+104 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
+23246 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.]
+2862 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
+104 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!
