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

Bxtterman Mar 23, 2020

#1**+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.]

geno3141 Mar 23, 2020

#2**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**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