\ a_{1,2} &=& x \pm (x+3) \\ a_{1} &=& x + (x+3) \\ \mathbf{a_{1}} & \mathbf{=} & \mathbf{2x+3} \\\\ a_{2} &=& x - (x+3) \\ \mathbf{a_{2}} & \mathbf{=} & \mathbf{ 3 } \end{array}\)

 

laugh

heureka  Nov 26, 2015
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6+0 Answers

 #1
avatar+17655 
+15

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.

geno3141  Nov 26, 2015
 #2
avatar+18715 
+30
Best Answer

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

 

\(\begin{array}{rcl} f(x) &=& \frac{ax}{2x+3} \\\\ f(f(x)) &=& f(\frac{ax}{2x+3}) =x\\\\ f(\frac{ax}{2x+3}) &=& \frac{ a\left( \frac{ax}{2x+3} \right) } { 2\left( \frac{ax}{2x+3} \right) +3} = x \\\\ \frac{ a\left( \frac{ax}{2x+3} \right) } { 2\left( \frac{ax}{2x+3} \right) +3} &=& x \\\\ a\left( \frac{ax}{2x+3} \right) &=& x \left[ 2\left( \frac{ax}{2x+3} \right) +3 \right] \\\\ \frac{a^2}{2x+3} &=& 2\left( \frac{ax}{2x+3} \right) +3 \qquad | \qquad \cdot (2x+3)\\\\ a^2 &=& 2ax +3(2x+3) \\ a^2 &=& 2ax +6x+9 \\ a^2-2x\cdot a -6x-9 &=& 0 \\\\ \boxed{~ \begin{array}{rcl} ax^2+bx+c &=& 0\\ x = {-b \pm \sqrt{b^2-4ac} \over 2a} \end{array} ~}\\ a_{1,2} &=& {2x \pm \sqrt{(2x)^2-4(-6x-9)} \over 2} \\ a_{1,2} &=& {2x \pm \sqrt{4x^2+4(6x+9)} \over 2}\\ a_{1,2} &=& {2x \pm 2\sqrt{x^2+6x+9} \over 2} \\ a_{1,2} &=& x \pm \sqrt{x^2+6x+9} \\ a_{1,2} &=& x \pm \sqrt{(x+3)^2} \\ a_{1,2} &=& x \pm (x+3) \\ a_{1} &=& x + (x+3) \\ \mathbf{a_{1}} & \mathbf{=} & \mathbf{2x+3} \\\\ a_{2} &=& x - (x+3) \\ \mathbf{a_{2}} & \mathbf{=} & \mathbf{ 3 } \end{array}\)

 

laugh

heureka  Nov 26, 2015
 #3
avatar
+5

Very kind thank you geno3141 and heureka for taking your time to help us!

We are digesting! smiley Have a nice day!

Guest Nov 26, 2015
 #4
avatar+91038 
0

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  :)

Melody  Nov 27, 2015
 #5
avatar+18715 
+30

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: blush

 

\(\begin{array}{rcl} a_{1} &=& x + (x+3) \\ \mathbf{a_{1}} & \mathbf{=} & \mathbf{2x+3} \\\\ a_{2} &=& x - (x+3) \\ \mathbf{a_{2}} & \mathbf{=} & \mathbf{ -3 } \end{array}\)

 

laugh

heureka  Nov 27, 2015
edited by heureka  Nov 27, 2015
edited by heureka  Nov 27, 2015
 #6
avatar
+5

Thank you heureka! That solves one of our question marks! cool

Guest Nov 27, 2015

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