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avatar+102 

Describe all constants a and b such that:

 

\(f\left(x\right)=\frac{2x+a}{bx-2}\)

 

and 

 

\(f\left(x\right)=f^{-1}\left(x\right)\)

 

for all x in the domain of f

 Mar 23, 2020
 #1
avatar+20981 
+1

If  f(x)  =  (2x + a) / (bx - 2)  we can find f-1(x) by following these steps:

 

1)  replace  f(x)  by  y:           y  =  (2x + a) / (bx - 2)

2)  interchange  x  and  y:     x  =  (2y + a) / (by - 2)

3)  solve for y:            x(by - 2)  =  2y + a

                                   bxy - 2x  =  2y + a

                                   bxy - 2y  =  2x + a

                                  y(bx - 2)  =  2x + a

                                             y  =  (2x + a) / (bx - 2)

 

Very interesting -- the function is its own inverse!

So, when does  (2x + a) / (bx - 2)  =  (2x + a) / (bx - 2)  for all values of x?

Well,  a  can be any number; but, it's different for  b  because the denominator can't be zero.

When the denominator equals zero,  bx  equals  2.

To prevent the denominator from ever equaling zero, get rid of the  bx  term  by making  b = 0.

[If  b  is any other number, there is a value of  x  that will make  bx  equal to  2.]

 Mar 23, 2020
 #2
avatar+2854 
0

oh im dumb.

 

I got (2x +a)/(bx-2) = (2x + a)/bx - 2) and I thought I did something wrong. I never learned inverse I only know how to switch out x and y after look up some google images XD

CalculatorUser  Mar 23, 2020
 #3
avatar+102 
0

Thank you for your answer. My Approach initially was that I saw f(x) = inverse is the same as f(f(x)) = x, so I tried plugging it in, but that was more complex (though I think I may have gotten the correct answer that way too). I think that if I tried this method and thought about it in a more logical perspective, it would have been much better. Thank you!

Bxtterman  Mar 23, 2020

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