Determine the sêcific Mobius transformation

[awkwardnerd]

1. The function f(x) is not defined for x = 0. It has the property that for all nonzero real numbers x, f(x) + 2f(1/x) = 3x. Find all values of a such that f(a) = f(-a)

2. The function f is defined by f(x) = (ax+b)/(cx+d), where a, b, c, and d are nonzero real numbers, and has the properties: f(19) = 19, f(97) = 97, and f(f(x) = x for all values of x except -d/c. Find the unique number that is not in the range of f.

First time seeing this. Somehow tell me how to approach it, I don't need the answer.

 

[Mark44]

1. The function f(x) is not defined for x = 0. It has the property that for all nonzero real numbers x, f(x) + 2f(1/x) = 3x. Find all values of a such that f(a) = f(-a)

Start by evaluating f(a) and f(-a), and setting these expressions equal.

2. The function f is defined by f(x) = (ax+b)/(cx+d), where a, b, c, and d are nonzero real numbers, and has the properties: f(19) = 19, f(97) = 97, and f(f(x) = x for all values of x except -d/c. Find the unique number that is not in the range of f.

Since f(f(x)) = x, f is its own inverse. How do you normally go about finding the inverse of a function?

First time seeing this. Somehow tell me how to approach it, I don't need the answer.

 

[╔(σ_σ)╝]

For part one notice that
f(x) + 2f(1/x)= 3x
f(1/x) + 2f(x) = 3/x

So it is possible to get an explicit equation for f(x) by adding both equations and some algebra.

a should be sqrt(2).

 

[HallsofIvy]

Very, very clever! But actually, there are two values of a, $\sqrt{2}$ and $-\sqrt{2}$.

 

[╔(σ_σ)╝]

Very, very clever! But actually, there are two values of a, $\sqrt{2}$ and $-\sqrt{2}$.

XD. Yeah I forgot about the +/- thing for square roots.